4: Equilibrium.

4: Equilibrium

Section I

You do not need to copy the outcome, but you should read it
No drill – just immediately start CW 1 Drill 1 12/18 (A) 12/19 (B) Outcome: I can explain chemical equilibrium in terms of the equilibrium constant. Goal: CW 1

CW 1: Equilibrium A(g) ⇌ B(g)
What is the significance of the double arrow in the equation? The reaction is reversible If the reaction starts with 100 molecules of A and zero molecules of B, will the concentration of A reach zero? Explain. No, some B is converted back into A If the reaction starts with zero molecules of A and 100 molecules of B, will the concentration of B increase or decrease as the reaction proceeds? Explain. [B] will decrease, some B is converted into A until [B] reaches a stable value

Consider an initial amount of 5.00 moles of A and moles of B. If 60% of the available molecules of A and 20% of available molecules of B react each minute, calculate the number of moles of A and B after one minute. 5.00 mol 2.00 mol 3.00 mol 3.00 mol 0.40 mol 0.40 mol 2.40 mol 4.60 mol

Obtain a set of initial conditions from the instructor and complete the chart below. Round your answers to the nearest hundredth of a mole. Use your data from Question 5 to answer the following: Identify the time when your reaction reached equilibrium. When equilibrium is reached, how do the moles of A lost and the moles of B gained per minute compare? Calculate the ratio of moles of B to moles of A at equilibrium for your reaction.

CW 1: Equilibrium Initial Moles A Moles B Percent Reacted Forward Reverse Equilibrium moles A moles B Ratio of Product to Reactant A 10.00 0.00 60.0% 20.0% 2.50 7.50 3.00 B 5.00 C D 8.00 2.00 6.00 E 4.00 1.00 F 25.0% 50.0% 6.67 3.33 0.50 G H I 5.33 2.67 J 1.33

CW 1: Equilibrium Use the class data to answer the following:
A(g) ⇌ B(g) Does the reaction reach equilibrium when the moles of A are equal to the moles of B in the container? No: none of the sets had equal moles of A and B Are the equilibrium concentrations of A and B always the same, regardless of initial concentrations? Note: Compare reactions with the same percent reacted for the forward and reverse reactions. Explain. For sets with the same percentages, the equilibrium concentrations of A and B are sometimes the same, but not always.

A(g) ⇌ B(g) Does the equilibrium ratio of product to reactant depend on the initial moles of product and reactant? Note: Compare reactions with the same percent reacted for the forward and reverse reactions. Explain. In both sets, even when the initial reactant and product amounts were different, the same ratio is obtained at equilibrium Does the equilibrium ratio of product to reactant depend on the percent of the molecules reacted in the forward and reverse reactions? Explain. The ratio at equilibrium depends on the percentages for the forward and reverse reactions. In each set, the same equilibrium ratio is obtained.

A(g) ⇌ B(g) Predict the final product to reactant ratio at equilibrium for the following initial conditions. The forward percent divided by the reverse percent gives the equilibrium ratio: What does the percent reacted forward and reverse represent in the reaction? The rate of the forward and reverse reactions Initial Moles A Initial Moles B Percent Reacted Forward Percent Reacted Reverse 15.0 5.0 80% 20% 80% 20% =4

CW 1: Equilibrium Considering the product/reactant ratio:
Which set(s) of reactions favor the products? Set A–E Do reactions that favor products have a faster forward or reverse rate? Reactions with faster forward rates will favor products Do reversible reactions that favor the products have an equilibrium product to reactant ratio greater then, less than, or equal to 1? Explain why. The equilibrium ratio will be greater than 1 for a reversible reaction that favors products. Products are on top in the ratio and should be larger then the bottom if products are favored. What is “equal” when a reaction reaches equilibrium? The change in the moles in the forward reaction is equal to the change in the moles in the reverse reaction.

CW 1: Equilibrium The Law of Mass Action
The law of mass action states that for a chemical system described by the chemical equation aA + bB ⇌ cC + dD the ratio [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏 is a constant at a given temperature. The ratio is called the equilibrium constant expression, and the numerical value of the ratio is called the equilibrium constant, Kc. Note that a, b, c, and d are the stoichiometric coefficients in the chemical equation. 𝐾 𝑐 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏 Note: Your textbook uses K instead of Kc when describing an equilibrium with concentrations. By convention, equilibrium constants are reported without units.

CW 1: Equilibrium Consider the following equations:
Provide an equilibrium constant expression for reaction 1. Provide an equilibrium constant expression for reaction 2. What is the relationship between reactions 1 and 2? They are the same reaction, forward and reverse PCl3(g) + Cl2(g) ⇌ PCl5(g) (1) PCl5(g) ⇌ PCl3(g) + Cl2(g) (2) 𝐾 𝑐 = 𝑃𝐶𝑙 𝑃𝐶𝑙 3 𝐶𝑙 2 𝐾 𝑐 = 𝑃𝐶𝑙 3 𝐶𝑙 𝑃𝐶𝑙 5

CW 1: Equilibrium Consider the following equations:
What is the relationship between the equilibrium constant expression for reaction 1 and the equilibrium constant expression for reaction 2? Their equilibrium constant expressions are flipped If the value of Kc for reaction 1 is 1.00x103, what is the value of Kc for reaction 2? The inverse of Kc for reaction 1: PCl3(g) + Cl2(g) ⇌ PCl5(g) (1) PCl5(g) ⇌ PCl3(g) + Cl2(g) (2) 𝐾 𝑐 = × =1.00× 10 −3

CW 1: Equilibrium The reaction below is known as the Haber process.
3H2(g) + N2(g) ⇌ 2NH3(g) Calculate Kc if equilibrium concentrations are [H2] = 1.5; [NH3] = 0.24; [N2] = 2.5. 𝐾 𝑐 = 𝑁𝐻 𝐻 𝑁 2 = =6.8× 10 −3

CW 1: Equilibrium The reaction below is known as the Haber process.
3H2(g) + N2(g) ⇌ 2NH3(g) A 1.00 L flask contains an equilibrium mixture of 24.9 g N2(g), g H2(g), and 2.15 g of NH3(g) at some temperature. Calculate the value of the equilibrium constant under these conditions for the forward and reverse reactions. 24.9 𝑔 𝑁 2 1 × 1 𝑚𝑜𝑙 𝑁 𝑔 𝑁 2 × 𝐿 =0.889 𝑀 𝑁 2 1.35 𝑔 𝐻 2 1 × 1 𝑚𝑜𝑙 𝐻 𝑔 𝐻 2 × 𝐿 =0.670 𝑀 𝐻 2 2.15 𝑔 𝑁𝐻 3 1 × 1 𝑚𝑜𝑙 𝑁𝐻 𝑔 𝑁𝐻 3 × 𝐿 =0.126 𝑀 𝑁𝐻 3 𝐾 𝑐 = 𝑁𝐻 𝐻 𝑁 2 = =5.93× 10 −2

CW 1: Equilibrium The following gases are added to a 1.00 L container: 2.0 mole of A; 4.0 mole B. At equilibrium, the container has 0.4 moles of D. A(g) + 3B(g) ⇌ C(g) + 2D(g) Calculate the moles of A, B, and C in the container at equilibrium. At equilibrium, there are 0.4 moles of D 0.4 𝑚𝑜𝑙 𝐷 1 × 1 𝑚𝑜𝑙 𝐴 2 𝑚𝑜𝑙 𝐷 =0.2 𝑚𝑜𝑙 𝐴 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 −2.0 𝑚𝑜𝑙 𝐴 =1.8 𝑚𝑜𝑙 𝐴 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 0.4 𝑚𝑜𝑙 𝐷 1 × 3 𝑚𝑜𝑙 𝐵 2 𝑚𝑜𝑙 𝐷 =0.6 𝑚𝑜𝑙 𝐵 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 −4.0 𝑚𝑜𝑙 𝐵 =3.4 𝑚𝑜𝑙 𝐵 𝑟𝑒𝑚𝑎𝑖𝑛𝑖𝑛𝑔 0.4 𝑚𝑜𝑙 𝐷 1 × 1 𝑚𝑜𝑙 𝐶 2 𝑚𝑜𝑙 𝐷 =0.2 𝑚𝑜𝑙 𝐶 𝑓𝑜𝑟𝑚𝑒𝑑

[A] = 1.8 M [B] = 3.4 M [C] = 0.2 M [D] = 0.4 M
CW 1: Equilibrium The following gases are added to a 1.00 L container: 2.0 mole of A; 4.0 mole B. At equilibrium, the container has 0.4 moles of D. A(g) + 3B(g) ⇌ C(g) + 2D(g) Calculate [A], [B], [C], and [D] at equilibrium. The volume is 1.00 L, so the moles are the molarities. [A] = 1.8 M [B] = 3.4 M [C] = 0.2 M [D] = 0.4 M Calculate the value of the equilibrium constant, Kc, for this reaction. 𝐾 𝑐 = [𝐶] 𝐷 𝐴 𝐵 3 = (0.2) =4.5× 10 −4

CW 1: Equilibrium One mole of A was placed in a 1.0 L flask and the reaction was monitored over time. 2A(g) ⇌ B(g) Determine the value of the equilibrium constant, Kc. 𝐾 𝑐 = [𝐵] 𝐴 2 = (0.4) =10

CW 1: Equilibrium One mole of A was placed in a 1.0 L flask and the reaction was monitored over time. 2A(g) ⇌ B(g) Which is larger at time "a", the forward rate or the reverse rate? The forward rate is larger the moles of B are still increasing, and the moles of A are still decreasing

CW 1: Equilibrium One mole of A was placed in a 1.0 L flask and the reaction was monitored over time. 2A(g) ⇌ B(g) Which is larger at time "b", the forward rate or the reverse rate? The forward rate is larger the moles of B are still increasing, and the moles of A are still decreasing

CW 1: Equilibrium One mole of A was placed in a 1.0 L flask and the reaction was monitored over time. 2A(g) ⇌ B(g) Which is larger at time "c", the forward rate or the reverse rate? They are equal; the reaction is at equilibrium

Equilibrium Expression Involving Pressures
CW 1: Equilibrium Equilibrium Expression Involving Pressures So far, we have been describing gaseous equilibria in terms of concentrations. Gaseous equilibria can also be described in terms of pressures, as pressure is related to concentration by the ideal gas law: This allows us to write gaseous equilibria much like Kc. The relationship between Kc and Kp is: 𝑃𝑉=𝑛𝑅𝑇 or 𝑃= 𝑛𝑅𝑇 𝑉 3H2(g) + N2(g) ⇌ 2NH3(g) 𝐾 𝑐 = 𝑁𝐻 𝐻 [ 𝑁 2 ] 𝐾 𝑃 = 𝑃 𝑁𝐻 𝑃 𝐻 𝑃 𝑁 2 𝑗𝐴+𝑘𝐵⇌𝑙𝐶+𝑚𝐷 𝐾 𝑝 = 𝐾 𝑐 𝑅𝑇 ∆𝑛 where ∆𝑛= 𝑙+𝑚 −(𝑗+𝑘)

CW 1: Equilibrium The reaction below was studied at 25°C. The equilibrium pressures are given. Calculate the value of Kp. Calculate the value of Kc. 2NO(g) + Cl2(g) ⇌ 2NOCl(g) PNOCl = 1.2 atm PNO = 5.0x10–2 atm PCl2 = 3.0x10–1 atm 𝐾 𝑝 = ( 𝑃 𝑁𝑂𝐶𝑙 ) 𝑃 𝑁𝑂 2 ( 𝑃 𝐶𝑙 2 ) = (1.2) × 10 −2 2 (3.0× 10 −1 ) =1.9× 10 3 Sum of product coefficients Sum of reactant coefficients 𝐾 𝑝 =𝐾 (𝑅𝑇) ∆𝑛 ∆𝑛=2− 2+1 =−1 𝐾= 𝐾 𝑝 (𝑅𝑇) ∆𝑛 = 1.9× 𝐿∙𝑎𝑡𝑚 𝑚𝑜𝑙∙𝐾 ×298 𝐾 −1 =4.6× 10 4

Heterogenous Equilibria
CW 1: Equilibrium Heterogenous Equilibria So far, we have only discussed equilibria for gaseous reactions. When all species involved in a reaction are in the same phase, the system is a homogenous equilibrium. Many equilibria involve species that are not all in the same phase and are known as heterogenous equilibria. Experimental results show that the position of a heterogenous equilibrium does not depend on the amounts of solids or liquids during a reaction. This is because the concentrations of solids and liquids do not change. Thus, when writing heterogenous equilibria, exclude solids and liquids from your expression. CaCO3(s) ⇌ CaO(s) + CO2(g) 𝐾 𝑐 = 𝐶𝑂 2 𝐾 𝑝 = 𝑃 𝐶𝑂 2

CuSO4∙5H2O(s) ⇌ CuSO4(s) + 5H2O(g)
CW 1: Equilibrium Write the expressions for Kc and Kp for the following processes: Solid phosphorus pentachloride decomposes into liquid phosphorus trichloride and chlorine gas. Solid copper(II) sulfate pentahydrate (bright blue) is heated to drive off water vapor to form solid copper(II) sulfate (light blue). PCl5(s) ⇌ PCl3(l) + Cl2(g) 𝐾 𝑐 =[ 𝐶𝑙 2 ] 𝐾 𝑝 = 𝑃 𝐶𝑙 2 CuSO4∙5H2O(s) ⇌ CuSO4(s) + 5H2O(g) 𝐾 𝑐 = [ 𝐻 2 𝑂] 5 𝐾 𝑝 = ( 𝑃 𝐻 2 𝑂 ) 5

Summary 1 12/18 (A) 12/19 (B) HW 1: Equilibrium Cornell Notes
Due: 12/20 (A) & 12/21 (B) Read sections 13.1 to 13.4 (Pages 511 to 522) and create Cornell Notes, hand in during class. Rate Law of a Crystal Violet Reaction Due: 12/20 (A Day) and 12/21 (B Day) Summary 1 12/18 (A) 12/19 (B) Outcome: I can explain chemical equilibrium in terms of the equilibrium constant. Goal: CW 1

How does the rate of the forward and reverse reactions determine the position of the equilibrium?
If the forward rate is faster, products are favored. If the reverse rate is faster, reactants are favored. What is equal when a reaction reaches equilibrium? Change forward (reactants  products) and change reverse (products  reactants) How do you write K for homogenous and heterogenous reactions? Products over reactants Coefficients become exponents What is Kc and Kp? Equilibrium expressions for concentration and pressure (gases) Drill 2 1/2 (A) 1/3 (B) Outcome: I can explain chemical equilibrium in terms of the equilibrium constant. Goal: CW 3 Hand In: HW 1: Cornell Notes 𝐾 𝑐 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏

CW 3: Using Equilibrium Constants
The Extent of a Reaction The tendency of a reaction to occur is related to the equilibrium constant. For large equilibrium constants, the positions of the equilibrium lie far on the products side, indicating that the reaction goes to completion. For small equilibrium constants, the equilibrium lies far on the reactants side, and the reaction does not occur. The size of the equilibrium constant does not convey how quickly the reaction proceeds – that is determined by the rate of reaction, which depends on the size of the required activation energy.

Reaction Quotient The reaction quotient is written in the same manner as the equilibrium constant and can be used to determine if a chemical system has reached equilibrium. If the reaction is not at equilibrium, it can also be used to predict the direction of the shift. aA + bB ⇌ cC + dD 𝐾 𝑐 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏 𝑄= 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏

For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0x10–2. Predict the direction of the shift by calculating Q for each of the following conditions. 3H2(g) + N2(g) ⇌ 2NH3(g) [NH3] = 1.0x10–3 M; [N2] = 1.0x10–5 M; [H2] = 2.0x10–3 M [NH3] = 2.00x10–4 M; [N2] = 1.50x10–5 M; [H2] = 3.54x10–1 M [NH3] = 1.0x10–4 M; [N2] = 1.0x10–5 M; [H2] = 1.0x10–2 M 𝑄= 𝑁𝐻 𝐻 [ 𝑁 2 ] = (1.0× 10 −3 ) 2 (2.0× 10 −3 ) 3 (1.0× 10 −5 ) =1.3× 10 7 Q too large, shift to reactants 𝑄= 𝑁𝐻 𝐻 [ 𝑁 2 ] = (2.00× 10 −4 ) 2 (3.54× 10 −1 ) 3 (1.50× 10 −5 ) =6.01× 10 −2 K = Q, equilibrium 𝑄= 𝑁𝐻 𝐻 [ 𝑁 2 ] = (1.0× 10 −4 ) 2 (1.0× 10 −2 ) 3 (1.0× 10 −5 ) =1000 Q too large, shift to reactants

Calculating Equilibrium Concentrations and Pressures Problem Solving Strategy Write the balanced chemical equation for the reaction. Write the equilibrium expression (using the law of mass action). List any concentrations or pressures (initial or equilibrium). Calculate Q to determine if the reaction is at equilibrium/ the direction of the shift. Define the change needed to reach equilibrium and define the equilibrium concentrations by applying the change to the initial concentrations (RICE). Substitute knowns (such as the equilibrium concentrations/ pressures, or K) into the equilibrium expression, solve for the unknown. Check that your answers are reasonable.

The reaction of gaseous N2O4 in a flask reached equilibrium at a temperature where the Kp = If the pressure of N2O4 was 2.71 atm, find the equilibrium pressure of NO2. N2O4(g) ⇌ 2NO2(g) 𝐾 𝑝 = ( 𝑃 𝑁𝑂 2 ) 2 ( 𝑃 𝑁 2 𝑂 4 ) =0.133 𝑃 𝑁𝑂 =𝐾 𝑝 × 𝑃 𝑁 2 𝑂 4 = × 2.71 𝑎𝑡𝑚 =0.360 𝑃 𝑁𝑂 2 = =0.600 𝑎𝑡𝑚

At a specific temperature, a 1.00 L flask initially contained moles of PCl3 and 8.70x10–3 moles of PCl5. Once equilibrium was reached, 2.00x10–3 moles of Cl2 were in the flask. Calculate the equilibrium concentrations of all species and the value of K. PCl5(g) ⇌ PCl3(g) + Cl2(g) R I C E PCl5(g) PCl3(g) Cl2(g) 8.70x10–3 0.298 –2.00x10–3 +2.00x10–3 +2.00x10–3 8.70x10–3–2.00x10–3 x10–3 2.00x10–3 𝐾 𝑐 = 𝑃𝐶𝑙 3 [ 𝐶𝑙 2 ] [ 𝑃𝐶𝑙 5 ] = [2.00× 10 −3 ] [6.70× 10 −3 ] =8.96× 10 −3

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask. CO(g) + H2O(g) ⇌ CO2(g) + H2(g) Q too small, shift to products 𝐾 𝑐 = 𝐶𝑂 2 𝐻 2 𝐶𝑂 𝐻 2 𝑂 =5.10 𝑄= =1.000 R I C E CO(g) H2O(g) CO2(g) H2(g) 1.000 1.000 1.000 1.000 –x –x +x +x 1.000 – x 1.000 – x x x

Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask. CO(g) + H2O(g) ⇌ CO2(g) + H2(g) 5.10= 𝐶𝑂 2 𝐻 2 𝐶𝑂 𝐻 2 𝑂 = 𝑥 𝑥 −𝑥 −𝑥 = (1.000+𝑥) 2 (1.000−𝑥) 2 = 𝑥 1.000−𝑥 2 5.10 = 𝑥 1.000−𝑥 𝑥=0.387 CO(g) H2O(g) CO2(g) H2(g) 1.000 – x x 0.613 M 1.387 M

Assume that the reaction for the formation of HF(g) from hydrogen and fluorine has an equilibrium constant of 1.15x102 at a certain temperature. If moles of each component were added to a L flask, calculate the equilibrium concentrations of all species. H2(g) + F2(g) ⇌ 2HF(g) Q too small, shift to products 𝐾 𝑐 = [𝐻𝐹] 2 𝐻 2 𝐹 2 =115 𝑄= [2.000] =1.000 R I C E H2(g) F2(g) 2HF(g) 2.000 2.000 2.000 –x –x +2x 2.000 – x 2.000 – x x

Assume that the reaction for the formation of HF(g) from hydrogen and fluorine has an equilibrium constant of 1.15x102 at a certain temperature. If moles of each component were added to a L flask, calculate the equilibrium concentrations of all species. H2(g) + F2(g) ⇌ 2HF(g) 115= [𝐻𝐹] 2 𝐻 2 𝐹 2 = [2.000+𝑥] −𝑥 −𝑥 = (2.000+𝑥) 2 (2.000−𝑥) 2 = 𝑥 2.000−𝑥 2 115 = 𝑥 2.000−𝑥 𝑥=1.528 H2(g) F2(g) 2HF(g) 2.000 – x x 0.472 M 5.056 M

Equilibria with Very Small Equilibrium Constants In the last few examples, we have seen that equilibrium problems are often mathematically difficult to solve – we had to use the quadratic formula. We can make simplifications to the math involved if the equilibrium constant is small.

For the reaction below, the value of the equilibrium constant is 1.6x10–5. If 1.0 moles of NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 2NOCl(g) ⇌ 2NO(g) + Cl2(g) Write the equilibrium expression. Calculate the initial concentrations for each species. 𝐾 𝑐 = [𝑁𝑂] 2 [ 𝐶𝑙 2 ] [𝑁𝑂𝐶𝑙] 2 𝑁𝑂𝐶𝑙 = 1.0 𝑚𝑜𝑙 2.0 𝐿 =0.5 𝑀 𝑁𝑂 =0 𝑀 𝐶𝑙 2 =0 𝑀

For the reaction below, the value of the equilibrium constant is 1.6x10–5. If 1.0 moles of NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 2NOCl(g) ⇌ 2NO(g) + Cl2(g) Define and apply conditions to achieve equilibrium. R I C E 2NO(g) Cl2(g) 2NOCl(g) 0.50 +2x +x –2x 2x x 0.50 – 2x

For the reaction below, the value of the equilibrium constant is 1.6x10–5. If 1.0 moles of NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 2NOCl(g) ⇌ 2NO(g) + Cl2(g) Substitute the known quantities into the equilibrium expression. Because the equilibrium constant is very small, the reaction will not proceed far towards completion, meaning the change to achieve equilibrium is small. Use this to simplify the math to solve this problem. 𝐾 𝑐 = [𝑁𝑂] 2 [ 𝐶𝑙 2 ] [𝑁𝑂𝐶𝑙] 2 = [2𝑥] 2 [𝑥] [0.50−2𝑥] 2 𝐾 𝑐 = [2𝑥] 2 [𝑥] [0.50] 2 If x is very small (because K is small): [0.50−2𝑥]≈0.50 1.6× 10 −5 = 4𝑥 ∴𝑥=1.0× 10 −2

For the reaction below, the value of the equilibrium constant is 1.6x10–5. If 1.0 moles of NOCl is placed in a 2.0 L flask, what are the equilibrium concentrations? 2NO(g) + Cl2(g) ⇌ 2NOCl(g) Check that your approximation is reasonable. If x = 1.0x10–2, then: 0.50−2𝑥 0.50−2 1.0× 10 −2 =0.48 The approximation gives values within 95%, so it is valid =96%

Summary 2 1/2 (A) 1/3 (B) HW 2: Equilibrium Problem Protein Synthesis
Due: 1/4 (A) & 1/7 (B) Complete the assigned problem set, hand in during class. Rate Law of a Crystal Violet Reaction Due: 12/20 (A Day) and 12/21 (B Day) Summary 2 1/2 (A) 1/3 (B) Outcome: I can explain chemical equilibrium in terms of the equilibrium constant. Goal: CW 3 Hand In: HW 1: Cornell Notes

H2O(g) + Cl2O(g) ⇌ 2HOCl(g)
At 25 °C, K = for the reaction H2O(g) + Cl2O(g) ⇌ 2HOCl(g) Calculate the equilibrium concentrations of each species if 1.0 mol of HOCl is placed in a 2.0 L flask. Drill 3 1/4 (A) 1/7 (B) Outcome: I can solve FRQ problems related to chemical equilibrium. Goal: CW 4 R I C E H2O(g) Cl2O(g) 2HOCl(g) 0.50 +x +x –2x x x 0.50 – 2x 0.22 M 0.22 M 0.066 M 𝐾=0.090= [𝐻𝑂𝐶𝑙] 2 𝐻 2 𝑂 [ 𝐶𝑙 2 𝑂] = [0.50−2𝑥] 2 𝑥 [𝑥] 0.090= −2𝑥 𝑥 2 ∴𝑥=0.217

CW 4: Equilibrium FRQ Problems
Choose two problems from Questions 1 to 3 and two problems from Questions 4 to 6. Complete all work on a separate sheet of paper in a small group. Use the answer keys to check your work.

Summary 3 1/4 (A) 1/7 (B) HW 3: Equilibrium Problems
Due: 1/8 (A) & 1/9 (B) Complete HW 3 through OWL. Summary 3 1/4 (A) 1/7 (B) Outcome: I can solve FRQ problems related to chemical equilibrium. Goal: CW 4

If 2. 5 mol NO2 of is placed in a 1
If 2.5 mol NO2 of is placed in a 1.0 L vessel, what are the equilibrium concentrations of all gases? Drill 4 1/8 (A) 1/9 (B) 2NO2(g) ⇌ 2NO(g) +O2(g) K = 6.8x10–7 Outcome: I can explain solution formation as an equilibrium process using the solubility product constant. Goal: CW 6

Check assumption later
Drill Solution R 2NO2 2NO O2 I 2.5 C –2x +2x +x E 2.5–2x [NO2] = 2.480 [NO] = [NO2] = 2.480 𝐾=6.8× 10 −7 = 2𝑥 2 (𝑥) (2.5−2𝑥) 2 Because K is small, x is small: Check assumption later 2.5−2𝑥≈2.5 2.5−2𝑥 2.5 ×100= ×100=99% Allows us to simplify and solve for x: Assumption OK! 6.8× 10 −7 = 2𝑥 2 (𝑥) (2.5) 2 ∴𝑥=0.0102

CW 6: Solubility Why is solubility an important phenomenon?
Salt and sugar dissolve to flavor foods CaSO4 is more soluble in hot water than cold water, coats boilers Teeth contain hydroxy-apatite, reacts with acids (formed from food) Fluoride replaces the hydroxide ion to form fluor-apatite and CaF2, both are less soluble in acids BaSO4 is ingested to improve the clarity of X-rays of the digestion track. Ba2+ is highly toxic by injection, but since the solubility of BaSO4 is so low, nearly none is formed.

CW 6: Solubility How is dissolving an equilibrium process?
When AgCl(s) is first added to H2O, no Ag+ or Cl– ions are present As the AgCl dissolves, the [Ag+] and [Cl–] increase Makes it more likely for these ions to collide and reform the solid

CW 6: Solubility What is the solubility product constant?
AaBb(s) ⇌ aA+(aq) + bB–(aq) Value of Ksp indicates how soluble a given compound is in H2O at a specific temperature Solubility product: equilibrium constant Solubility: equilibrium position CaF2(s) ⇌ Ca2+(aq) + 2F–(aq) 𝐾 𝑠𝑝 = 𝐴 + 𝑎 × 𝐵 − 𝑏 𝐾 𝑠𝑝 =[ 𝐶𝑎 2+ ]× 𝐹 − 2

CW 6: Solubility Copper(I) bromide has a measured solubility of 2.0x10–4 mol/L at 25°C. Calculate its Ksp value. CuBr(s) ⇌ Cu+(aq) + Br–(aq) Solubility CuBr = 2.0x10–4 mol/L If you have 1 L of H2O, 2.0x10–4 moles of CuBr will dissolve 𝐶𝑢 + = 𝐵𝑟 − =2.0× 10 −4 𝐾 𝑠𝑝 = 𝐶𝑢 + × 𝐵𝑟 − = [2.0× 10 −4 ] 2 𝐾 𝑠𝑝 =4.0× 10 −8

CW 6: Solubility Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0x10–15 mol/L at 25°C. Bi2S3(s) ⇌ 2Bi3+(aq) + 3S2– (aq) Solubility Bi2S3 = 1.0x10–15 mol/L If you have 1 L of H2O, 1.0x10–15 moles of Bi2S3 will dissolve 1.0× 10 −15 𝑚𝑜𝑙 𝐵𝑖 2 𝑆 3 1 𝐿 × 2 𝑚𝑜𝑙 𝐵𝑖 1 𝑚𝑜𝑙 𝐵𝑖 2 𝑆 3 =2.0× 10 −15 𝑀 𝐵𝑖 3+ 1.0× 10 −15 𝑚𝑜𝑙 𝐵𝑖 2 𝑆 3 1 𝐿 × 3 𝑚𝑜𝑙 𝑆 1 𝑚𝑜𝑙 𝐵𝑖 2 𝑆 3 =3.0× 10 −15 𝑀 𝑆 2− 𝐾 𝑠𝑝 = [ 𝐵𝑖 3+ ] 2 × [ 𝑆 2− ] 3 = [2.0× 10 −15 ] 2 [3.0× 10 −15 ] 3 𝐾 𝑠𝑝 =1.1× 10 −73

x = amount of Cu(IO3)2(s) that dissolves
CW 6: Solubility The Ksp value for copper(II) iodate, Cu(IO3)2, is 1.4x10–7 at 25°C. Calculate its solubility at 25°C. Cu(IO3)2(s) ⇌ Cu2+(aq) + 2IO3– (aq) Ksp = 1.4x10–7 R I C E Cu(IO3)2(s) Cu2+(aq) 2IO3– (aq) 𝐾 𝑠𝑝 =[ 𝐶𝑢 2+ ]× [ 𝐼𝑂 3 − ] 2 1.4× 10 −7 = 𝑥 × 2𝑥 2 = 4𝑥 3 –x +x +2x –x x 2x ∴𝑥=3.3× 10 −3 𝑀 x = amount of Cu(IO3)2(s) that dissolves Solubility Cu(OI3)2 = 3.3x10–3 mol/L

CW 6: Solubility How are Ksp values used to predict solubility?
Only salts that form the same number of ions can be compared Otherwise, the Ksp expressions are different: PbCl2(s) ⇌ Pb2+(aq) + 2Cl–(aq) AgCl(s) ⇌ Ag+(aq) + Cl–(aq) 𝐾 𝑠𝑝 = 𝑃𝑏 2+ × 𝐶𝑙 − 2 𝐾 𝑠𝑝 = 𝐴𝑔 + × 𝐶𝑙 − 𝐾 𝑠𝑝 = 𝑥 × 2𝑥 2 𝐾 𝑠𝑝 = 𝑥 ×[𝑥] 𝐾 𝑠𝑝 = 4𝑥 3 𝐾 𝑠𝑝 = 𝑥 2

CW 6: Solubility What is the common ion effect?
If the water already contains a common ion with the salt being added, the solubility of the salt will decrease Less AgCl(s) will dissolve in the solution because it contains 0.10 M Cl– (aq) already Limits how much of the Cl in AgCl(s) can dissolve

x = amount CaF2(s) that dissolves
CW 6: Solubility Calculate the solubility of solid CaF2 (Ksp = 4.0x10–11) in a M NaF solution. CaF2(s) ⇌ Ca2+(aq) + 2F– (aq) Ksp = 4.0x10–11 R I C E CaF2(s) Ca2+(aq) 2F– (aq) 0.025  From M NaF –x +x +2x –x x 2x x = amount CaF2(s) that dissolves 𝐾 𝑠𝑝 =[ 𝐶𝑎 2+ ]× [ 𝐹 − ] 2 4.0× 10 −11 = 𝑥 × 2𝑥

CW 6: Solubility Calculate the solubility of solid CaF2 (Ksp = 4.0x10–11) in a M NaF solution. 𝐾 𝑠𝑝 =[ 𝐶𝑎 2+ ]× [ 𝐹 − ] 2 4.0× 10 −11 = 𝑥 × 2𝑥 Assuming that 2x is negligible compared with (because Ksp is small) 2𝑥+0.025≈0.025 4.0× 10 −11 = 𝑥 × ∴𝑥=6.4× 10 −8 𝑀 6.4x10–8 mol/L of solid CaF2 dissolves per liter of the M NaF solution

CW 6: Solubility How does pH effect solubility?
Acids and bases may react with the chemical species involved in the equilibrium, removing them from the reaction The solubility of Mg(OH)2 is higher in acid: Also true for salts: Ag3PO4 is more soluble in acid: In general, if the anion X– is a strong base (that is HX is a weak acid), a salt MX will show increased solubility in acid. Acid reacts with OH–, causes equilibrium to shift to products Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq) Acid reacts with PO43–, causes equilibrium to shift to products Ag3PO4(s) ⇌ 3Ag+(aq) + PO43–(aq)

CaF2(s) ⇌ Ca2+(aq) + 2F–(aq)
CW 6: Solubility How is the ion product related to the reaction quotient? Ion product: Q for dissolving, calculated the same way If Q > Ksp, precipitation occurs until the concentrations are reduced to satisfy Ksp If Q < Ksp, no precipitation occurs CaF2(s) ⇌ Ca2+(aq) + 2F–(aq) 𝐾 𝑠𝑝 =[ 𝐶𝑎 2+ ]× [ 𝐹 − ] 2 𝑄=[ 𝐶𝑎 2+ ]× [ 𝐹 − ] 2

CW 6: Solubility Create a summary of your notes, explaining:
Write dissociation equations and Ksp expressions Common ion effect Ion product (Q) and Ksp Calculating Ksp, solubility, concentrations, RICE

Summary 4 1/8 (A) 1/9 (B) HW 4: Review for Unit test
Due: 1/14 (A) & 1/15 (B) Complete the review guide using your notes/ unit packet/ resources, hand in during class. Complete Pre-Lab (See CW 5) for next class Summary 4 1/8 (A) 1/9 (B) Outcome: I can explain solution formation as an equilibrium process using the solubility product constant. Goal: CW 6

Determine if MgCO3 will be more soluble, less soluble, or equally as soluble as in pure water in each of the following solutions: Drill 5 1/10 (A) 1/11 (B) Outcome: I can use Le Châtelier's Principle to create a rainbow of equilibria. Goal: CW 5 0.10 M Mg(NO3)2 0.10 M K2CO3 0.10 M NH4NO3 0.10 M HCl Less soluble Less soluble Equally soluble More soluble

CW 5: Le Châtelier's Rainbow
Le Châtelier’s principle: if a stress applied to a chemical system at equilibrium, the equilibrium will shift to counteract the change and establish a new equilibrium. We can predict the direction of the shift based on the stress that was applied. Changing Concentrations Changing Pressures Changing Temperatures Competing Reactions

Changing Concentrations A + 2B ⇌ C + D Stress Resulting Shift Add A Products Remove C Add D Reactants Remove B Add = equilibrium shifts Away Remove = equilibrium shifts to Replace

Changing Pressures A(g) + 2B(g) ⇌ C(g) + D(g) 3 moles gas 2 moles gas Stress Resulting Shift Increase P Products Decrease P Reactants Increase P = reactions responds by shifting to the side with less moles of gas (to reduce pressure) Decrease P = reactions responds by shifting to the side with more moles of gas (to increasing pressure)

Changing Temperatures Exothermic A + 2B ⇌ C + D + heat Endothermic A + 2B + heat ⇌ C + D Add = equilibrium shifts Away Remove = equilibrium shifts to Replace Stress Resulting Shift Add heat Reactants Remove heat Products Stress Resulting Shift Add heat Products Remove heat Reactants

Competing Reactions Pb(NO3)2(aq) + Na2CO3(aq) ⇌ PbCO3(s) + NaNO3(aq) Pb2+(aq) + CO32–(aq) ⇌ PbCO3(s) Na2CO3(aq) Pb(OH)2(s) What if we add NaOH? Pb2+ ions are consumed by OH–, equilibrium shifts to reactants

Summary 5 1/10 (A) 1/11 (B) HW 4: Review for Unit test
Due: 1/14 (A) & 1/15 (B) Complete the review guide using your notes/ unit packet/ resources, hand in during class. Le Châtelier’s Rainbow Lab Report Due: 1/23 (A) & 1/22 (B) Summary 5 1/10 (A) 1/11 (B) Outcome: I can use Le Châtelier's Principle to create a rainbow of equilibria. Goal: CW 5

Section II

Drill 6 1/18 (A) 1/17 (B) What is the pH of a 0.25 M HCl?
What is the pH of 0.15 M NaOH? Drill 6 1/18 (A) 1/17 (B) 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 0.25 𝑝𝐻=0.60 Outcome: I can calculate the pH of a weak acid or base Goal: CW 7 𝑝𝑂𝐻=−𝑙𝑜𝑔 𝑂𝐻 − 𝑝𝑂𝐻=−𝑙𝑜𝑔 0.15 𝑝𝑂𝐻=0.82 𝑝𝐻=14−0.82 𝑝𝐻=13.18

CW 7: A Better View of Acids and Bases
The Brønsted-Lowry Model The Arrhenius concept of acids and bases states that acids produce hydrogen ions in aqueous solution, while bases produce hydroxide ions. While this concept was a major step forward, it only applies to aqueous solutions and only allows one type of base (OH–). A more complete understanding is given by the Brønsted-Lowry model, which states that acids are proton (H+) donors and bases are proton acceptors.

The HCl acts as an acid (H+ donor), while H2O acts as a base (H+ acceptor). When the proton is transferred to the H2O molecule, the hydronium ion (H3O+) is formed. In this reaction, H3O+ acts as the conjugate acid, as it can donate a proton in the reverse reaction. The Cl– ion is the conjugate base, as it can accept a proton in the reverse reaction. HCl + H2O ⇌ H3O+ Cl– acid base conjugate

The two species in a conjugate acid-base pair differ by a proton only. Conjugate Acid Conjugate Base H2CO3 HCO3– CO3–2 H3O+ H2O H2S HS–

Why is the charge on the hydrogen sulfide (HS–) –1? Write the balanced chemical reaction for the reaction of H2S, an acid, with water. Write the balanced chemical reaction for CO32–, as a base, with water. H2S + H2O ⇌ H3O+ + HS– H2S donates H+ H2O accepts H+ H2O + CO32– ⇌ HCO3– + OH– H2O donates H+ CO32– accepts H+

Ammonia (NH3) can react as an acid or a base. What is the conjugate acid of ammonia? What is the conjugate base of ammonia? Conjugate acid: gain a proton = NH4+ Conjugate base: lose a proton = NH2– Complete the following acid-base reaction in which NH3(l) acts as both an acid and a base. NH3(l) + NH3(l) ⇌ NH4+(aq) + NH2–(aq)

The data below gives H3O+ and OH– concentrations for several solutions at 25°C. Label each solution as acidic, basic, or neutral. Which of the following, if any, of these expressions is a constant? Moles HOCl Added [H3O+] [OH–] 0.00 1.0x10–7 0.30 9.3x10–5 1.1x10–10 0.75 1.5x10–4 6.8x10–11 1.00 1.7x10–4 5.9x10–11 N A A A 𝐻 3 𝑂 𝑂𝐻 − 𝐻 3 𝑂 + +[ 𝑂𝐻 − 𝐻 3 𝑂 + −[ 𝑂𝐻 − 𝐻 3 𝑂 + ×[ 𝑂𝐻 −

The data below gives H3O+ and OH– concentrations for several solutions at 25°C. The value of 𝐻 3 𝑂 + ×[ 𝑂𝐻 − ] is known as the water dissociation equilibrium constant, Kw. Based on the data above, what is the value of Kw? Moles HOCl Added [H3O+] [OH–] 0.00 1.0x10–7 0.30 9.3x10–5 1.1x10–10 0.75 1.5x10–4 6.8x10–11 1.00 1.7x10–4 5.9x10–11 1.0× 10 −14

Water can react as an acid or a base. What is the conjugate acid of water? What is the conjugate base of water? Complete the following acid-base reaction in which H2O(l) acts as both an acid and a base. Write the Kw expression and use it to determine the [H3O+] and [OH–] for pure water. How does this relate to the pH of pure water? H3O+ OH– H2O(l) + H2O(l) ⇌ H3O+ + OH– 𝐾 𝑤 = 𝐻 3 𝑂 + 𝑂𝐻 − −𝑙𝑜𝑔 𝐾 𝑤 =−𝑙𝑜𝑔 𝐻 3 𝑂 + 𝑂𝐻 − 𝑝 𝐾 𝑤 =𝑝𝐻+𝑝𝑂𝐻 𝐾 𝑤 =1.0× 10 −14 14=𝑝𝐻+𝑝𝑂𝐻

The Acid Dissociation Constant In general, when an acid dissolves in water, it follows the general equation below. HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) This equation represents a competition for the proton between the two bases H2O and A–. If H2O is a much stronger base than A–, more products are made. If A– is a much stronger base than H2O, more reactants are made. The position of this equilibrium is known as the acid dissociation constant, Ka.

The Acid Dissociation Constant H2O is not included in the equilibrium expression because it is a pure liquid. In dilute solution, the H2O(l) is present in such large excess that we can assume it remains constant and is excluded H2O(l) from the equilibrium expression. 𝐾 𝑎 = 𝐻 3 𝑂 + [ 𝐴 − ] [𝐻𝐴] = 𝐻 + [ 𝐴 − ] [𝐻𝐴] Both H3O+ and H+ are commonly used to represent the hydrated proton.

Ka % Dissociation in 1.00 M Solution Hydroiodic acid, HI 3x109 100 Perchloric acid, HOClO3 1x108 Nitric acid, HNO3 28 96.7 Nitrous acid, HNO2 5.1x10–4 2.3 Acetic acid, CH3COOH 1.75x10–5 0.42 Hydrosulfuric acid, H2S 1.0x10–7 0.03 𝐾 𝑎 = 𝐻 3 𝑂 + [ 𝐴 − ] [𝐻𝐴] = 𝐻 + [ 𝐴 − ] [𝐻𝐴] HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)

Write the Ka expression for HNO3. When 1.00 mole HNO3 is added to enough water to make 1.00 L solution, [H3O+] = [NO3–] = 0.967 What is the [HNO3] in this solution? Show that Ka = 28 for HNO3. 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑁𝑂 3 − 𝐻𝑁𝑂 3 HNO3 + H2O ⇌ H3O+ + NO3– 1.00 mol – mol = mol HNO3 𝐾 𝑎 = =28

Calculate [H3O+], [NO2–], and [HNO2] for a M solution of HNO2. HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2–(aq) Dissociates 2.3% (from data table) [H3O+] = [NO2–] = 2.3% of M = M [HNO2] = M – M = M

Strong and Weak Acids A strong acid dissociates nearly completely when added to water. This means that the equilibrium lies far to the products side and Ka is a large value. A strong acid will create a conjugate base weaker than water – one with a low affinity for a proton. Conversely, a weak acid shows very little dissociation when added to water. The equilibrium lies far on the reactants side and Ka is a small value. A weak acid will produce a conjugate base stronger than water.

Based on the information presented below, Is HCl(aq) a strong acid or weak acid? Is HF(aq) a strong acid or weak acid? Label the solutions with the acid they best represent. Acid Moles of acid added [H3O+] M [OH–] M HCl 0.10 1.0x10–13 HF 8.5x10–3 1.2x10–12 HCl HF

Based on the information presented below, What do each of the following represent? Identify the species, other than water, that is not shown by the model. Provide a reason for why it is not shown. HCl – it has all dissociated, so none is present Acid Moles of acid added [H3O+] M [OH–] M HCl 0.10 1.0x10–13 HF 8.5x10–3 1.2x10–12 H3O+ Cl– HF F–

Consider the reaction for the weak acid HOCl with water: HOCl(aq) + H2O(l) ⇌ Complete and balance the equation. Write the equilibrium expression for the Ka or HOCl. H3O+(aq) + OCl–(aq) 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑂𝐶𝑙 − 𝐻𝑂𝐶𝑙

Consider the reaction for the weak acid HOCl with water: HOCl(aq) + H2O(l) ⇌ Complete the following table if 0.30 moles of HOCl are added to enough water to make 1.00 L of solution at 25°C. H3O+(aq) + OCl–(aq) Reagents HOCl H3O+ OCl– Initial moles 0.30 Chane in moles −𝑥 Equilibrium moles 0.30−𝑥 Equilibrium concentration 0.30−𝑥 1.0 +𝑥 +𝑥 𝑥 𝑥 𝑥 1.0 𝑥 1.0

Consider the reaction for the weak acid HOCl with water: HOCl(aq) + H2O(l) ⇌ The Ka of HOCl, at 25°C, is 2.9x10–8. Find “x” and determine the equilibrium [HOCl], [H3O+], and [OCl–]. Then determine the pH of the solution. H3O+(aq) + OCl–(aq) 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑂𝐶𝑙 − 𝐻𝑂𝐶𝑙 𝑝𝐻=−𝑙𝑜𝑔 9.33× 10 −5 2.9× 10 −8 = 𝑥 𝑥 −𝑥 𝑝𝐻=4.03 2.9× 10 −8 = 𝑥 K is small! ∴𝑥=9.33× 10 −5

Consider the data below: Check that the value you calculated for [H3O+] in Question 9 is correct and fill in the missing items. Explain why the number of moles of HOCl added is equal to the number of moles of HOCl at equilibrium although some HOCl reacts. Very little HOCl reacts. To two significant figures, the concentrations before and after reaction are the same Moles HOCl Added [H3O+] [OH–] [HOCl] [OCl–] 0.00 1.0x10–7 0.30 9.3x10–5 1.1x10–10 9.3x10–5 0.75 1.5x10–4 6.8x10–11 1.00 1.7x10–4 5.9x10–11

Consider the data below: Explain why the equilibrium concentration of H3O+ is equal to the equilibrium concentration of OCl–. One molecule of H3O+ and one molecule of OCl– are produced for every molecule of HOCl that reacts Explain how finding the pH of a strong acid compares to finding the pH of a weak acid. Not all of the weak acid dissociates, so we have to use an equilibrium expression to find [H+] to find pH Moles HOCl Added [H3O+] [OH–] [HOCl] [OCl–] 0.00 1.0x10–7 0.30 9.3x10–5 1.1x10–10 9.3x10–5 0.75 1.5x10–4 6.8x10–11 1.00 1.7x10–4 5.9x10–11

Strong and Weak Bases A strong base behaves similarly to a strong acid, while a weak base behaves like a weak acid. When a base (B) is added to water, OH– ions are produced according to the reaction below. The base dissociation constant, Kb, is also shown. B(aq) + H2O ⇌ BH+(aq) + OH–(aq) 𝐾 𝑏 = 𝐵𝐻 + [ 𝑂𝐻 − ] [𝐵]

Based on the information presented below, Is pyridine, C5H5N(aq), a strong base or weak base? What do each of the following represent? Moles C5H5N Added [H3O+] [OH–] 0.00 1.0x10–7 0.30 4.4x10–10 2.3x10–5 0.75 2.8x10–10 3.6x10–5 1.00 2.4x10–10 4.1x10–5 OH– HC5H5N+ C5H5N

Based on the information presented below, Identify the species, other than water, that is not shown by the model. Provide a reason for why it is not shown. H+: none is formed when the base is added to water How was the species BH, , produced? Formed when the base accepts a proton Moles C5H5N Added [H3O+] [OH–] 0.00 1.0x10–7 0.30 4.4x10–10 2.3x10–5 0.75 2.8x10–10 3.6x10–5 1.00 2.4x10–10 4.1x10–5

Consider the reaction of the weak base C5H5N with H2O: C5H5N(aq) + H2O(l) ⇌ Complete and balance the equation. Write the equilibrium expression for the Kb for C5H5N. C5H5NH+(aq) + OH–(aq) 𝐾 𝑏 = 𝐻 𝐶 5 𝐻 5 𝑁 + [ 𝑂𝐻 − ] [ 𝐶 5 𝐻 5 𝑁]

Consider the reaction of the weak base C5H5N with H2O: C5H5N(aq) + H2O(l) ⇌ Complete the following table if 0.30 moles of C5H5N are added to enough water to make 1.00 L of solution at 25°C. C5H5NH+(aq) + OH–(aq) Reagents C5H5N HC5H5N+ OH– Initial moles 0.30 Chane in moles −𝑥 Equilibrium moles 0.30−𝑥 Equilibrium concentration 0.30−𝑥 1.0 +𝑥 +𝑥 𝑥 𝑥 𝑥 1.0 𝑥 1.0

Consider the reaction of the weak base C5H5N with H2O: C5H5N(aq) + H2O(l) ⇌ The Kb of C5H5N, at 25°C, is 1.7x10–9. Find “x” and determine the equilibrium [C5H5N], [OH–], and [C5H5NH+]. Then determine the pH of the solution. C5H5NH+(aq) + OH–(aq) 𝑝𝑂𝐻=−𝑙𝑜𝑔 𝑂𝐻 − 𝐾 𝑏 = 𝐻 𝐶 5 𝐻 5 𝑁 + [ 𝑂𝐻 − ] [ 𝐶 5 𝐻 5 𝑁] 𝑝𝑂𝐻=−𝑙𝑜𝑔 2.26× 10 −5 1.7× 10 −9 = 𝑥 𝑥 −𝑥 𝑝𝑂𝐻=4.65 𝑝𝐻=14−4.65=9.35 1.7× 10 −9 = 𝑥 K is small! ∴𝑥=2.26× 10 −5

Consider the data below: Check that the value you calculated for [OH–] in Question 13 is correct and fill in the missing items. Explain why the number of moles of C5H5N added is equal to the number of moles of C5H5N at equilibrium although some C5H5N reacts. Very little C5H5N reacts Moles C5H5N Added [H3O+] [OH–] [C5H5N] [C5H5NH+] 0.00 1.0x10–7 0.30 4.4x10–10 2.3x10–5 0.30 0.75 2.8x10–10 3.6x10–5 1.00 2.4x10–10 4.1x10–5

Consider the data below: Explain why the equilibrium concentration of OH– is equal to the equilibrium concentration of C5H5NH+. One molecule of OH– and one molecule of C5H5NH+ are produced for every molecule of C5H5NH that reacts Explain how finding the pH of a strong base compares to finding the pH of a weak base. Not all of the weak base dissociates, so we have to use an equilibrium expression to find [OH –] to find pOH then pH Moles C5H5N Added [H3O+] [OH–] [C5H5N] [C5H5NH+] 0.00 1.0x10–7 0.30 4.4x10–10 2.3x10–5 0.30 0.75 2.8x10–10 3.6x10–5 1.00 2.4x10–10 4.1x10–5

Finding pH The water dissociation equilibrium constant is (at 25°C): 𝐾 𝑤 = 𝐻 3 𝑂 + 𝑂𝐻 − =1.0× 10 −14 There are several useful definitions that you should be familiar with: For 𝑝𝑋 expressions involving concentrations, the concentration units are always mole/liter, but they are omitted in the calculation. Thus, for example, values for pH are unitless. For logs, the number of sig figs is determined by the number of digits right of the decimal point. If pH = 2.15, then the corresponding [H3O+] = 7.1 × 10–3 (2 sig figs). 𝑝𝐻=− log 𝐻 3 𝑂 + 𝑝𝑂𝐻=−log 𝑂𝐻 − 𝑝 𝐾 𝑤 =−𝑙𝑜𝑔 𝐾 𝑤 In general, 𝑝𝑋=−𝑙𝑜𝑔𝑋

𝑝𝐻=−𝑙𝑜𝑔 𝐻 + Considering the information below, Show that the pH reading shown above is correct, given that [H3O+] = 5.0x10–4 M. If the acid shown is HCl, how many Cl– ions should be present? 8 Cl–(aq) ions If the acid shown is HC2H3O2, how many C2H3O2– ions should be present? 1 C2H3O2–(aq), 1 H3O+(aq), 7 HC2H3O2(aq) 𝑝𝐻=−𝑙𝑜𝑔 5.0× 10 −4 𝑝𝐻=3.30

For a neutral aqueous solution: What is the pH? What is the pOH? What is the numerical value of pKw? Recall that 𝑙𝑜𝑔 𝐴𝐵 = 𝑙𝑜𝑔 𝐴 + 𝑙𝑜𝑔 𝐵 . What is the relationship between pH, pOH, and pKw? 7 7 𝐾 𝑤 =1.0× 10 −14 𝑝 𝐾 𝑤 =14 𝐾 𝑤 = 𝐻 3 𝑂 + 𝑂𝐻 − −𝑙𝑜𝑔 𝐾 𝑤 =−𝑙𝑜𝑔 𝐻 3 𝑂 + 𝑂𝐻 − 𝑝 𝐾 𝑤 =𝑝𝐻+𝑝𝑂𝐻 14=𝑝𝐻+𝑝𝑂𝐻

Summary 6 1/18 (A) 1/17 (B) Le Châtelier’s Rainbow Lab Report
Due: 1/23 (A) & 1/22 (B) HW 5: Describing Solutions of Weak Acids Complete the assigned problem set, hand in during class. HW 6: Review for the Quarterly Assessment Due: 1/25 (1A, 2A), 1/28 (1B), 1/29 (3A) Complete the review guide using your notes/ unit packet/ resources, hand in during class. Summary 6 1/18 (A) 1/17 (B) Outcome: I can calculate the pH of a weak acid or base Goal: CW 7

Complete Question 10 from CW 7 in your unit packet.
Drill 7 1/23 (A) 1/22 (B) Outcome: I can calculate the pH of a polyprotic acid. Goal: CW 8 Hand In: Le Châtelier’s Rainbow Lab

CW 8: Polyprotic Acids Acids that can donate more than one proton, such as carbonic acid (H2CO3), are known as polyprotic acids. Polyprotic acids dissociate in a step wise fashion: H2CO3(aq) ⇌ H+(aq) + HCO3–(aq) 𝐾 𝑎 1 = 𝐻 + 𝐻𝐶𝑂 3 − 𝐻 2 𝐶𝑂 3 =4.3× 10 −7 HCO3–(aq) ⇌ H+(aq) + CO32–(aq) 𝐾 𝑎 2 = 𝐻 + 𝐶𝑂 3 2− 𝐻𝐶𝑂 3 − =5.6× 10 −11

CW 8: Polyprotic Acids The Ka1 and Ka2 values represent the acid dissociation constant for the first and second protons respectively. For most polyprotic acids, the value of Ka decreases as each successive proton is removed, and the acid involved in each step is successively weaker. It is easier to remove the first proton than the second and so on. This makes sense because as the negative charge on the acid increases, it becomes more difficult to remove the positively charged proton. Finding the pH of a solution of a polyprotic acid involves determining the [H+] contributed by each step of the dissociation process. To illustrate, we will consider a typical case, phosphoric acid, and a unique case, sulfuric acid.

CW 8: Polyprotic Acids Phosphoric Acid
Phosphoric acid behaves like most polyprotic acids in that the Ka value for the removal of each successive proton decreases. For a solution of phosphoric acid in water, only the first dissociation step makes an important contribution to [H+]. This greatly simplifies pH calculations for phosphoric acid.

CW 8: Polyprotic Acids Write the Ka expression for the stepwise loss of each proton below. H3PO4(aq) ⇌ H+(aq) + H2PO4–(aq) Ka1 = 7.5x10–3 = H2PO4–(aq) ⇌ H+(aq) + HPO4–2(aq) Ka2 = 6.2x10–8 = HPO4–2(aq) ⇌ H+(aq) + PO4–3(aq) Ka3 = 4.8x10–13 = 𝐻 + 𝐻 2 𝑃𝑂 4 − 𝐻 3 𝑃𝑂 4 𝐻 + 𝐻 𝑃𝑂 4 − 𝐻 2 𝑃𝑂 4 − 𝐻 + 𝑃𝑂 4 −3 𝐻𝑃𝑂 4 −2

H3PO4(aq) ⇌ H+(aq) + H2PO4–(aq)
CW 8: Polyprotic Acids Calculate the pH of a 5.0 M H3PO4 solution: Find the equilibrium concentrations of H3PO4, H+, and H2PO4 after the loss of the first proton. H3PO4(aq) ⇌ H+(aq) + H2PO4–(aq) 𝐾 𝑎1 =7.5× 10 −3 = 𝐻 + 𝐻 2 𝑃𝑂 4 − 𝐻 3 𝑃𝑂 4 R H3PO4(aq) H+(aq) H2PO4–(aq) I C E [E] 7.5× 10 −3 = 𝑥 𝑥 5.0−𝑥 5.0 −𝑥 +𝑥 +𝑥 5.0−𝑥 𝑥 𝑥 7.5× 10 −3 = 𝑥 𝐾 𝑎1 =7.5× 10 −3 = 𝐻 + 𝐻 2 𝑃𝑂 4 − 𝐻 3 𝑃𝑂 4 4.81 0.19 0.19 K is small 𝑥=0.19

H2PO4–(aq) ⇌ H+(aq) + HPO42– (aq)
CW 8: Polyprotic Acids Calculate the pH of a 5.0 M H3PO4 solution: Calculate the additional H+ that comes from the loss of the second proton. Does this additional H+ cause a significant change in the [H+]? H2PO4–(aq) ⇌ H+(aq) + HPO42– (aq) 𝐾 𝑎2 =6.2× 10 −8 = 𝐻 + 𝐻 𝑃𝑂 4 − 𝐻 2 𝑃𝑂 4 − R H2PO4(aq) H+(aq) HPO42– (aq) I C E [E] 6.2× 10 −8 = 𝑥 𝑥 −𝑥 0.19 0.19 −𝑥 +𝑥 +𝑥 0.19−𝑥 0.19+𝑥 𝑥 6.2× 10 −8 = 𝑥 0.19 0.19 6.2× 10 −8 K is small 𝑥=6.2× 10 −8

CW 8: Polyprotic Acids Calculate the pH of a 5.0 M H3PO4 solution:
Given the value of Ka3, would you expect the loss of the third proton to significantly impact the [H+]? Explain. The Ka is very small, so few additional H+ ionize. Ka1 = 7.5x10–3, [H+]1 = 0.19 Ka2 = 6.2x10–8, [H+]2 = 6.2x10–8 Ka3 = 4.8x10–13, [H+]3 = very small Use the [H+] to find the pH. 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 0.19 𝑝𝐻=−𝑙𝑜𝑔 𝑝𝐻=0.72 𝑝𝐻=0.72

CW 8: Polyprotic Acids Sulfuric Acid
Sulfuric acid is unique amongst the common acids in that it is a strong acid in its first dissociation step and a weak acid in its second dissociation step. H2SO4(aq) ⇌ H+(aq) + HSO4–(aq) Ka1 = very large HSO4–(aq) ⇌ H+(aq) + SO42–(aq) Ka2 = 1.2x10–2

CW 8: Polyprotic Acids Calculate the pH of a 1.0 M H2SO4 solution:
List the major species which will impact the pH. H+ HSO4– H2O Write the Ka expression for the loss of each H+. 𝐾 𝑎1 = 𝐻 + 𝐻 𝑆𝑂 4 − 𝐻 2 𝑆𝑂 4 𝐾 𝑎2 = 𝐻 + 𝑆𝑂 4 2− 𝐻𝑆𝑂 4 −

Calculate the [H+] due to the dissociation of the first proton. Calculate the [H+] due to the dissociation of the second proton. H2SO4(aq) ⇌ H+(aq) + HSO4– (aq) 𝐾 𝑎2 =𝑣𝑒𝑟𝑦 𝑙𝑎𝑟𝑔𝑒 1.0 M 1.0 M 1.0 M Treat it as a strong acid! HSO4–(aq) ⇌ H+(aq) + SO42– (aq) 𝐾 𝑎2 =1.2× 10 −2 = 𝐻 + 𝑆𝑂 4 −2 𝐻𝑃𝑂 4 − R HSO4– (aq) H+(aq) SO42– (aq) I C E [E] 1.2× 10 −2 = 1.0+𝑥 𝑥 1.0−𝑥 1.0 1.0 −𝑥 +𝑥 +𝑥 1.2× 10 −2 = 𝑥 1.0 1.0−𝑥 1.0+𝑥 𝑥 K is small 0.988 1.012 0.012 𝑥=1.2× 10 −2

What is the total [H+] and the pH of the solution? [H+]1 = 1.0 M [H+]2 = 1.2x10–2 M [H+]total = M 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 1.012 𝑝𝐻=0.0

CW 8: Polyprotic Acids Determine the pH for a 0.10 M H2SO4.
Calculate the [H+] due to the dissociation of the first proton. Calculate the [H+] due to the dissociation of the second proton. H2SO4(aq) ⇌ H+(aq) + HSO4– (aq) 𝐾 𝑎2 =𝑣𝑒𝑟𝑦 𝑙𝑎𝑟𝑔𝑒 0.10 M 0.10 M 0.10 M Treat it as a strong acid! HSO4–(aq) ⇌ H+(aq) + SO42– (aq) 𝐾 𝑎2 =1.2× 10 −2 = 𝐻 + 𝑆𝑂 4 −2 𝐻𝑃𝑂 4 − R HSO4– (aq) H+(aq) SO42– (aq) I C E [E] 1.2× 10 −2 = 𝑥 𝑥 −𝑥 0.10 0.10 −𝑥 +𝑥 +𝑥 K is NOT small Solve quadratic 0.10−𝑥 0.10+𝑥 𝑥 − 𝑥 2 −0.112𝑥 0.0902 0.1099 𝑥=9.85× 10 −3

CW 8: Polyprotic Acids Determine the pH for a 0.10 M H2SO4.
What is the total [H+] and the pH of the solution? [H+]1 = 0.10 M [H+]2 = M [H+]total = M Does the loss of the second proton impact [H+] in dilute solutions (< 1.0 M)? Yes – because the concentration of acid is smaller, the value of x becomes important in solving for [H+], and a quadratic is required Another way to say this: dissociation from HSO4– impacts [H+] 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 𝑝𝐻=0.96

Finding the pH of a Polyprotic Acid
CW 8: Polyprotic Acids Finding the pH of a Polyprotic Acid In general, the same procedures used above may be applied to determine the pH of a polyprotic acid. Typically, successive Ka value are so much smaller than the first value that only the first dissociation step makes a significant contribution to the equilibrium concentration of H+. This means that the calculation of the pH for a solution of a typical weak polyprotic acid is identical to that for a solution of a weak monoprotic acid.

CW 8: Polyprotic Acids Calculate the pH and [S2–] in a 0.10 M H2S solution. Ka1 = 1.0x10–7 Ka2 = 1.0x10–19 H2S (aq) ⇌ H+(aq) + HS– (aq) HS– (aq) ⇌ H+(aq) + S2–(aq) R H2S(aq) H+(aq) HS– (aq) I C E [E] 1.0× 10 −7 = 𝑥 𝑥 −𝑥 0.10 −𝑥 +𝑥 +𝑥 K is small 0.10−𝑥 𝑥 𝑥 𝑥=1.0× 10 −4 0.10 0.10 1.0× 10 −4 𝑝𝐻=−𝑙𝑜𝑔 1.0× 10 −4 𝑝𝐻=4 By similar math, [S2–] = 1.0x10–4

CW 8: Polyprotic Acids A typical vitamin C tablet (containing pure ascorbic acid, H2C6H6O6) weighs about 500 mg. Calculate the pH if one vitamin C tablet is dissolved into enough water to make 200 mL of solution. Ka1 = 7.9x10–5 Ka2 = 1.6x10–12 H2C6H6O6(aq) ⇌ H+(aq) + HC6H6O6– (aq) HC6H6O6– (aq) ⇌ H+(aq) + C6H6O62–(aq) 500 𝑚𝑔 𝐻 2 𝐶 6 𝐻 6 𝑂 𝑚𝐿 × 1 𝑚𝑜𝑙 𝐻 2 𝐶 6 𝐻 6 𝑂 𝑔 𝐻 2 𝐶 6 𝐻 6 𝑂 6 =0.014 𝑀

CW 8: Polyprotic Acids A typical vitamin C tablet (containing pure ascorbic acid, H2C6H6O6) weighs about 500 mg. Calculate the pH if one vitamin C tablet is dissolved into enough water to make 200 mL of solution. Ka1 = 7.9x10–5 Ka2 = 1.6x10–12 H2C6H6O6(aq) ⇌ H+(aq) + HC6H6O6– (aq) HC6H6O6– (aq) ⇌ H+(aq) + C6H6O62–(aq) R H2C6H6O6 (aq) H+(aq) HC6H6O6– (aq) I C E [E] 7.9× 10 −5 = 𝑥 𝑥 −𝑥 0.014 −𝑥 +𝑥 +𝑥 K is NOT small Solve quadratic 0.014−𝑥 𝑥 𝑥 − 𝑥 2 −7.9× 10 −5 𝑥+1.106× 10 −6 0.013 0.0010 0.0010 𝑥=0.0010

Summary 7 1/23 (A) 1/22 (B) CV Rate Law Post Lab Assessment
Due 1/25 for all classes HW 6: Review for the Quarterly Assessment Due: 1/25 (1A, 2A), 1/28 (1B), 1/29 (3A) Complete the review guide using your notes/ unit packet/ resources, hand in during class. Summary 7 1/23 (A) 1/22 (B) Outcome: I can calculate the pH of a polyprotic acid. Goal: CW 8 Hand In: Le Châtelier’s Rainbow Lab

Estimate the pH of the following solutions (NO CALCULATOR).
Drill 8 1/23 (A) 1/24 (B) -log(1.0x10–2) = -log(5.0x10–3) = -log(1.0x10–3) = -log(8.7x10–4) = -log(2.3x10–4) = -log(1.0x10–4) = pH = 2.00 pH = 2.40 Outcome: I can predict how the addition of a salt impacts the pH of a solution. Goal: CW 9 pH = 3.00 pH = 3.10 pH = 3.70 pH = 4.00

CW 9: Acid–Base Properties: Salts
A salt is an ionic compound. When a salt dissolves in water, cations and anions are produced. These ions may act as acids or bases, and thus have an impact on the pH of their solutions.

Salts That Produce Neutral Solutions Strong acids are defined by their complete dissociation in water. HCl  H+ + Cl– Ka = very large The size of Ka is very large, meaning that the reverse reaction does not occur. Thus, when anions of strong acids (such as Cl–) are placed in water, they do not combine with H+, and have no effect on the pH.

Salts That Produce Neutral Solutions Similar logic can be applied to strong bases. Because they have large Kb values, the reverse reaction does not occur. Thus, when cations of strong bases (such as Na+) are placed in water, they do not combine with OH–, and have no effect on pH. NaOH  Na+ + OH– Kb = very large

Salts That Produce Neutral Solutions If a salt consists of a cation of a strong base and an anion of a strong acid, neither ion will impact the pH of the solution, so the pH only depends on the water (neutral, pH 7). Cations of strong bases want to be dissociated Anions of strong acids want to be dissociated These ions DO NOT react with water (they want to be dissociated) and therefore do not impact the pH

Using the list of strong acids and bases, list five neutral salts. Choose any cation from the bases Choose any anion from the acids Strong Acids Strong Bases HCl HBr HI HNO3 H2SO4 HClO3 HClO4 LiOH NaOH KOH RbOH CsOH Mg(OH)2 Ca(OH)2 Sr(OH)2 Ba(OH)2

Salts That Produce Basic Solutions If one of the ions formed by the salt can act as a base, it will react with water to form OH–. For example, if sodium acetate is added to water, the acetate can act as a base: C2H3O2–(aq) + H2O(l) ⇌ HC2H3O2(aq) + OH–(aq) If the anion formed by a salt is a conjugate base of a weak acid (such as C2H3O2–), the aqueous solution will be basic.

Calculate the pH of a 0.30 M NaF solution. The Ka for HF is 7.2x10–4. Identify the major species present in the solution and determine if they have acidic or basic properties. Na+: neutral (NaOH = strong base) F–: basic (HF = weak acid) If an ion can act as an acid, write the chemical equation for its reaction with water and Ka expression. If an ion acts as a base, write the chemical equation for its reaction with water and Kb expression. F–(aq) + H2O ⇌ HF(aq) + OH–(aq) 𝐾 𝑏 = 𝐻𝐹 [ 𝑂𝐻 − ] [ 𝐹 − ]

Calculate the pH of a 0.30 M NaF solution. The Ka for HF is 7.2x10–4. Create a RICE chart to solve for the resulting concentration of H+ or OH– due to the addition of the salt. The value of Ka is given. To determine Kb, use 𝐾 𝑤 = 𝐾 𝑎 × 𝐾 𝑏 . R F–(aq) HF(aq) OH– (aq) I C E 0.30 −𝑥 +𝑥 +𝑥 0.30−𝑥 𝑥 𝑥 𝐾 𝑏 = 𝐾 𝑤 𝐾 𝑎 = 1.0× 10 − × 10 −4 =1.4× 10 −11

Calculate the pH of a 0.30 M NaF solution. The Ka for HF is 7.2x10–4. Solve for the value of x. Is this the [OH–] or [H+]? Find the pH. 1.4× 10 −11 = 𝑥 𝑥 −𝑥 1.4 × 10 −11 = 𝑥 K is small 𝑥=2.0× 10 −6 This is [OH–] 𝑝𝑂𝐻=−𝑙𝑜𝑔 𝑂𝐻 − 𝑝𝑂𝐻=−𝑙𝑜𝑔 2.0× 10 −6 𝑝𝑂𝐻=5.69 𝑝𝐻=14−5.69=8.31

Salts That Produce Acidic Solutions If one of the ions formed by the salt can act as an acid, it will react with water to form H3O+. For example, if ammonium chloride is added to water, the ammonium can act as an acid: NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) If the cation formed by a salt is a conjugate acid of a weak base (such as NH4+), the aqueous solution will be acidic.

Calculate the pH of a 0.10 M NH4Cl solution. The Kb for NH3 is 1.8x10–5. 1. Does this salt show acid-base properties? NH4Cl(s)  NH4+(aq) + Cl–(aq) NH4+: Acidic Cl–: Neutral 2. Write the acid-base reaction with water, write Ka or Kb 𝐾 𝑎 = 𝐻 3 𝑂 + [ 𝑁𝐻 3 ] [ 𝑁𝐻 4 + ] NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) 3. Solve for Ka or Kb if necessary 𝐾 𝑎 = 𝐾 𝑤 𝐾 𝑏 = 1.0× 10 − × 10 −5 =5.6× 10 −10

Calculate the pH of a 0.10 M NH4Cl solution. The Kb for NH3 is 1.8x10–5. 4. RICE and solve for x 𝐾 𝑎 = 𝐻 3 𝑂 + [ 𝑁𝐻 3 ] [ 𝑁𝐻 4 + ] R NH4+ H3O+ NH3 I C E 0.10 5.6× 10 −10 = 𝑥 [𝑥] [0.10−𝑥] −𝑥 +𝑥 +𝑥 0.10−𝑥 𝑥 𝑥 K is small 𝑥=7.5× 10 −6 5. Find the pH 𝑝𝐻=−𝑙𝑜𝑔 𝐻 + 𝑝𝐻=−𝑙𝑜𝑔 7.5× 10 −6 =5.13

A second type of salt that produces an acidic solution is one that contains a highly charged metal ion, such as Al3+ or Fe3+. These ions become hydrated, forming a weak acid. Al(H2O)63+(aq) ⇌ Al(OH)(H2O)52+(aq) + H+(aq) The high charge of the metal ion polarizes the O–H bonds in the attached water molecules, making their hydrogens more acidic than those in free water molecules. Typically, the higher the charge, the more acidic the hydrated ion.

Calculate the pH of a 0.10 M AlCl3 solution. The Ka for Al(H2O)63+ is 1.4x10–5. Al(H2O)63+(aq) ⇌ Al(OH)(H2O)52+(aq) + H+(aq) Ka for Al(H2O)63+ = 1.4x10–5 Al(H2O)63+ is a stronger acid than H2O 1.4× 10 −5 = 𝑥 𝑥 −𝑥 R Al(H2O)63+ H3O+ Al(OH)(H2O)52+ I C E 0.10 K is small −𝑥 +𝑥 +𝑥 𝑥=0.0012 0.10−𝑥 𝑥 𝑥 𝑝𝐻=−𝑙𝑜𝑔 1.2× 10 −3 𝑝𝐻=2.92

Predicting pH In the examples so far, only one ion behaves as an acid or base. Many salts produce two ions which can act as an acid or base, such as ammonium acetate (NH4C2H3O2). To predict if the solution will be acidic or basic, compare the value of Ka to Kb: Ka > Kb pH < 7 (acidic) Kb > Ka pH > 7 (basic) Ka = Kb pH = 7 (neutral)

Predict whether an aqueous solution made from the following salts will be acidic, basic, or neutral. If the solution is not neutral, write the salt hydrolysis reaction. NaNO2 Na+: from strong base (NaOH): neutral NO2–: from weak acid (HNO2): basic NO2– + H2O ⇌ HNO2 + OH – NaNO3 NO3–: from strong acid (HNO3): neutral

Predict whether an aqueous solution made from the following salts will be acidic, basic, or neutral. If the solution is not neutral, write the salt hydrolysis reaction. KClO K+: from strong base (KOH): neutral ClO–: from weak acid (HOCl): basic ClO– + H2O ⇌ HOCl + OH – CoCl3 Co3+: highly charged metal ion: acidic Co(H2O)63+(aq) ⇌ Co(OH)(H2O)52+(aq) + H+(aq) Cl–: from strong acid (HCl): neutral

Summarize salts with acid-base properties. Cation Anion Examples Effect of each ion on pH From a strong base From a strong acid From a weak acid Conjugate acid of a weak base Conjugate base of a weak acid Highly charged metal atom KNO3 NaCl Cation: no effect Anion: no effect KCN NaF Cation: no effect Anion: H+ acceptor, basic NH4Cl NH4NO3 Cation: H+ donor, acidic Anion: no effect NH4F NH4CN Cation: H+ donor, acidic Anion: H+ acceptor, basic FeCl3 Al(NO3)3 Cation: hydrated ion is H+ donor, acidic Anion: no effect

If the ion comes from a strong acid or strong base, it wants to remain dissociated, and cannot donate/accept H+ from water. Cations: Li, Na, K, Rb, Cs, Mg, Ca, Sr, Ba Anions: Cl, Br, I, NO3, SO4, ClO3, ClO4 If the ion is anything else then those listed above, it will react with water to form a weak acid or weak base. These have small Ka or Kb values, and will not dissociate in the reverse reaction. Forms either H3O+ (if ion donates H+ to H2O) or OH– (if ion accepts H+ from H2O).

Summary 8 1/23 (A) 1/24 (B) CV Rate Law Post Lab Assessment
Due 1/25 for all classes HW 6: Review for the Quarterly Assessment Due: 1/25 (1A, 2A), 1/28 (1B), 1/29 (3A) Complete the review guide using your notes/ unit packet/ resources, hand in during class. Summary 8 1/23 (A) 1/24 (B) Outcome: I can predict how the addition of a salt impacts the pH of a solution. Goal: CW 9

A student performed calculations to find the pH of two acetic acid (Ka = 7.2x10–4) solutions.
Determine if the “K is small” shortcut is acceptable for each calculation. Explain. Drill 9 2/1 (A) 2/4 (B) Outcome: I can explain relative acidity of a molecule based on its chemical structure and identify liable protons. Goal: CW 10 7.2× 10 −4 = 𝑥 𝑥 −𝑥 𝑥=0.0084 7.2× 10 −4 = 𝑥 𝑥 1.0−𝑥 𝑥=0.027 0.973 ×100=97.3% Shortcut OK! 0.0916 ×100=91.6% Shortcut NOT OK!

Le Châtelier's Rainbow Feedback
Explain equilibrium shift in terms of reaction rates (forward and reverse)

Some Stuff Safety quiz Papers to hand back
New lab policies and flip book: install in lab notebook. Goggles: place into correct basket. Bathroom passes CW 9 and CW 10: complete on your own if not done. Expect a quiz on this content soon. Le Châtelier's Rainbow will be on Q3. Projects: poster and crystal

CW 10: Relative Acid-Base Strength
The Effect of Bond Strength In order to break a chemical bond, energy must be absorbed. The amount of energy required to break (or form) a chemical bond is the bond enthalpy, a measure of bond strength. Group Acid Bond Bond Enthalpy (kJ/ mol) Ka A H2O H–O 463 1.8x10–16 H2S H–S 367 1.0x10–7 B NH4+ H–N 390 5.6x10–10 PH4+ H–P 325 ≈1014 C HF H–F 586 7.2x10–4 HCl H–Cl 432 1x106 HBr H–Br 366 1x109 HI H–I 298 3x109

Comparing the acids in Group A: How does bond enthalpy impact the strength of the acid? Explain. Higher bond enthalpy = stronger bond, harder to break = acid is less likely to dissociate. Is this trend consistent with Group B and Group C? Explain. Within a group, but not group to group, the same trend applies.

Comparing the acids in Group C: What is the difference in bond enthalpy for HBr and HI? How does relate to the difference in Ka for these acids? The enthalpy difference is 68 kJ/mol, and the acids have similar strengths (similar Ka). What is the difference in bond enthalpy for HF and HCl? How does relate to the difference in Ka for these acids? The enthalpy difference is 154 kJ/mol, and one acid is strong while the other is weak.

Chemical bonds with large bond enthalpy values are difficult to break. How does this relate to the Ka value of the acid? Use a dissociation equation to explain. H–Br(g)  H+(aq) + Br–(aq) This bond is easily to break (low bond enthalpy), so HBr dissociates to a large extent, and the acid is strong (large Ka).

The Effect of Polarity Molecule Bond Bond Enthalpy (kJ/ mol) Electronegativity of Atoms Ka H–C 413 H: 2.1 C: 2.5 3.2x10–16 H–Cl 432 Cl: 3.0 1x106

Compare the bond strength of H–C and H–Cl. Based one bond strength alone, which acid do you expect the have the higher Ka value? Based on ONLY bond strength, H–C should be stronger due to its lower bond enthalpy. Compare the difference in electronegativity values for the H–C and H–Cl bonds. Based on this, which bond is more polar, and thus more likely to dissociate in water? Electronegativity difference H–C: 0.4 Electronegativity difference H–Cl: 0.9 Based on the electronegativity difference, the –Cl bond is more polar and more likely to dissociate in water.

Use your answers to explain why the H in HCl is acidic, and the H in CHCl3 is not. The H–C bond is very weakly polar, so water has a hard time interacting with the H and cannot “grab” an H+. When the bond strength is similar, what effect does electronegativity difference have on acid strength? The greater the electronegativity difference, the more polar the bond, the easier it is to lose H+, and the stronger the acid.

A Quick Note on Electronegativity: Electronegativity is how well a nucleus attracts electrons shared in a bond. This depends on: The number of protons in the nucleus (more protons = more positive = stronger attraction) The distance of the electrons from the nucleus (further away = weaker attraction) The amount of shielding between outer electrons and the nucleus (if inner electrons “block” outer electrons from the nucleus, the attraction is weaker) In this example, you were given values for electronegativity. In most cases, you will not be given these values and should infer relative electronegativity using periodic trends. An excellent review of this concept may be found here.

Polyprotic Acid Strength For molecules with more than one hydrogen atom, the hydrogen atom with the largest partial positive charge tends to be the acidic hydrogen.

What is the sum of the partial charges on each of the molecules above? Sum for both is zero (electronically neutral). Circle the most acidic hydrogen in each acid. Explain your choice. The H atoms with the highest partial positive charge. These are the most polar bonds easily broken by interaction with water. Which acid is most likely stronger? Explain your reasoning. CCl3COOH: the H+ has a larger partial charge. The Cl atoms are highly electronegative, and “pull” electron density away from the H, making it more positive.

Acidity of Molecules with X–O–H Bonds For molecules that contain one or more oxygen atoms, the most acidic hydrogen atom is always bonded to an oxygen atom.

Circle the most acidic hydrogen in formic acid, HCOOH. ethanol, CH3CH2OH. Explain how your answers to Questions 10 and 11 consistent with Question 9. These will have the largest partial positive charge due to high electron drawing capacity of the O atom (highly electronegative).

Consider the data below. Group Acid X (in X–O–H) Partial Charge on Acidic Hydrogen Ka A CH3COOH CH3CO– 0.298 1.8x10–5 CH2ClCOOH CH2ClCO– 0.308 1.4x10–3 CHCl2COOH CHCl2CO– 0.317 5.1x10–2 CCl3COOH CCl3CO– 0.325 0.22 B HOCl Cl– 0.280 2.9x10–8 HOBr Br– 0.275 2.4x10–9 HOI I– 0.270 2.3x10–11

Consider the data below. In Group A, why does the partial charge on the acidic hydrogen increase as more chlorine atoms are added? Additional Cl atoms draw electrons away from the H atom, making its partial positive charge larger. In Group B, what structural feature accounts for the decrease in the partial charge on the acidic hydrogen? As you move down the halogen column on the periodic table, the halogen becomes less electronegative, and has less electron drawing ability.

Consider the data below. If the bond strength between the acidic hydrogen and the atom to which it is attached are roughly comparable, what is the relationship between acid strength and the partial positive charge on the acidic hydrogen? The more partial positive charge on the acidic H, the more acidic it is. These H atoms are easily broken because they are highly polar.

Acidic and Basic Oxides We have just seen that molecules with X–O–H structures can behave as acids and that the acid strength depends on the electron-withdrawing ability of X. Molecules like these can also behave as bases if the hydroxide ion is formed rather than producing a proton. X–O–H If this bond breaks, H+ is formed OH– is formed.

X–O–H If this bond breaks, H+ is formed OH– is formed. If H+ or OH– is formed depends on the X–O bond. If X has a high electronegativity, the X–O bond will be covalent and strong. When added to water, the X–O bond remains intact while the O–H bond breaks, forming H+. If X has a low electronegativity, the X–O bond will be ionic, and will break when added to water, forming OH–.

Acidic Oxides When SO3 is dissolved in water, sulfuric acid is formed. Thus, when a covalent oxide dissolves in water, an acidic solution is formed. These are called acidic oxides. SO3(g) + H2O(l)  H2SO4(aq) Since X = S in H–O–X, the bond X–O bond is covalent and remains intact. The H–O bond breaks instead, producing H+.

Basic Oxides When an ionic oxide dissolves in water, the X–O bonds breaks, forming OH–. CaO(s) + H2O(l)  Ca(OH)2(aq) These are called basic oxides and are typically formed from oxides of group 1 and 2 metals.

Refer to the data given in Question 12 to predict the value of Ka for CF3COOH. Explain your reasoning. The Ka for CF3COOH should be slightly higher than the Ka for CCl3COOH, as F is more electronegative than Cl. Ka for CF3COOH = 0.30 Rank the following oxyacids in order of increasing strength: HClO4, HClO3, HClO2, HClO. As more O atoms are added, they are better able to draw electron density away from the H, leading to a more positive partial charge on H.

For each of the following pairs of acids, predict which will have the larger value of Ka. H2S and H2Se HONO and HOPO NH4+ and Cl3NH+ H2S and H2Te HONO2 and HONO To answer, consider bond strength and electronegativity.

Salicylic acid is a weak acid with Ka = 3.0x10–4. Complete the reaction below with water by drawing the products. + H3O+

Salicylic acid is a weak acid with Ka = 3.0x10–4. Calculate the pH of 50 mL of 0.15 M salicylic acid. SA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq) 3.0× 10 −4 = 𝑥 𝑥 −𝑥 R SA H3O+ A– I C E K is small 0.15 𝑥=4.5× 10 −5 −𝑥 +𝑥 +𝑥 𝑝𝐻=−𝑙𝑜𝑔 4.5× 10 −5 0.15−𝑥 𝑥 𝑥 𝑝𝐻=4.35

Predict if the following will result in an acidic or basic solution when added to water. Include the acid or base formed. SO2 K2O Li2O CO2 acid SO2 + H2O  H2SO4 base K2O + H2O  2KOH base Li2O + H2O  2LiOH acid CO2 + H2O  H2CO3

Summary 9 2/1 (A) 2/4 (B) Complete CW 9 and CW 10
A video going over these will be posted. OWL needs to be updated – don’t worry about OWL until I update you next class. Summary 9 2/1 (A) 2/4 (B) Outcome: I can explain relative acidity of a molecule based on its chemical structure and identify liable protons. Goal: CW 10

Write the reaction with water for any acidic or basic ions:
NaNO3 LiC2H3O2 KI NH4NO2 Drill 10 2/5 (A) 2/6 (B) N N C2H3O2– + H2O ⇌ HC2H3O2 + OH– Outcome: I can identify buffer systems and calculate their pH. Goal: CW 11 N B N N NH4+ + H2O ⇌ NH3 + H3O+ A B NO2– + H2O ⇌ HNO2 + OH–

CW 11: Buffer Systems Common Ions
The presence of a common ion can limit the dissociation of an acid, much like how it can limit the solubility of a salt. Consider a mixture of the weak acid hydrofluoric acid (Ka = 7.2x10–4) and its salt sodium fluoride. How will this impact the acid dissociation constant?

SKIP (for now?) CW 11: Buffer Systems
In a 1.0 M solution of HF, the [H+] = 2.7x10–2 M, and a percent dissociation of 2.7% Calculate the [H+] and the percent dissociation of HF (Ka = 7.2x10–4) in a solution containing 1.0 M NaF. Compare the pH and percent dissociation of M HONH2 (Kb = 1.1x10–8) in pure water and in a M HONH3Cl solution. SKIP (for now?)

CW 11: Buffer Systems Buffered Solutions
Buffers are solutions that contain components to allow the solution to resist changes in pH when an acid or base is added. Buffers are used in medicine, industry and manufacturing when pH stable solutions are required. Eating, breathing, and exercising can all alter the pH of your blood, but the buffers in your blood help to control pH and keep enzymes working properly.

CW 11: Buffer Systems For each beaker, label each reaction equation with SA (strong acid), SB (strong base), WA (weak acid), or WB (weak base) for the reactants; and write CA (conjugate acid) or CB (conjugate base) for the products. For Beaker A and B, fill in the RICE/RICF charts to determine all equilibrium/final concentrations of the molecules or ions. RICF (“final”) is used for reactions that go to completion, such as reactions with strong acids or bases. For Beaker C and D, fill in the RICE charts, using “x” as needed. Write the Ka expression and show calculations to solve for the equilibrium concentrations.

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH–(aq)
CW 11: Buffer Systems Beaker A: 100 mL pure water H2O(l) + H2O(l) ⇌ H3O+(aq) + OH–(aq) R 2 H2O ⇌ H3O+ OH– I --- C E WB WA CA CB +1.0x10–7 +1.0x10–7 1.0x10–7 1.0x10–7

HCl(aq) + H2O(l) ⇌ H3O+(aq) + Cl–(aq)
CW 11: Buffer Systems Beaker B: 100 mL of 1.00 M HCl HCl(aq) + H2O(l) ⇌ H3O+(aq) + Cl–(aq) R HCl ⇌ H3O+ Cl– I --- C F SA WB CA CB 1.00 –1.00 +1.00 +1.00 0.00 1.00 1.00

CW 11: Buffer Systems Beaker C: 100 mL of 1.00 M HNO2 (Ka = 4.6x10–4)
HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2–(aq) R HNO2 ⇌ H3O+ NO2– I --- C E solve WA WB CA CB 1.00 –x +x +x 1.00 – x x x 0.979 2.1x10–2 2.1x10–2 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑁𝑂 2 − 𝐻𝑁𝑂 2 4.6× 10 −4 = 𝑥 𝑥 −𝑥 K is small! 𝑥=2.1× 10 −2

CW 11: Buffer Systems HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2–(aq) R HNO2 ⇌
Beaker D: 100 mL of 1.00 M HNO2 & 100 mL of 1.00 M NaNO2 (Be careful – dilution!) HNO2(aq) + H2O(l) ⇌ H3O+(aq) + NO2–(aq) R HNO2 ⇌ H3O+ NO2– I --- C E Solve WA WB CA CB 0.500 0.500 –x +x +x 0.500 – x x x 0.500 4.6x10–4 0.500 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑁𝑂 2 − 𝐻𝑁𝑂 2 4.6× 10 −4 = 𝑥 𝑥 −𝑥 K is small! 𝑥=4.6× 10 −4

CW 11: Buffer Systems none H3O+
Imagine that 1.00 M sodium hydroxide was added dropwise to each of the beakers. List the species in each beaker that would react with this added base and neutralize it. If neutralization is not likely to occur, write none. Beaker A Pure Water Beaker B HCl Beaker C HNO2 Beaker D HNO2 & NaNO2 none (H2O is a very weak acid) H3O+

CW 11: Buffer Systems none none NO2–
Imagine that 1.00 M hydrochloric acid was added dropwise to each of the beakers. List the species in each beaker that would react with this added base and neutralize it. If neutralization is not likely to occur, write none. Beaker A Pure Water Beaker B HCl Beaker C HNO2 Beaker D HNO2 & NaNO2 none none NO2–

CW 11: Buffer Systems A buffer solution is one that can neutralize small quantities of acid and base. This is possible because the solution contains both a weak acid and a weak base (the weak base is usually the conjugate base of the weak acid), allowing the solution to keep a constant pH.

CW 11: Buffer Systems D Which beaker contains a buffer? _____ For this buffer… What species is the weak acid? What species is the weak base? WA: HNO2 WB: NO2– Write the neutralization reaction if 1.0 M NaOH is added to the buffer. NaOH + HNO2  H2O + NaNO2 The weak acid reacts with NaOH; reaction goes to completion because NaOH is a strong base Write the neutralization reaction if 1.0 M HCl is added to this buffer. HCl + NO2–  HNO2 + Cl– The weak base reacts with HCl; reaction goes to complete because HCl is a strong acid Suppose that you added 101 mL of 1.0 M HCl to the buffer. The solution will become very acidic. Why is this the case? All of the weak base has reacted, so when the last mL of HCl is added, nothing is present to neutralize it.

CW 11: Buffer Systems Fill in the chart below to determine which of the following solutions are buffer solutions. Determine the three most dominant species in the solution. Label each dominant species with either strong acid (SA), strong base (SB), weak acid (WA), weak base (WB) or neutral ion (N). State whether the solution is a buffer solution.

CW 11: Buffer Systems Substances mixed in equal volumes
List the 3 dominant species and label as SA, SB, WA, WB, or N Buffer? 1.0 M HCl 1.0 M NaCl _________ _________ _________ 1.0 M HNO2 1.0 M HNO3 1.0 M NH4Cl 1.0 M NH3 1.0 M CH3COOH H+ Cl– Na+ No SA N N HNO2 H+ NO3– No WA SA N NH4+ Cl– NH3 Yes WA N WB CH3COOH Cl– Na+ No WA N N

CW 11: Buffer Systems Substances mixed in equal volumes
List the 3 dominant species and label as SA, SB, WA, WB, or N Buffer? 1.0 M HF 1.0 M NaF _________ _________ _________ 1.0 M H2SO4 1.0 M Na2SO4 1.0 M KOH 1.0 M KCl 1.0 M Ca(OH)2 1.0 M HCl _______ _______ _______ _______ HF F– Na+ Yes WA WB N H+ Na+ HSO4– No SA N WA K+ Cl– OH– No N N SB Ca2+ Cl– OH– H+ No N N SB SA

CW 11: Buffer Systems Determine the pH in each of the following buffer systems by calculating [H3O+]. Beaker 1 100 mL 1.0 M NH4Cl & 100 mL 1.0 M NH3 Ka for NH4+ = 5.6x10–10 Net ionic: NH4Cl(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) R NH4+ ⇌ H3O+ NH3 I 0.500 --- 0.500 C –x +x E 0.500–x x 0.500+x Beaker 2 100 mL 1.0 M HF & 100 mL 1.0 M NaF Ka for HF = 3.5x10–4 Net ionic: HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq) R HF ⇌ H3O+ F– I 0.500 --- 0 C –x +x E 0.500–x x 0.500+x 𝐾 𝑎 = 𝐻 3 𝑂 + 𝑁𝐻 𝑁𝐻 4 + 𝐾 𝑎 = 𝐻 3 𝑂 + 𝐹 − 𝐻𝐹 𝑥=5.6× 10 −10 𝑥=3.5× 10 −4 𝑝𝐻=−𝑙𝑜𝑔 5.6× 10 −10 =9.25 𝑝𝐻=−𝑙𝑜𝑔 3.5× 10 −4 =3.46

CW 11: Buffer Systems Determine the pH in each of the following buffer systems by calculating [H3O+]. Beaker 3 100 mL 1.0 M H2CO3 & 100 mL 1.0 M NaHCO3 Ka for H2CO3 = 4.4x10–7 Net ionic: R H2CO3 ⇌ H3O+ HCO3– I 0.500 --- 0 0.500 C –x +x +x E 0.500–x x 0.500+x Beaker 4 100 mL 1.0 M HF & 300 mL 1.0 M NaF Ka for HF = 3.5x10–4 Net ionic: HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq) R HF ⇌ H3O+ F– I 0.250 --- 0 0.750 C –x +x +x E 0.250–x x 0.750+x H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3–(aq) 𝐾 𝑎 = 𝐻 3 𝑂 + 𝐻𝐶𝑂 3 − 𝐻 2 𝐶𝑂 3 𝐾 𝑎 = 𝐻 3 𝑂 + 𝐹 − 𝐻𝐹 𝑥=4.4× 10 −7 𝑥=1.2× 10 −4 𝑝𝐻=−𝑙𝑜𝑔 4.4× 10 −7 =6.36 𝑝𝐻=−𝑙𝑜𝑔 1.2× 10 −4 =3.93

CW 11: Buffer Systems Do all buffers keep solutions at a neutral pH? Explain. No, depending on the components the “buffering zone” occurs at different pH values, some acidic, some basic. Calculate the pKa by taking the –log of the Ka. Beaker 1 Ka = 5.6x10–10 pKa = 2 Ka = 3.5x10–4 3 Ka = 4.4x10–7 4 −𝑙𝑜𝑔 5.6× 10 −10 =9.25 −𝑙𝑜𝑔 3.5× 10 −4 =3.46 −𝑙𝑜𝑔 4.4× 10 −7 =6.36 −𝑙𝑜𝑔 3.5× 10 −4 =3.46

CW 11: Buffer Systems How are the pKa values of the weak acids related to the pH of the buffer system that is made when equal molarities of acid and conjugate base are combined? The pH = pKa When unequal molarities of weak acid and conjugate base are combined to make a buffer (Beaker 4), does the pH change significantly from the pKa of the weak acid? The pH ≈ pKa Beaker 2: 3.46 Beaker 4: 3.93 (a little more basic due to extra F– from NaF)

CW 11: Buffer Systems An industrial process requires a constant pH of The weak acids available in the warehouse are listed below. Which acid would be best to make a buffer with pH 3.00? Choose the acid with a pKa closest to 3.00 HNO2 Ka = 7.2x10–4 pKa = 3.14 What other chemical(s) are needed to prepare the buffer? A neutral cation with NO2–, such as NaNO2 or KNO2 HC7H3O2 (benzoic acid) HNO2 (nitrous acid) H3PO3 (phosphorous acid) HCN (hydrocyanic acid) Ka = 6.3x10–5 Ka = 7.2x10–4 Ka = 3.7x10–2 Ka = 6.2x10–10

The Henderson-Hasselbalch Equation
CW 11: Buffer Systems The Henderson-Hasselbalch Equation The equation below represents typical acid dissociation. Rearranging, we have: We can take the log of each side of the above equation: And rearrange to yield the Henderson-Hasselbalch equation: HA(aq) ⇌ H+(aq) + OH–(aq) 𝐾 𝑎 = 𝐻 + 𝐴 − 𝐻𝐴 𝐻 + = 𝐾 𝑎 × 𝐻𝐴 𝐴 − −𝑙𝑜𝑔 𝐻 + =−𝑙𝑜𝑔 𝐾 𝑎 × 𝐻𝐴 𝐴 − 𝑝𝐻=𝑝 𝐾 𝑎 −𝑙𝑜𝑔 𝐻𝐴 𝐴 − 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 𝐻𝐴 or 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 𝑏𝑎𝑠𝑒 𝑎𝑐𝑖𝑑

CW 11: Buffer Systems Return to Question 10 and calculate the pH of each buffer using the Henderson-Hasselbalch equation. 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 𝐻𝐴 Beaker 1 𝑝𝐻=9.25+𝑙𝑜𝑔 =9.25 Beaker 2 𝑝𝐻=3.46+𝑙𝑜𝑔 =3.46 Beaker 3 𝑝𝐻=6.36+𝑙𝑜𝑔 =6.36 Beaker 4 𝑝𝐻=3.46+𝑙𝑜𝑔 =3.94

CW 11: Buffer Systems Buffering Capacity
Buffer capacity is defined as the moles of an acid or base necessary to change the pH of a solution by 1, divided by the pH change and the volume of buffer in liters; it is a unitless number. It represents the number of protons or hydroxide ions the buffer can absorb without a significant change in pH. Buffer capacity depends on the magnitudes of [A–] and [HA]. The pH depends on the ratio [HA]/[A–].

CW 11: Buffer Systems Buffering Capacity
Large changes to the ratio [HA]/[A–] will produce large changes in pH. This can be avoided by having equal concentrations of [A–] and [HA]. In this case, [HA]/[A–] = 1, the log(1) = 0, and the pH = pKa: 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 𝐻𝐴 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 1 𝑝𝐻=𝑝 𝐾 𝑎

CW 11: Buffer Systems A chemist needs a solution buffered at pH 4.30.
Acid Ka [HA]/[A–] pKa Chloroacetic acid 1.35x10–3 3.7x10–2 2.9 Propanoic acid 1.3x10–5 3.8 4.9 Benzoic acid 6.4x10–5 0.78 4.2 Hypochlorous acid 3.5x10–8 1.4x103 7.5 A pH of 4.30 corresponds to 5.0x10–5: Then solve: 𝑝𝐻=−𝑙𝑜𝑔⁡[ 𝐻 + ] 𝐻 + = 𝐾 𝑎 × 𝐻𝐴 𝐴 − [ 𝐻 + ]= 10 −𝑝𝐻 𝐻 + 𝐾 𝑎 = 𝐻𝐴 𝐴 − = 10 −4.30 =5.0× 10 −5 𝑀

CW 11: Buffer Systems A chemist needs a solution buffered at pH 4.30.
Calculate the ratio [HA]/[A–] at pH = 4.30 for each acid and fill in to the data table. Calculate the pKa for each acid and fill in to the data table. Which system will work best? Explain. Benzoic acid, because the ratio is close to 1 and the pKa is closest to the desired pH.

Drill 10 2/5 (A) 2/6 (B) HW 7: Relative Acid-Base Strength
Due: 2/7 (A) and 2/8 (B) Complete CW 11, Questions 1 to 10 Drill 10 2/5 (A) 2/6 (B) Outcome: I can identify buffer systems and calculate their pH. Goal: CW 11

Drill 11 2/7 (A) 2/8 (B) Immediately continue working on CW 11.
Outcome: I can create a buffer and test buffer capacity. Goal: CW 12

CW 12: Designing a Buffer 4 3 2 1 No HC2H3O2 Left
What’s 1, 2, 3, 4? 4 At any point: 𝑝𝐻= 𝑝𝐾 𝑎 +𝑙𝑜𝑔 𝑏𝑎𝑠𝑒 𝑎𝑐𝑖𝑑 Equivalence Point pH = 10.00 3 pH = pKa At the ½ equivalence point: Exactly half of the acid has been neutralized, so [HA] = [A– ] 𝐻 + = 𝐾 𝑎 × [𝐻𝐴] [ 𝐴 − ] ∴ 𝐻 + = 𝐾 𝑎 ∴𝑝𝐻=− log 𝐾 𝑎 = 𝑝𝐾 𝑎 2 1 No NaOH Added Equivalence Point @ 21.2 mL

CW 12: Designing a Buffer Give a definition of a buffer.
Buffer: an equilibrium system that can neutralize small amounts of added acid or base to keep pH constant. Two components: Weak acid, neutralizes added base Weak base (usually the conjugate base of the weak acid), neutralizes added acid

The pH is relatively stable, despite the addition of NaOH
CW 12: Designing a Buffer The titration curve below was obtained by titrating acetic acid with sodium hydroxide. Based on your definition, circle the region on the graph that shows buffering. Justify your choice. The pH is relatively stable, despite the addition of NaOH

CW 12: Designing a Buffer The titration curve below was obtained by titrating acetic acid with sodium hydroxide. What is the buffering range of pH values? pH 5 to pH 6.5 (roughly) Around mL of NaOH added, the pH increases rapidly. Explain this in terms of the species available to react before and after the sharp increase. Before the sharp increase, there is still some acetic acid left. Once this buffer component is used up, there is nothing to neutralize the added NaOH, so the pH shoots up.

CW 12: Designing a Buffer The pH curve of two titrations are shown below. Which one represents the titration of a weak acid with NaOH? Of a strong acid with NaOH? In what ways are they different? Do either show a buffering effect? Weak: Shows buffering effect around pH 4.5 Less angry snake Weak Acid Strong: No buffering effect More angry snake Strong Acid

CW 12: Designing a Buffer The pH curve of two titrations are shown below. What is meant by the term “equivalence point”? What are the approximate equivalence points for both curves? Does the equivalence point always occur at pH 7? Explain. Point at which [H+] = [OH–], middle of the sharp increase Does not always occur at 7 due to equilibrium At what point along a titration curve is pKa = pH for a weak acid? What is the pKa and Ka of the acid? pKa = pH at halfway to the equivalence point pKa = 4.5 Ka = 3.5x10–5 Why do titration curves of strong acids and bases “flatten out” around pH 1 and pH 13? Around 1: acid is large in excess Around 13: base is large in excess

CW 12: Designing a Buffer When can salts act as buffers? What are some families of salts that can act as buffers? A salt of a weak acid or base can act as a buffer Acetates Sulfites Nitrites Cyanides Etc.

CW 12: Designing a Buffer A buffer system was created with a target pH of 4 ± 1. The buffer was titrated with 0.10 M NaOH and 0.10 M HCl to determine its buffer capacity.

CW 12: Designing a Buffer A buffer system was created with a target pH of 4 ± 1. The buffer was titrated with 0.10 M NaOH and 0.10 M HCl to determine its buffer capacity. Determine the volume of HCl that the buffer neutralize before the pH changes by 1 unit. Use the concentration to find the moles of HCl that were neutralized. Determine the volume of NaOH that the buffer neutralize before the pH changes by 1 unit. Use the concentration to find the moles of NaOH that were neutralized. 7 𝑚𝐿 1 × 0.10 𝑚𝑜𝑙 𝐻𝐶𝑙 1000 𝑚𝐿 =7× 10 −4 𝑚𝑜𝑙 𝐻𝐶𝑙 5 𝑚𝐿 1 × 0.10 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1000 𝑚𝐿 =5× 10 −4 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻

CW 12: Designing a Buffer A buffer system was created with a target pH of 4 ± 1. The buffer was titrated with 0.10 M NaOH and 0.10 M HCl to determine its buffer capacity. If the buffer system was created with 5.1x10–4 moles of weak acid and 7.1x10–4 moles of its conjugate base, calculate the following ratios: 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝑙 𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑏𝑎𝑠𝑖𝑐 𝑏𝑢𝑓𝑓𝑒𝑟 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 = 7× 10 −4 7.1× 10 −4 =0.99 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑧𝑒𝑑 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑎𝑐𝑖𝑑𝑖𝑐 𝑏𝑢𝑓𝑓𝑒𝑟 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 = 5× 10 −4 5.1× 10 −4 =0.98

CW 12: Designing a Buffer A buffer system was created with a target pH of 4 ± 1. The buffer was titrated with 0.10 M NaOH and 0.10 M HCl to determine its buffer capacity. Explain how these ratios relate to buffer capacity using the Henderson-Hasselbalch equation (HINT: CW 11). For effective buffers, the ratio should be close to 1. When this ratio is plugged into HHB, the log(1) = 0, and the desired pH of the buffer is equal to the pKa (or pKb). 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 𝐴 − 𝐻𝐴 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 1 𝑝𝐻=𝑝 𝐾 𝑎

CW 12: Designing a Buffer Write a hypothesis in “If we combine… then our buffer will…” format. The masses and volumes of all materials used should be recorded in your lab notebook. Include supporting evidence for your choice in components.

CW 12: Designing a Buffer Your employer has requested several buffers to be used in ongoing research. Each group will design a buffer system to meet the criteria listed on a “mission card,” which you will obtain from Ms. L. You must prepare 100 mL of a buffer solution that stays within ±0.5 units of the target pH that can maintain a relatively constant pH (within one pH unit of the initial value) with the addition of up to 20 mL of 0.20 M HCl or NaOH. The materials available to create your buffer are listed below. Solutions 0.10 M acetic acid (CH3COOH), Ka = 1.8x10–5 0.10 M ammonia (NH3) 0.10 M sodium dihydrogen phosphate (NaH2PO4), Ka of H2PO4 = 6.3x10–8 0.10 M citric acid (H3C6H5O7), Ka = 7.8x10–4 0.10 M monohydrogen citrate (Na2HC6H5O7), Ka = 4.1x10–7 Solids Ammonium chloride (NH4Cl), Ka of NH4+ = 5.7x10–10 Sodium acetate (NaCH3COO) Sodium hydrogen phosphate (Na2HPO4) Sodium citrate (Na3C6H5O7) Sodium chloride (NaCl) Sodium dihydrogen citrate (NaH2C6H5O7)

Summary 11 2/7 (A) 2/8 (B) Complete Pre-Lab work from Page 46
Collecting the pre-lab on the day of the experiment! Summary 11 2/7 (A) 2/8 (B) Outcome: I can create a buffer and test buffer capacity. Goal: CW 12

Buffer Design Lab Begin testing buffer capacity immediately
Group member 1: Logon to a laptop and set up pH probe Group member 2: Measure 50.0 mL of the buffer into each of the Erlenmeyer flasks Group member 3: Adjust burettes to 0.00 mL using the NaOH and HCl beakers To find the equivalence point: Click on Data in the top ribbon, then select New Calculated Column In the Name box, type in 1st Dev. In the Short Name box, type D1. In the expression box, type: delta(“pH”) and click done. On the graph, on the Y-axis, click where it says pH and select 1D. Rescale your graph. The peak is the equivalence point. To save: Click File, Export As, and select CSV. Give the file a meaningful name and save it where you can find it later/ share with your group members.

Drill 12 2/15 (A) 2/19 (B) Lab Assessment!
Outcome: I can explain features of titration curve and calculate the pH at any point during a titration. Goal: CW 13

CW 13: Acid-Base Titrations
An acid-base titration is a method to determine an unknown concentration of acid or base. A titration determines the volume of a standard solution with known concentration that is required to completely react the acid or base in the sample. An acid-base indicator changes color at the end point of a titration. Ideally, the indicator is selected so that it changes color when [H+] = [OH–], known as the equivalence point.

A titration curve shows pH versus mL of titrant (in the tube) added. The shape of the curve is often characteristic of the type of acid and base (strong, weak, polyprotic) that react. Another special point occurs halfway to the equivalence point. At this point, half of the acid/base present in the analyte has reacted, and the concentration of the acid/base is equal to the concentration of its conjugate. At the halfway point, [H+] = Ka and pH = pKa.

Problem Solving Strategy for Titrations Involving Weak Acids or Bases Solve the stoichiometry first for any reactions with strong acids or bases – these are assumed to go to completion. Determine the concentrations of each species before and after the reaction occurs. Handle the equilibrium second. List all species present in the solution with the concentrations (from the stoichiometry step you already solved) and any K values. The pH depends on the K of the strongest acid or base.

Some Tricks Before the equivalence point, you have a buffer system, so you can use the Henderson-Hasselbalch equation or the rearranged Ka or Kb expressions: At the equivalence point, [H+] = [OH–], so this is just an equilibrium problem due to reaction of any conjugate acid or base formed during the titration. After the equivalence point, the pH depends on the acid or base in excess. You can also determine Ka or Kb from Kw: 𝐻 + = 𝐾 𝑎 × 𝐻𝐴 𝐴 − 𝑂𝐻 − = 𝐾 𝑏 × 𝐵 𝐻𝐵 + 𝑝𝐻=𝑝 𝐾 𝑎 −𝑙𝑜𝑔 𝐻𝐴 𝐴 − 𝑝𝑂𝐻=𝑝 𝐾 𝑏 −𝑙𝑜𝑔 𝐵 𝐻𝐵 + 𝐾 𝑎 = 𝐾 𝑤 𝐾 𝑏 𝐾 𝑏 = 𝐾 𝑤 𝐾 𝑎

Strong Acid/ Strong Base Titrant: 0.10 M NaOH Analyte: 50.0 mL of 0.10 M HNO3 Initial Present: H+, NO3–, H2O pH: Depends on [H+] due to complete dissociation of strong acid Before Equivalence Point Present: H+ (less than initially due to reaction with OH–), Na+, NO3–, H2O, Added OH– is neutralized pH: Depends on [H+] that wasn’t neutralized At Equivalence Point Present: Na+, NO3–, H2O, All H+ and OH– is neutralized pH: 7 After Equivalence Point Present: Na+ NO3–, OH–, H2O pH: Depends on excess [OH–]

Strong Acid/ Strong Base

Weak Acid/ Strong Base Titrant: 0.10 M NaOH Analyte: 50.0 mL of 0.10 M HC2H3O2 Initial Present: HC2H3O2, H+, C2H3O2–, H2O pH: Depends on Ka of HC2H3O2 (RICE) Before Equivalence Point Present: H+ (less than initially due to reaction with OH–), Na+, C2H3O2–, H2O, Added OH– is neutralized pH: Depends on Ka of HC2H3O2 (RICE) and amount of HC2H3O2 that wasn’t neutralized (determined from stoich) At Equivalence Point Present: Na+, C2H3O2–, H2O, All H+ and OH– is neutralized pH: >7, Depends on Kb of C2H3O2– due to reaction with water (RICE) After Equivalence Point Present: Na+, C2H3O2–, H2O, OH– pH: Depends on excess [OH–], contribution from C2H3O2– is small (small Kb)

Weak Acid/ Strong Base

Weak Base/ Strong Acid Titrant: 0.10 M HCl Analyte: 50.0 mL of 0.10 M NH3 Initial Present: NH3, NH4+, OH–, H2O pH: Depends on Kb of NH3 (RICE) Before Equivalence Point Present: OH– (less than initially due to reaction with HCl), NH3, NH4+, Cl–, H2O, Added H+ is neutralized pH: Depends on Kb of NH3 (RICE) and amount of NH3 that wasn’t neutralized (determined from stoich) At Equivalence Point Present: NH4+, Cl–, H2O, All H+ and OH– is neutralized pH: <7, Depends on Ka of NH4+ due to reaction with water (RICE) After Equivalence Point Present: H+, NH4+, Cl–, H2O, pH: Depends on excess [H+], contribution from NH4+ is small (small Ka)

Weak Acid/ Strong Base

Mark the halfway point and the write Ka or Kb for each acid or base on its graph. Strong acid, so no need to find Ka 10–4 = Ka 10–9 = Kb ½ EP = pKa 10–pH = Ka

HCN is a very weak acid (Ka = 6.2x10–10). If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution: After 8.00 mL of M NaOH has been added. 1. Stoichiometry HCN + NaOH  NaCN + HOH Reaction HCN NaOH  NaCN H2O Before 50.0 mL × M = 5.00 mmol 8.00 mL × M = mmol -- 0 mmol Used/ Formed –0.800 mmol mmol After 4.20 mmol 0.800 mmol 𝐻𝐶𝑁 = 4.20 𝑚𝑚𝑜𝑙 58 𝑚𝐿 = 𝑀 𝐶𝑁 − = 𝑚𝑚𝑜𝑙 58 𝑚𝐿 = 𝑀

HCN is a very weak acid (Ka = 6.2x10–10). If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution: After 8.00 mL of M NaOH has been added. 2. Equilibrium 𝐾 𝑎 = 𝐶𝑁 − [ 𝐻 3 𝑂 + ] [𝐻𝐶𝑁] HCN + H2O ⇌ H3O+ + CN– 6.2× 10 −10 = 𝑥 [ 𝑥] [0.0724−𝑥] R HCN H2O ⇌ H3O+ CN– I 0.0724 -- 0.0138 C –x +x E 0.0724–x x x 𝐾 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 6.2× 10 −10 = 𝑥 [0.0138] [0.0724] ∴𝑥=3.3× 10 −9 Note: if you wrote the CN– + H2O ⇌ HCN + OH– equilibrium, you’d solve this problem using Kb, find [OH–], pOH, then pH. You get the same answer! 𝑝𝐻=− log 𝑥 =8.49

HCN is a very weak acid (Ka = 6.2x10–10). If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution: At the halfway point of the titration. At the halfway point, pH = pKa and [H+] = Ka pH = –log[H+] = –log(Ka) = –log(6.2×10–10) = 9.21

HCN is a very weak acid (Ka = 6.2x10–10). If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution: At the equivalence point of the titration. What’s present in solution? How much CN– is present? In what volume? What molarity? Na+ OH– H+ CN– H2O N All EP WB 50.0 𝑚𝐿 1 × 𝑚𝑜𝑙 𝐻𝐶𝑁 1 𝐿 × 1 𝑚𝑜𝑙 𝐶𝑁 − 1 𝑚𝑜𝑙 𝐻𝐶𝑁 =5.00 𝑚𝑚𝑜𝑙 𝐶𝑁 − 50.0 mL HCN mL NaOH = mL total 5.00 𝑚𝑚𝑜𝑙 𝐶𝑁 − 𝑚𝐿 = 𝑀 𝐶𝑁 −

HCN is a very weak acid (Ka = 6.2x10–10). If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution: At the equivalence point of the titration. Now RICE 𝐾 𝑏 = [𝐻𝐶𝑁] 𝑂𝐻 − [ 𝐶𝑁 − ] CN– + H2O ⇌ HCN + OH– 1.6× 10 −5 = 𝑥 [𝑥] [0.050−𝑥] R CN– H2O ⇌ HCN OH– I 0.050 -- C –x +x E 0.050 – x x 𝐾 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 1.6× 10 −5 = 𝑥 [𝑥] [0.050] ∴𝑥=8.9× 10 −4 𝑝𝑂𝐻=− log 𝑥 =3.05 𝑝𝐻=14−𝑝𝑂𝐻=10.95

The graph below shows the titration curves for the titration of several mL samples of 0.10 M acids of varying strengths with 0.10 M NaOH. What is the equivalence point for all graphs? Does the strength of the acid or the amount of acid determine the equivalence point? The equivalence point occurs at 50 mL of added 0.10 M NaOH for all acids Amount of acid or base determines equivalence point How does the strength of the acid impact the shape of the pH curve? The weaker the acid, the greater the pH at the equivalence point

If a mL sample of M NH3 is titrated with M HCl, calculate the pH of the solution: Initially Only NH3 and H2O are initially present, so pH depends on NH3. 𝐾 𝑏 = [ 𝑁𝐻 4 + ] 𝑂𝐻 − [ 𝑁𝐻 3 ] NH3 + H2O ⇌ NH4+ + OH– R NH3 H2O ⇌ NH4+ OH– I 0.050 -- C –x +x E 0.050 – x x 1.8× 10 −5 = 𝑥 [𝑥] [0.050−𝑥] 𝐾 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 1.8× 10 −5 = 𝑥 [𝑥] [0.050] ∴𝑥=9.5× 10 −4 𝑝𝑂𝐻=− log 𝑥 =3.02 𝑝𝐻=14−𝑝𝑂𝐻=10.98

If a mL sample of M NH3 is titrated with M HCl, calculate the pH of the solution: After mL of 0.10 M HCl has been added 1. Stoichiometry HCl + NH3  NH4+ + Cl– Reaction HCl NH3  NH4+ H2O Before 50.00 mL × 0.10 M = 5.00 mmol 100.0 mL × M = 5.00 mmol -- 0 mmol Used/ Formed –5.00 mmol +5.00 mmol After 5.00 mmol All NH3 has been neutralized = equivalence point 𝑁𝐻 4 + = 5.00 𝑚𝑚𝑜𝑙 𝑚𝐿 = 𝑀 pH only depends on NH4+ formed during reaction.

If a mL sample of M NH3 is titrated with M HCl, calculate the pH of the solution: After mL of 0.10 M HCl has been added 2. Equilibrium 𝐾 𝑎 = [ 𝑁𝐻 3 ] 𝐻 3 𝑂 + [ 𝑁𝐻 4 + ] NH4+ + H2O ⇌ NH3 + H3O+ 5.6× 10 −10 = 𝑥 [𝑥] [0.033−𝑥] R NH4+ H2O ⇌ NH3 H3O+ I 0.033 -- C –x +x E 0.033 – x x 𝐾 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 5.6× 10 −10 = 𝑥 [𝑥] [0.033] ∴𝑥=4.3× 10 −6 𝑝𝐻=− log 𝑥 =5.37

Selecting an Indicator There are two common methods for determining the equivalence point during a titration: Use a pH probe to monitor pH, plot the curve and find the center of the vertical region Use an acid-base indicator that changes at the pH of the equivalence point Acid-base indicators are often weak acids themselves (represented by HIn), meaning that the concentration of the indicator components are defined by an equilibrium. The color of the indicator is different in its protonated form (HIn) and deprotonated form (In–). For the color change to be visible in an acid, the [In–]/[HIn] ratio should be 1/10. In bases, the [In–]/[HIn] ratio should be 10/1. HIn ⇌ H+(aq) + In–(aq) 𝐾 𝑎 = 𝐻 + 𝐼𝑛 − 𝐻𝐼𝑛

Bromothymol blue, an indicator with Ka = 1.0x10–7, is yellow in its HIn form and blue in its In– form. If this indicator is added to an acidic or basic solution, at what pH will the color change first be visible? ACID HIn ⇌ H+(aq) + In–(aq) BASE 𝐾 𝑎 = 𝐻 + [ 𝐼𝑛 − ] [𝐻𝐼𝑛] 𝐾 𝑎 = 𝐻 + [ 𝐼𝑛 − ] [𝐻𝐼𝑛] The color change is visible when The color change is visible when [ 𝐼𝑛 − ] [𝐻𝐼𝑛] = 1 10 [ 𝐼𝑛 − ] [𝐻𝐼𝑛] = 10 1 1.0× 10 −7 = 𝐻 + [1] [10] 1.0× 10 −7 = 𝐻 + [10] [1] 𝐻 + =1.0× 10 −6 𝐻 + =1.0× 10 −8 𝑝𝐻=6 𝑝𝐻=8 So the color change is visible at pH 6 So the color change is visible at pH 8

If we plug in the ideal [In–]/[HIn] ratio for an acid into the Henderson-Hasselbalch equation: 𝑝𝐻=𝑝 𝐾 𝑎 +𝑙𝑜𝑔 1 10 𝑝𝐻=𝑝 𝐾 𝑎 −1 This means that an indicator with a pKa value of 8 will change color at pH = 7.

Potassium hydrogen phthalate (KHP) is used to determine the concentrations of strong bases by the reaction: If an experiment begins with 0.5 g KHP (molar mass g/mol) and has a final volume of 100 mL, what is an appropriate indicator to use? (HINT: chart on Page 632 of textbook.) We need an indicator with a pKa that is 1 unit less than the pH at the equivalence point; determine this pH At the EP, all 0.5 g KHP are consumed, and have been converted to the conjugate base (P2–) (stoich). The pH depends on the [P2–] at equilibrium (RICE). HP–(aq) + OH–(aq) ⇌ H2O(l) + P2–(aq) pKa for HP– = 5.51

Use phenolphthalein (changes pH around 8) CW 13: Acid-Base Titrations If an experiment begins with 0.5 g KHP (molar mass g/mol) and has a final volume of 100 mL, what is an appropriate indicator to use? (HINT: chart on Page 632 of textbook.) HP–(aq) + OH–(aq) ⇌ H2O(l) + P2–(aq) pKa for HP– = 5.51 1. Stoichiometry 0.5 𝑔 𝐾𝐻𝑃 𝐿 × 1 𝑚𝑜𝑙 𝐾𝐻𝑃 𝑔 𝐾𝐻𝑃 × 1 𝑚𝑜𝑙 𝑃 2− 1 𝑚𝑜𝑙 𝐾𝐻𝑃 = 𝑀 𝑃 2− 2. Equilibrium 𝐾 𝑎 = [ 𝑃 2− ] 𝐻𝑃 − [ 𝑂𝐻 − ] R HP– OH– ⇌ H2O P2– I -- 0.0245 C +x –x E x 0.0245–x 𝐾 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 ∴𝑥=8.8× 10 −6 𝑝𝑂𝐻=− log 𝑥 =5.1 𝑝𝐻=14−𝑝𝑂𝐻=8.9

Return to the chart on Page 51. Recommend an indicator for each titration. EP = 7.0 EP = 8.9 EP = 5.3 Bromothymol blue Phenolphthalein Bromcresol Purple Choose an indicator with an end point (color change) near the EP

HW 9: Titrations Involving Weak Acid and Bases
Due: 2/20 (A) and 2/21 (B) HW 10: Review for Unit Test II Due: 2/22 (A) and 2/25 (B) Buffer Design Lab Report Due 2/26 (A) and 2/27 (B) Summary 12 2/15 (A) 2/19 (B) Outcome: I can explain features of titration curve and calculate the pH at any point during a titration. Goal: CW 13

Drill 13 2/22 (A) 2/21 (B) Copy down outcome and take out CW 13
Outcome: I can explain features of titration curve and calculate the concentration of analyte. Goal: Titration Curve Analysis

How to Interpret Titration Curves
Find the equivalence point The steepest part of the curve where the pH rises the fastest Used to find concentration of analyte Find the mid point Located in the center of the buffer region Geometrically halfway between the equivalence point and the beginning of the titration The midpoint determines the pKa of the acid

How to Interpret Titration Curves
Things to do first Graph your data Turn on major and minor tick marks Right click the axis, select “minor tick marks” Right click the axis, select “format axis,” set the bounds and units Do this for both axes Is there enough precision in the tick marks? You should have at least 1 mL or smaller for the minor tick mark on the x-axis You should have at least 0.2 pH units or smaller for the minor tick mark on the y-axis

Two Methods Geometric method First derivative method
requires a ruler, a pencil, and the titration graph First derivative method requires a spreadsheet and some formula entries gives you cool graphs with the 1st derivative pointing to the equivalence and mid points scores you brownie points with the instructors Pick your method (either will work)

A Typical Titration Curve
mid point equivalence point

Equivalence Point (Geometric Method)
1) Use a ruler to draw lines that follow the flat part of the curve equivalence point 2) Draw a line that follows the vertical part of the curve

3) using a ruler, measure the distance between the top intersection and the bottom intersection equivalence point 4) the geometric center of this line segment is the equivalence point

5) draw a vertical line from the equivalence point to the x-axis equivalence point 6) where the line crosses the x-axis is the volume at the equivalence point (28.7 mL in this case)

Mid Point (Geometric Method)
1) If there is a steep rise in the pH at the beginning of the graph, draw a line that follows the steep part of the curve mid point

2) Using a ruler, measure the distance between the far left and right intersections equivalence point 3) The center between these points is the mid point mid point

4) Draw a horizontal line from the mid point to the y-axis equivalence point 5) Where the line crosses the x-axis is the volume at the mid point (pH = 7.2 in this case) mid point

Using Excel Equivalence point Mid point
Use the first derivative  d pH / d Vol The spike in the graph points to the equiv. pt. Mid point Reverse the axes for the pH curve x axis = pH values; y-axis = Vol values Use the first derivative  d Vol / d pH The spike in the graph points to the mid point Use spreadsheet to make these calculations: 1st derivative (d pH / d Vol) = (pH2 - pH1)/(Vol2 - Vol1) 1st derivative (d Vol / d pH) = (Vol2 - Vol1)/ (pH2 - pH1) Or just (the first 1st derivative)-1

Equivalence Point (Derivative Method)
Identify volume value at the peak (28.5 mL in this case)

Mid Point (Derivative Method)
Identify pH value at the peak (pH = 7.3 in this case)

Titration Curve Analysis

For each titration curve Identify the titrant (in the tube) and the analyte (solution being tested) as a strong base (SB), weak base (WB), strong acid (SA), weak acid (WA), or polyprotic acid (PPA). You should consider the pH at the equivalence point, the starting and ending pH values, and the shape of the curve. For each labeled point on the titration curve, write the chemical species that are present in solution after any reaction has occurred. Use the following acids and bases: SA: hydrochloric acid SB: sodium hydroxide WA: acetic acid WB: ammonia PPA: sulfuric acid

For each titration curve Use a ruler to analyze each curve and find the equivalence point(s). Record the volume of titrant to reach the equivalence point and the pH at the equivalence point. Using the equivalence point(s); Determine the Ka or Kb when the analyte is a weak acid or base. For strong acids or bases, use the space to explain why you do not need to find the Ka or Kb. Use your text to identify an effective indicator that could be used to show the equivalence point. Determine the concentration of the sample being titrated, assuming the titrant is M and that 10.0 mL of sample was titrated each time. Show all work neatly on the provided pages.

Initially Present in Flask: SB: KOH K OH– Before Equivalence Point: HClO in excess At Equivalence Point: No H+ or OH– After Equivalence Point: NaOH in excess WA: HClO H ClO–

Drill 13 2/22 (A) 2/21 (B) HW 10: Review for Unit Test II
Due: 2/26 (A) and 2/25 (B) Buffer Design Lab Report Due 2/28 (A) and 2/27 (B) Drill 13 2/22 (A) 2/21 (B) Outcome: I can explain features of titration curve and calculate the concentration of analyte. Goal: Titration Curve Analysis

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4: Equilibrium.

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