Chapter 16 Aqueous Ionic Equilibrium

Chapter 16 Aqueous Ionic Equilibrium
Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

The Danger of Antifreeze
Each year, thousands of pets and wildlife species die from consuming antifreeze Most brands of antifreeze contain ethylene glycol sweet taste initial effect drunkenness Metabolized in the liver to glycolic acid HOCH2COOH glycolic acid (aka a-hydroxyethanoic acid) ethylene glycol (aka 1,2–ethandiol) Tro: Chemistry: A Molecular Approach, 2/e

Why Is Glycolic Acid Toxic?
If present in high enough concentration in the bloodstream, glycolic acid overwhelms the buffering ability of the HCO3− in the blood, causing the blood pH to drop When the blood pH is low, its ability to carry O2 is compromised acidosis HbH+(aq) + O2(g)  HbO2(aq) + H+(aq) One treatment is to give the patient ethyl alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol Tro: Chemistry: A Molecular Approach, 2/e

Buffers Buffers are solutions that resist changes in pH when an acid or base is added They act by neutralizing acid or base that is added to the buffered solution But just like everything else, there is a limit to what they can do, and eventually the pH changes Many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion blood has a mixture of H2CO3 and HCO3− Tro: Chemistry: A Molecular Approach, 2/e

Making an Acid Buffer Tro: Chemistry: A Molecular Approach, 2/e

HA(aq) + OH−(aq) → A−(aq) + H2O(l)
How Acid Buffers Work: Addition of Base HA(aq) + H2O(l)  A−(aq) + H3O+(aq) Buffers work by applying Le Châtelier’s Principle to weak acid equilibrium Buffer solutions contain significant amounts of the weak acid molecules, HA These molecules react with added base to neutralize it HA(aq) + OH−(aq) → A−(aq) + H2O(l) you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium Tro: Chemistry: A Molecular Approach, 2/e

 + How Buffers Work H2O A− HA A− HA new A− H3O+ Added HO−
Tro: Chemistry: A Molecular Approach, 2/e

H+(aq) + A−(aq) → HA(aq)
How Acid Buffers Work: Addition of Acid HA(aq) + H2O(l)  A−(aq) + H3O+(aq) The buffer solution also contains significant amounts of the conjugate base anion, A− These ions combine with added acid to make more HA H+(aq) + A−(aq) → HA(aq) After the equilibrium shifts, the concentration of H3O+ is kept constant Tro: Chemistry: A Molecular Approach, 2/e

 + How Buffers Work H2O HA HA A− A− new HA H3O+ Added H3O+

Common Ion Effect HA(aq) + H2O(l)  A−(aq) + H3O+(aq)
Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left This causes the pH to be higher than the pH of the acid solution lowering the H3O+ ion concentration Tro: Chemistry: A Molecular Approach, 2/e

Common Ion Effect Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 1: What is the pH of a buffer that is 0
Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A−] [H3O+] initial 0.100 ≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A−] [H3O+] initial 0.100 change equilibrium x +x +x 0.100 x x x Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init , then solve for x [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium x 0.100 x x Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium x x = 1.8 x 10−5 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 1.8E-5 0.100 x x x x = 1.8 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 1.8E−5 substitute [H3O+] into the formula for pH and solve Tro: Chemistry: A Molecular Approach, 2/e

Example 16.1: What is the pH of a buffer that is M HC2H3O2 and M NaC2H3O2? Ka for HC2H3O2 = 1.8 x 10−5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 1.8E−5 the values match Tro: Chemistry: A Molecular Approach, 2/e

Practice − What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? Tro: Chemistry: A Molecular Approach, 2/e

Practice − What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0 HF + H2O  F + H3O+ [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0. 14 M HF (pKa = 3
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? HF + H2O  F + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HA] [A−] [H3O+] initial 0.14 0.071 change equilibrium x +x +x 0.14 x x x Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF?
Ka for HF = 7.0 x 10−4 pKa for HF = 3.15 determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change −x +x equilibrium 0.012 0.100 x 0.14 x x Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? Ka for HF = 7.0 x 10−4 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change −x +x equilibrium x x = 1.4 x 10−3 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? substitute x into the equilibrium concentration definitions and solve [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change −x +x equilibrium 0.072 1.4E-3 0.14 x x x x = 1.4 x 10−3 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change −x +x equilibrium 0.072 1.4E−3 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? Ka for HF = 7.0 x 10−4 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.14 0.071 ≈ 0 change −x +x equilibrium 0.072 1.4E−3 the values are close enough Tro: Chemistry: A Molecular Approach, 2/e

Henderson-Hasselbalch Equation
Calculating the pH of a buffer solution can be simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base as long as the “x is small” approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

Deriving the Henderson-Hasselbalch Equation

Example 16.2: What is the pH of a buffer that is M HC7H5O2 and M NaC7H5O2? HC7H5O2 + H2O  C7H5O2 + H3O+ assume the [HA] and [A−] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch Equation check the “x is small” approximation Ka for HC7H5O2 = 6.5 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and M KF? find the pKa from the given Ka assume the [HA] and [A−] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation check the “x is small” approximation HF + H2O  F + H3O+ Tro: Chemistry: A Molecular Approach, 2/e

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?
The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable Generally, the “x is small” approximation will work when both of the following are true: the initial concentrations of acid and salt are not very dilute the Ka is fairly small For most problems, this means that the initial acid and salt concentrations should be over 100 to 1000x larger than the value of Ka Tro: Chemistry: A Molecular Approach, 2/e

How Much Does the pH of a Buffer Change When an Acid or Base Is Added?
Though buffers do resist change in pH when acid or base is added to them, their pH does change Calculating the new pH after adding acid or base requires breaking the problem into two parts a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A− to make more HA added base reacts with the HA to make more A− an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−] Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 3: What is the pH of a buffer that has 0
Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−. construct a stoichiometry table for the reaction HC2H3O2 + OH−  C2H3O2 + H2O HA A− OH− mols before 0.100 mols added ─ 0.010 mols after Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? HC2H3O2 + OH−  C2H3O2 + H2O fill in the table – tracking the changes in the number of moles for each component HA A− OH− mols before 0.100 ≈ 0 mols added ─ 0.010 mols after 0.090 0.110 Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−. construct a stoichiometry table for the reaction enter the initial number of moles for each HC2H3O2 + OH−  C2H3O2 + H2O HA A– OH− mols before 0.100 0.010 mols change mols end new molarity Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities HC2H3O2 + OH−  C2H3O2 + H2O HA A− OH− mols before 0.100 0.010 mols change mols end new molarity −0.010 +0.010 −0.010 0.090 0.110 0.090 0.110 Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−] HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change equilibrium Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? HC2H3O2 + H2O  C2H3O2 + H3O+ represent the change in the concentrations in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change equilibrium x +x +x 0.090 x x x Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 determine the value of Ka because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x [HA] [A−] [H3O+] initial 0.100 ≈ 0 change −x +x equilibrium 0.090 0.110 x 0.090 x x Tro: Chemistry: A Molecular Approach, 2/e

the approximation is valid
Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change −x +x equilibrium x x = 1.47 x 10−5 the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? substitute x into the equilibrium concentration definitions and solve [HA] [A-] [H3O+] initial 0.090 0.110 ≈ 0 change −x +x equilibrium 1.5E-5 0.090 x x x x = 1.47 x 10−5 Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? substitute [H3O+] into the formula for pH and solve [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change −x +x equilibrium 1.5E−5 Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? Ka for HC2H3O2 = 1.8 x 10−5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change −x +x equilibrium 1.5E−5 the values match Tro: Chemistry: A Molecular Approach, 2/e

or, by using the Henderson-Hasselbalch Equation

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? write the reaction for the acid with water construct an ICE table for the reaction enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−] fll in the ICE table HC2H3O2 + H2O  C2H3O2 + H3O+ [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change equilibrium x +x +x x 0.090 x x Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? find the pKa from the given Ka assume the [HA] and [A−] equilibrium concentrations are the same as the initial HC2H3O2 + H2O  C2H3O2 + H3O+ Ka for HC2H3O2 = 1.8 x 10−5 [HA] [A−] [H3O+] initial 0.090 0.110 ≈ 0 change −x +x equilibrium x Tro: Chemistry: A Molecular Approach, 2/e

Example 16.3: What is the pH of a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L that has mol NaOH added to it? substitute into the Henderson-Hasselbalch equation check the “x is small” approximation HC2H3O2 + H2O  C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 3: Compare the effect on pH of adding 0
Example 16.3: Compare the effect on pH of adding mol NaOH to a buffer that has mol HC2H3O2 and mol NaC2H3O2 in 1.00 L to adding mol NaOH to 1.00 L of pure water HC2H3O2 + H2O  C2H3O2 + H3O+ pKa for HC2H3O2 = 4.745 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has moles HF (pKa = 3.15) and moles KF in 1.00 L of solution when moles of HCl is added? (The “x is small” approximation is valid) Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has 0. 140 moles HF and 0
Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−. construct a stoichiometry table for the reaction F− + H3O+  HF + H2O F− H3O+ HF mols before 0.071 0.140 mols added – 0.020 mols after Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? F− + H3O+  HF + H2O fill in the table – tracking the changes in the number of moles for each component F− H3O+ HF mols before 0.071 0.140 mols added – 0.020 mols after 0.051 0.160 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−. construct a stoichiometry table for the reaction enter the initial number of moles for each F− + H3O+  HF + H2O F− H3O+ HF mols before 0.071 0.020 0.140 mols change mols after new molarity Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? using the added chemical as the limiting reactant, determine how the moles of the other chemicals change add the change to the initial number of moles to find the moles after reaction divide by the liters of solution to find the new molarities F− + H3O+  HF + H2O F− H3O+ HF mols before 0.071 0.020 0.140 mols change mols after new molarity −0.020 −0.020 +0.020 0.051 0.160 0.051 0.160 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the pH of a buffer that has moles HF and moles KF in 1.00 L of solution when moles of HCl is added? HF + H2O  F + H3O+ write the reaction for the acid with water construct an ICE table. assume the [HA] and [A−] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation [HF] [F−] [H3O+] initial 0.160 0.051 ≈ 0 change −x +x equilibrium x Tro: Chemistry: A Molecular Approach, 2/e

Basic Buffers B:(aq) + H2O(l)  H:B+(aq) + OH−(aq)
Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl− H2O(l) + NH3 (aq)  NH4+(aq) + OH−(aq) Tro: Chemistry: A Molecular Approach, 2/e

Henderson-Hasselbalch Equation for Basic Buffers
The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H2O  H:B+ + OH− To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reaction H:B+ + H2O  B: + H3O+ this does not affect the concentrations, just the way we are looking at the reaction Tro: Chemistry: A Molecular Approach, 2/e

Relationship between pKa and pKb
Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base Ka  Kb = Kw = 1.0 x 10−14 −log(Ka  Kb) = −log(Kw) = 14 −log(Ka) + −log(Kb) = 14 pKa + pKb = 14 Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 4: What is the pH of a buffer that is 0. 50 M NH3 (pKb = 4
Example 16.4: What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? NH3 + H2O  NH4+ + OH− find the pKa of the conjugate acid (NH4+) from the given Kb assume the [B] and [HB+] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation check the “x is small” approximation Tro: Chemistry: A Molecular Approach, 2/e

Henderson-Hasselbalch Equation for Basic Buffers
The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product B: + H2O  H:B+ + OH− We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 4: What is the pH of a buffer that is 0. 50 M NH3 (pKb = 4
Example 16.4: What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl? NH3 + H2O  NH4+ + OH− find the pKb if given Kb assume the [B] and [HB+] equilibrium concentrations are the same as the initial substitute into the Henderson-Hasselbalch equation base form, find pOH check the “x is small” approximation calculate pH from pOH alternative method Tro: Chemistry: A Molecular Approach, 2/e

Buffering Effectiveness
A good buffer should be able to neutralize moderate amounts of added acid or base However, there is a limit to how much can be added before the pH changes significantly The buffering capacity is the amount of acid or base a buffer can neutralize The buffering range is the pH range the buffer can be effective The effectiveness of a buffer depends on two factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base Tro: Chemistry: A Molecular Approach, 2/e

Effect of Relative Amounts of Acid and Conjugate Base
A buffer is most effective with equal concentrations of acid and base Buffer 1 0.100 mol HA & mol A− Initial pH = 5.00 Buffer 2 0.18 mol HA & mol A− Initial pH = 4.05 pKa (HA) = 5.00 HA + OH−  A + H2O after adding mol NaOH pH = 5.09 after adding mol NaOH pH = 4.25 HA A− OH− mols before 0.100 mols added ─ 0.010 mols after 0.090 0.110 ≈ 0 HA A− OH− mols before 0.18 0.020 mols added ─ 0.010 mols after 0.17 0.030 ≈ 0 63

Effect of Absolute Concentrations of Acid and Conjugate Base
A buffer is most effective when the concentrations of acid and base are largest Buffer 1 0.50 mol HA & 0.50 mol A− Initial pH = 5.00 Buffer 2 0.050 mol HA & mol A− Initial pH = 5.00 pKa (HA) = 5.00 HA + OH−  A + H2O after adding mol NaOH pH = 5.02 after adding mol NaOH pH = 5.18 HA A− OH− mols before 0.50 0.500 mols added ─ 0.010 mols after 0.49 0.51 ≈ 0 HA A− OH− mols before 0.050 mols added ─ 0.010 mols after 0.040 0.060 ≈ 0 64

Buffering Capacity a concentrated buffer can neutralize more added acid or base than a dilute buffer Tro: Chemistry: A Molecular Approach, 2/e

Effectiveness of Buffers
A buffer will be most effective when the [base]:[acid] = 1 equal concentrations of acid and base A buffer will be effective when < [base]:[acid] < 10 A buffer will be most effective when the [acid] and the [base] are large Tro: Chemistry: A Molecular Approach, 2/e

0.1 < [base]:[acid] < 10
Buffering Range We have said that a buffer will be effective when 0.1 < [base]:[acid] < 10 Substituting into the Henderson-Hasselbalch equation we can calculate the maximum and minimum pH at which the buffer will be effective Lowest pH Highest pH Therefore, the effective pH range of a buffer is pKa ± 1 When choosing an acid to make a buffer, choose one whose is pKa closest to the pH of the buffer Tro: Chemistry: A Molecular Approach, 2/e

Chlorous acid, HClO2 pKa = 1.95
Example 16.5a: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25? Chlorous acid, HClO2 pKa = 1.95 Nitrous acid, HNO2 pKa = 3.34 Formic acid, HCHO2 pKa = 3.74 Formic acid, HCHO2 pKa = 3.74 Hypochlorous acid, HClO pKa = 7.54 The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range Tro: Chemistry: A Molecular Approach, 2/e

Example 16.5b: What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?
Formic acid, HCHO2, pKa = 3.74 To make a buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75?
Benzoic acid, HC7H5O2, pKa = 4.19 Tro: Chemistry: A Molecular Approach, 2/e

Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75?
Benzoic acid, HC7H5O2, pKa = 4.19 to make a buffer with pH 3.75, you would use times as much NaC7H5O2 as HC7H5O2 Tro: Chemistry: A Molecular Approach, 2/e 71

Buffering Capacity Buffering capacity is the amount of acid or base that can be added to a buffer without causing a large change in pH The buffering capacity increases with increasing absolute concentration of the buffer components As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves Buffers that need to work mainly with added acid generally have [base] > [acid] Buffers that need to work mainly with added base generally have [acid] > [base] Tro: Chemistry: A Molecular Approach, 2/e

Titration In an acid-base titration, a solution of unknown concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete when the reaction is complete we have reached the endpoint of the titration An indicator may be added to determine the endpoint an indicator is a chemical that changes color when the pH changes When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point Tro: Chemistry: A Molecular Approach, 2/e

Titration Tro: Chemistry: A Molecular Approach, 2/e

Titration Curve A plot of pH vs. amount of added titrant
The inflection point of the curve is the equivalence point of the titration Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH The pH of the equivalence point depends on the pH of the salt solution equivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7 Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH Tro: Chemistry: A Molecular Approach, 2/e

Titration Curve: Unknown Strong Base Added to Strong Acid

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes After Equivalence (excess base) Equivalence Point equal moles of HCl and NaOH pH = 7.00  Before Equivalence (excess acid) Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10−3 Before equivalence point added 5.0 mL NaOH 5.0 x 10−4 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 5.0E-4 mols change −5.0E-4 +5.0E-4 mols end 2.00E-3 molarity, new 0.0667 0.017 HCl NaCl NaOH mols before 2.50E-3 5.0E-4 mols change mols end molarity, new HCl NaCl NaOH mols before 2.50E-3 5.0E-4 mols change −5.0E-4 +5.0E-4 mols end 2.00E-3 molarity, new Tro: Chemistry: A Molecular Approach, 2/e 78

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(aq) To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10−3 moles At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over Because the NaCl is a neutral salt, the pH at equivalence = 7.00 Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10−3 At equivalence point added 25.0 mL NaOH 2.5 x 10−3 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 2.5E-3 mols change −2.5E-3 +2.5E-3 mols end molarity, new HCl NaCl NaOH mols before 2.50E-3 2.5E-3 mols change −2.5E-3 +2.5E-3 mols end molarity, new 0.050 HCl NaCl NaOH mols before 2.50E-3 2.5E-3 mols change mols end molarity, new Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Initial pH = −log(0.100) = 1.00 Initial mol of HCl = L x mol/L = 2.50 x 10−3 After equivalence point added 30.0 mL NaOH 3.0 x 10−3 mole NaOH added HCl NaCl NaOH mols before 2.50E-3 3.0E-3 mols change −2.5E-3 +2.5E-3 mols end 2.5E-3 5.0E-4 molarity, new 0.045 0.0091 HCl NaCl NaOH mols before 2.50E-3 3.0E-3 mols change −2.5E-3 +2.5E-3 mols end 2.5E-3 5.0E-4 molarity, new HCl NaCl NaOH mols before 2.50E-3 3.0E-3 mols change mols end molarity, new

Adding 0.100 M NaOH to 0.100 M HCl added 5.0 mL NaOH 0.00200 mol HCl
pH = 1.18 added 35.0 mL NaOH mol NaOH pH = 12.22 added 30.0 mL NaOH mol NaOH pH = 11.96 added 10.0 mL NaOH mol HCl pH = 1.37 added 25.0 mL NaOH equivalence point pH = 7.00 25.0 mL M HCl mol HCl pH = 1.00 added 40.0 mL NaOH mol NaOH pH = 12.36 added 15.0 mL NaOH mol HCl pH = 1.60 added 50.0 mL NaOH mol NaOH pH = 12.52 added 20.0 mL NaOH mol HCl pH = 1.95 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the pH of the solution that results when 10
Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 Tro: Chemistry: A Molecular Approach, 2/e

HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l)
Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO3= L x 0.25 mol/L=1.25 x 10−2 Before equivalence point added 10.0 mL NaOH HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-3 mols change −1.5E-3 +1.5E-3 mols end 1.1E-3 molarity, new 0.018 0.025 HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-3 mols change −1.5E-3 +1.5E-3 mols end 1.1E-3 molarity, new Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the amount of 0
Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO3= L x 0.25 mol/L=1.25 x 10−2 At equivalence point: moles of NaOH = 1.25 x 10−2 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the pH of the solution that results when 100
Practice – Calculate the pH of the solution that results when mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Calculate the pH of the solution that results when mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) Initial pH = −log(0.250) = 0.60 Initial mol of HNO3= L x 0.25 mol/L=1.25 x 10−2 After equivalence point added mL NaOH HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-2 mols change −1.25E-2 +1.25E-2 mols end 0.0025 molarity, new 0.0833 0.017 HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-2 mols change −1.25E-2 +1.25E-2 mols end 0.0025 molarity, new

Practice – Calculate the pH of the solution that results when mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M HNO3 HNO3(aq) + NaOH(aq)  NaNO3(aq) + H2O(l) Initial pH = −log(0.250) = 0.60 initial mol of HNO3= L x 0.25 mol/L=1.25 x 10−2 After equivalence point added mL NaOH HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-2 mols change −1.25E-2 +1.25E-2 mols end 0.0025 molarity, new HNO3 NaNO3 NaOH mols before 1.25E-2 1.5E-2 mols change −1.25E-2 +1.25E-2 mols end 0.0025 molarity, new 0.0833 0.017 Tro: Chemistry: A Molecular Approach, 2/e

Titration of a Strong Base with a Strong Acid
If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown Tro: Chemistry: A Molecular Approach, 2/e

Titration of a Weak Acid with a Strong Base
Titrating a weak acid with a strong base results in differences in the titration curve at the equivalence point and excess acid region The initial pH is determined using the Ka of the weak acid The pH in the excess acid region is determined as you would determine the pH of a buffer The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid The pH after equivalence is dominated by the excess strong base the basicity from the conjugate base anion is negligible Tro: Chemistry: A Molecular Approach, 2/e

Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq) Initial pH Ka = 1.8 x 10−4 [HCHO2] [CHO2−] [H3O+] initial 0.100 0.000 ≈ 0 change −x +x equilibrium 0.100 − x x Tro: Chemistry: A Molecular Approach, 2/e

HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq) Initial mol of HCHO2 = L x mol/L = 2.50 x 10−3 Before equivalence added 5.0 mL NaOH HA A− OH− mols before 2.50E-3 mols added – 5.0E-4 mols after 2.00E-3 ≈ 0 Tro: Chemistry: A Molecular Approach, 2/e

HCHO2(aq) + NaOH(aq)  NaCHO2 (aq) + H2O(aq) Initial mol of HCHO2 = L x mol/L = 2.50 x 10−3 At equivalence CHO2−(aq) + H2O(l)  HCHO2(aq) + OH−(aq) added 25.0 mL NaOH [HCHO2] [CHO2−] [OH−] initial 0.0500 ≈ 0 change +x −x equilibrium x 5.00E-2-x Kb = 5.6 x 10−11 HA A− OH− mols before 2.50E-3 mols added – mols after ≈ 0 [OH−] = 1.7 x 10−6 M 94

HCHO2(aq) + NaOH(aq)  NaCHO2(aq) + H2O(aq) Initial mol of HCHO2 = L x mol/L = 2.50 x 10−3 After equivalence added 30.0 mL NaOH 3.0 x 10−3 mole NaOH added HA A− NaOH mols before 2.50E-3 3.0E-3 mols change −2.5E-3 +2.5E-3 mols end 2.5E-3 5.0E-4 molarity, new HA A− NaOH mols before 2.50E-3 3.0E-3 mols change −2.5E-3 +2.5E-3 mols end 2.5E-3 5.0E-4 molarity, new 0.045 0.0091 HA A− NaOH mols before 2.50E-3 3.0E-3 mols change mols end molarity, new 95 95

Adding NaOH to HCHO2 added 25.0 mL NaOH equivalence point
mol CHO2− [CHO2−]init = M [OH−]eq = 1.7 x 10−6 pH = 8.23 added 30.0 mL NaOH mol NaOH xs pH = 11.96 added 5.0 mL NaOH mol HCHO2 pH = 3.14 added 10.0 mL NaOH mol HCHO2 pH = 3.56 initial HCHO2 solution mol HCHO2 pH = 2.37 added 35.0 mL NaOH mol NaOH xs pH = 12.22 added 12.5 mL NaOH mol HCHO2 pH = 3.74 = pKa half-neutralization added 40.0 mL NaOH mol NaOH xs pH = 12.36 added 50.0 mL NaOH mol NaOH xs pH = 12.52 added 15.0 mL NaOH mol HCHO2 pH = 3.92 added 20.0 mL NaOH mol HCHO2 pH = 4.34 Tro: Chemistry: A Molecular Approach, 2/e

Titrating Weak Acid with a Strong Base
The initial pH is that of the weak acid solution calculate like a weak acid equilibrium problem e.g., 15.5 and 15.6 Before the equivalence point, the solution becomes a buffer calculate mol HAinit and mol A−init using reaction stoichiometry calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−init Half-neutralization pH = pKa Tro: Chemistry: A Molecular Approach, 2/e

Titrating Weak Acid with a Strong Base
At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established mol A− = original mole HA calculate the volume of added base as you did in Example 4.8 [A−]init = mol A−/total liters calculate like a weak base equilibrium problem e.g., 15.14 Beyond equivalence point, the OH is in excess [OH−] = mol MOH xs/total liters [H3O+][OH−]=1 x 10−14 Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 7a: A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Example 16.7a: A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the volume of KOH at the equivalence point. write an equation for the reaction for B with HA use stoichiometry to determine the volume of added B HNO2 + KOH  NO2 + H2O Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 7b: A 40. 0 mL sample of 0. 100 M HNO2 is titrated with 0
Example 16.7b: A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH. write an equation for the reaction for B with HA determine the moles of HAbefore & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A made HNO2 + KOH  NO2 + H2O HNO2 NO2− OH− mols before ≈ 0 mols added – mols after Tro: Chemistry: A Molecular Approach, 2/e

Example 16.7b: A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH after adding 5.00 mL KOH. HNO2 + H2O  NO2 + H3O+ write an equation for the reaction of HA with H2O determine Ka and pKa for HA use the Henderson-Hasselbalch equation to determine the pH Table Ka = 4.6 x 10−4 HNO2 NO2− OH− mols before ≈ 0 mols added – mols after Tro: Chemistry: A Molecular Approach, 2/e

Example 16.7b: A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point. write an equation for the reaction for B with HA determine the moles of HAbefore & moles of added B make a stoichiometry table and determine the moles of HA in excess and moles A made HNO2 + KOH  NO2 + H2O at half-equivalence, moles KOH = ½ mole HNO2 HNO2 NO2− OH− mols before ≈ 0 mols added – mols after Tro: Chemistry: A Molecular Approach, 2/e

Example 16.7b: A 40.0 mL sample of M HNO2 is titrated with M KOH. Calculate the pH at the half-equivalence point. HNO2 + H2O  NO2 + H3O+ write an equation for the reaction of HA with H2O determine Ka and pKa for HA use the Henderson-Hasselbalch equation to determine the pH Table Ka = 4.6 x 10-4 HNO2 NO2− OH− mols before ≈ 0 mols added – mols after Tro: Chemistry: A Molecular Approach, 2/e

Titration Curve of a Weak Base with a Strong Acid

Practice – Titration of 25. 0 mL of 0. 10 M NH3 (pKb = 4. 75) with 0
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the initial pH of the NH3 solution. Tro: Chemistry: A Molecular Approach, 2/e

NH3(aq) + HCl(aq)  NH4Cl(aq)
Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq) pKb = 4.75 Kb = 10−4.75 = 1.8 x 10−5 NH3(aq) + HCl(aq)  NH4Cl(aq) Initial: NH3(aq) + H2O(l)  NH4+(aq) + OH−(aq) [HCl] [NH4+] [NH3] initial 0.10 change +x −x equilibrium x 0.10−x Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH3(aq) + HCl(aq)  NH4Cl(aq) Initial mol of NH3 = L x mol/L = 2.50 x 10−3 Before equivalence: after adding 5.0 mL of HCl NH4+(aq) + H2O(l)  NH4+(aq) + H2O(l) pKb = 4.75 pKa = − 4.75 = 9.25 NH3 NH4Cl HCl mols before 2.50E-3 5.0E-4 mols change −5.0E-4 mols end 2.00E-3 molarity, new 0.0667 0.017 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl. NH3(aq) + HCl(aq)  NH4Cl(aq) Initial mol of NH3 = L x mol/L = 2.50 x 10−3 Before equivalence: after adding 5.0 mL of HCl NH3 NH4Cl HCl mols before 2.50E-3 5.0E-4 mols change −5.0E-4 mols end 2.00E-3 molarity, new 0.0667 0.017 Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. NH3(aq) + HCl(aq)  NH4Cl(aq) Initial mol of NH3 = L x mol/L = 2.50 x 10−3 At equivalence mol NH3 = mol HCl = 2.50 x 10−3 NH3 NH4Cl HCl mols before 2.50E-3 2.5E-3 mols change −2.5E-3 +2.5E-3 mols end molarity, new 0.050 added 25.0 mL HCl Tro: Chemistry: A Molecular Approach, 2/e

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence. NH3(aq) + HCl(aq)  NH4Cl(aq) at equivalence [NH4Cl] = M NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq) [NH3] [NH4+] [H3O+] initial 0.050 ≈ 0 change +x −x equilibrium x 0.050−x Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. Tro: Chemistry: A Molecular Approach, 2/e

Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl. NH3(aq) + HCl(aq)  NH4Cl(aq) Initial mol of NH3 = L x mol/L = 2.50 x 10−3 After equivalence: after adding 30.0 mL HCl NH3 NH4Cl HCl mols before 2.50E-3 3.0E-3 mols change −2.5E-3 +2.5E-3 mols end 2.5E-3 5.0E-4 molarity, new 0.045 0.0091 when you mix a strong acid, HCl, with a weak acid, NH4+, you only need to consider the strong acid Tro: Chemistry: A Molecular Approach, 2/e

Titration of a Polyprotic Acid
If Ka1 >> Ka2, there will be two equivalence points in the titration the closer the Ka’s are to each other, the less distinguishable the equivalence points are titration of 25.0 mL of M H2SO3 with M NaOH Tro: Chemistry: A Molecular Approach, 2/e

Monitoring pH During a Titration
The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+] using a probe that specifically measures just H3O+ The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator Tro: Chemistry: A Molecular Approach, 2/e

Monitoring pH During a Titration

HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq)
Indicators Many dyes change color depending on the pH of the solution These dyes are weak acids, establishing an equilibrium with the H2O and H3O+ in the solution HInd(aq) + H2O(l)  Ind(aq) + H3O+(aq) The color of the solution depends on the relative concentrations of Ind:HInd when Ind:HInd ≈ 1, the color will be mix of the colors of Ind and HInd when Ind:HInd > 10, the color will be mix of the colors of Ind when Ind:HInd < 0.1, the color will be mix of the colors of HInd Tro: Chemistry: A Molecular Approach, 2/e

Phenolphthalein Tro: Chemistry: A Molecular Approach, 2/e

Methyl Red Tro: Chemistry: A Molecular Approach, 2/e

Monitoring a Titration with an Indicator
For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH pKa of HInd ≈ pH at equivalence point Tro: Chemistry: A Molecular Approach, 2/e

Acid-Base Indicators Tro: Chemistry: A Molecular Approach, 2/e

Solubility Equilibria
All ionic compounds dissolve in water to some degree however, many compounds have such low solubility in water that we classify them as insoluble We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water Tro: Chemistry: A Molecular Approach, 2/e

Solubility Product The equilibrium constant for the dissociation of a solid salt into its aqueous ions is called the solubility product, Ksp For an ionic solid MnXm, the dissociation reaction is: MnXm(s)  nMm+(aq) + mXn−(aq) The solubility product would be Ksp = [Mm+]n[Xn−]m For example, the dissociation reaction for PbCl2 is PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) And its equilibrium constant is Ksp = [Pb2+][Cl−]2 Tro: Chemistry: A Molecular Approach, 2/e

Tro: Chemistry: A Molecular Approach, 2/e

for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq)
Molar Solubility Solubility is the amount of solute that will dissolve in a given amount of solution at a particular temperature The molar solubility is the number of moles of solute that will dissolve in a liter of solution the molarity of the dissolved solute in a saturated solution for the general reaction MnXm(s)  nMm+(aq) + mXn−(aq) Tro: Chemistry: A Molecular Approach, 2/e

PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) Ksp = [Pb2+][Cl−]2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro: Chemistry: A Molecular Approach, 2/e

Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C
substitute into the Ksp expression find the value of Ksp from Table 16.2, plug into the equation, and solve for S Ksp = [Pb2+][Cl−]2 Ksp = (S)(2S)2 [Pb2+] [Cl−] Initial Change +S +2S Equilibrium S 2S Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M

PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid PbBr2(s)  Pb2+(aq) + 2 Br−(aq) Ksp = [Pb2+][Br−]2 [Pb2+] [Br−] initial change +(1.05 x 10−2) +2(1.05 x 10−2) equilibrium (1.05 x 10−2) (2.10 x 10−2) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M
substitute into the Ksp expression plug into the equation and solve Ksp = [Pb2+][Br−]2 Ksp = (1.05 x 10−2)(2.10 x 10−2)2 [Pb2+] [Br−] initial change +(1.05 x 10−2) +2(1.05 x 10−2) equilibrium (1.05 x 10−2) (2.10 x 10−2) Tro: Chemistry: A Molecular Approach, 2/e

Ksp and Relative Solubility
Molar solubility is related to Ksp But you cannot always compare solubilities of compounds by comparing their Ksps To compare Ksps, the compounds must have the same dissociation stoichiometry Tro: Chemistry: A Molecular Approach, 2/e

The Effect of Common Ion on Solubility
Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2 PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) addition of Cl− shifts the equilibrium to the left Tro: Chemistry: A Molecular Approach, 2/e

CaF2(s)  Ca2+(aq) + 2 F−(aq)
Example 16.10: Calculate the molar solubility of CaF2 in M NaF at 25 C write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid CaF2(s)  Ca2+(aq) + 2 F−(aq) Ksp = [Ca2+][F−]2 [Ca2+] [F−] initial 0.100 change +S +2S equilibrium S S Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 10: Calculate the molar solubility of CaF2 in 0
Example 16.10: Calculate the molar solubility of CaF2 in M NaF at 25 C substitute into the Ksp expression, assume S is small find the value of Ksp from Table 16.2, plug into the equation, and solve for S Ksp = [Ca2+][F−]2 Ksp = (S)( S)2 Ksp = (S)(0.100)2 [Ca2+] [F−] initial 0.100 change +S +2S equilibrium S S Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M Ksp of AgCl = 1.77 x 10−10 Tro: Chemistry: A Molecular Approach, 2/e

AgCl(s)  Ag+(aq) + Cl−(aq)
Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M write the dissociation reaction and Ksp expression create an ICE table defining the change in terms of the solubility of the solid AgCl(s)  Ag+(aq) + Cl−(aq) Ksp = [Ag+][Cl−] [Ag+] [Cl−] initial 0.55 change +S equilibrium S S Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M
substitute into the Ksp expression, assume S is small find the value of Ksp from Table 16.2, plug into the equation, and solve for S Ksp = [Ag+][Cl−] Ksp = (S)( S) Ksp = (S)(0.55) [Ag+] [Cl−] Initial 0.55 Change +S Equilibrium S S Tro: Chemistry: A Molecular Approach, 2/e

The Effect of pH on Solubility
For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxide and the lower the pH, the higher the solubility higher pH = increased [OH−] M(OH)n(s)  Mn+(aq) + nOH−(aq) For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility M2(CO3)n(s)  2 Mn+(aq) + nCO32−(aq) H3O+(aq) + CO32− (aq)  HCO3− (aq) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

Precipitation Precipitation will occur when the concentrations of the ions exceed the solubility of the ionic compound If we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur Q = Ksp, the solution is saturated, no precipitation Q < Ksp, the solution is unsaturated, no precipitation Q > Ksp, the solution would be above saturation, the salt above saturation will precipitate Some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions Tro: Chemistry: A Molecular Approach, 2/e

a supersaturated solution will precipitate if a seed crystal is added
precipitation occurs if Q > Ksp Tro: Chemistry: A Molecular Approach, 2/e

Selective Precipitation
A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different Tro: Chemistry: A Molecular Approach, 2/e

PbBr2(s)  Pb2+(aq) + 2 Br−(aq)
Example 16.12: Will a precipitate form when we mix Pb(NO3)2(aq) with NaBr(aq) if the concentrations after mixing are M and M respectively? write the equation for the reaction determine the ion concentrations of the original salts determine the Ksp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to Ksp. If Q > Ksp, precipitation Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq) Pb(NO3)2 = M Pb2+ = M, NO3− = 2( M) NaBr = M Na+ = M, Br− = M Ksp of PbBr2 = 4.67 x 10–6 PbBr2(s)  Pb2+(aq) + 2 Br−(aq) Q < Ksp, so no precipitation Tro: Chemistry: A Molecular Approach, 2/e

Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both M? Ksp of Ca(OH)2 = 4.68 x 10−6 Tro: Chemistry: A Molecular Approach, 2/e

Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq)
Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both M? write the equation for the reaction determine the ion concentrations of the original salts determine the Ksp for any “insoluble” product write the dissociation reaction for the insoluble product calculate Q, using the ion concentrations compare Q to Ksp. If Q > Ksp, precipitation Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq) Ca(NO3)2 = M Ca2+ = M, NO3− = 2( M) NaOH = M Na+ = M, OH− = M Ksp of Ca(OH)2 = 4.68 x 10–6 Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq) Q > Ksp, so precipitation Tro: Chemistry: A Molecular Approach, 2/e

Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)
Example 16.13: What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with [0.059]) from seawater? Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq) precipitating may just occur when Q = Ksp Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from M NaOH(aq)? Ksp of Ca(OH)2 = 4.68 x 10−6 Tro: Chemistry: A Molecular Approach, 2/e

precipitating may just occur when Q = Ksp
Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from M NaOH(aq)? Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq) Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq) precipitating may just occur when Q = Ksp [Ca(NO3)2] = [Ca2+] = M Tro: Chemistry: A Molecular Approach, 2/e

precipitating may just occur when Q = Ksp
Example 16.14: What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? Ca(OH)2(s)  Ca2+(aq) + 2 OH−(aq) precipitating may just occur when Q = Ksp Tro: Chemistry: A Molecular Approach, 2/e

Example 16. 14: What is the [Mg2+] when Ca2+ (with [0
Example 16.14: What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from seawater? precipitating Mg2+ begins when [OH−] = 1.9 x 10−6 M precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq) when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from M to 4.8 x 10−10 M Tro: Chemistry: A Molecular Approach, 2/e

Practice – A solution is made by mixing Pb(NO3)2(aq) with AgNO3(aq) so both compounds have a concentration of M. NaCl(s) is added to precipitate out both AgCl(s) and PbCl2(aq). What is the [Ag+] concentration when the Pb2+ just begins to precipitate? Tro: Chemistry: A Molecular Approach, 2/e

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Practice – What is the [Ag+] concentration when the Pb2+( M) just begins to precipitate? AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) AgCl(s)  Ag+(aq) + Cl−(aq) precipitating may just occur when Q = Ksp Tro: Chemistry: A Molecular Approach, 2/e

PbCl2(s)  Pb2+(aq) + 2 Cl−(aq)
Practice – What is the [Ag+] concentration when the Pb2+( M) just begins to precipitate? Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq) PbCl2(s)  Pb2+(aq) + 2 Cl−(aq) precipitating may just occur when Q = Ksp Tro: Chemistry: A Molecular Approach, 2/e

AgCl(s)  Ag+(aq) + Cl−(aq)
Practice – What is the [Ag+] concentration when the Pb2+( M) just begins to precipitate precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M AgCl(s)  Ag+(aq) + Cl−(aq) when Pb2+ just begins to precipitate out, the [Ag+] has dropped from M to 1.6 x 10−9 M Tro: Chemistry: A Molecular Approach, 2/e

Qualitative Analysis An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme wet chemistry A sample containing several ions is subjected to the addition of several precipitating agents Addition of each reagent causes one of the ions present to precipitate out Tro: Chemistry: A Molecular Approach, 2/e

Qualitative Analysis Tro: Chemistry: A Molecular Approach, 2/e

Group 1 Group one cations are Ag+, Pb2+ and Hg22+
All these cations form compounds with Cl− that are insoluble in water as long as the concentration is large enough PbCl2 may be borderline molar solubility of PbCl2 = 1.43 x 10−2 M Precipitated by the addition of HCl Tro: Chemistry: A Molecular Approach, 2/e

Group 2 Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+ All these cations form compounds with HS− and S2− that are insoluble in water at low pH Precipitated by the addition of H2S in HCl Tro: Chemistry: A Molecular Approach, 2/e

Group 3 Group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides All these cations form compounds with S2− that are insoluble in water at high pH Precipitated by the addition of H2S in NaOH Tro: Chemistry: A Molecular Approach, 2/e

Group 4 Group four cations are Mg2+, Ca2+, Ba2+
All these cations form compounds with PO43− that are insoluble in water at high pH Precipitated by the addition of (NH4)2HPO4 Tro: Chemistry: A Molecular Approach, 2/e

Group 5 Group five cations are Na+, K+, NH4+
All these cations form compounds that are soluble in water – they do not precipitate Identified by the color of their flame Tro: Chemistry: A Molecular Approach, 2/e

Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq)
Complex Ion Formation Transition metals tend to be good Lewis acids They often bond to one or more H2O molecules to form a hydrated ion H2O is the Lewis base, donating electron pairs to form coordinate covalent bonds Ag+(aq) + 2 H2O(l)  Ag(H2O)2+(aq) Ions that form by combining a cation with several anions or neutral molecules are called complex ions e.g., Ag(H2O)2+ The attached ions or molecules are called ligands e.g., H2O Tro: Chemistry: A Molecular Approach, 2/e

Complex Ion Equilibria
If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand Ag(H2O)2+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) + 2 H2O(l) generally H2O is not included, because its complex ion is always present in aqueous solution Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Tro: Chemistry: A Molecular Approach, 2/e

Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)
Formation Constant The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) The equilibrium constant for the formation reaction is called the formation constant, Kf Tro: Chemistry: A Molecular Approach, 2/e

Formation Constants Tro: Chemistry: A Molecular Approach, 2/e

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Example 16.15: mL of 1.5 x 10−3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Write the formation reaction and Kf expression. Look up Kf value determine the concentration of ions in the diluted solutions Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) Tro: Chemistry: A Molecular Approach, 2/e

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Example 16.15: mL of 1.5 x 10−3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Create an ICE table. Because Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium. Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) [Cu2+] [NH3] [Cu(NH3)22+] initial 6.7E-4 0.11 change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4 equilibrium x Tro: Chemistry: A Molecular Approach, 2/e

Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq)
Example 16.15: mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with mL of 0.20 M NH3. What is the [Cu2+] at equilibrium? Cu2+(aq) + 4 NH3(aq)  Cu(NH3)22+(aq) substitute in and solve for x confirm the “x is small” approximation [Cu2+] [NH3] [Cu(NH3)22+] initial 6.7E-4 0.11 change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4 equilibrium x 2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is [HgI42−] when 125 mL of 0
Practice – What is [HgI42−] when 125 mL of M KI is reacted with 75 mL of M HgCl2? 4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq) Tro: Chemistry: A Molecular Approach, 2/e

Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
Practice – What is [HgI42−] when 125 mL of M KI is reacted with 75 mL of M HgCl2? 4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq) Write the formation reaction and Kf expression. Look up Kf value determine the concentration of ions in the diluted solutions Hg2+(aq) + 4 I−(aq)  HgI42−(aq) Tro: Chemistry: A Molecular Approach, 2/e 172

Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
Practice – What is [HgI42−] when 125 mL of M KI is reacted with 75 mL of M HgCl2? 4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq) Create an ICE table. Because Kf is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium. Hg2+(aq) + 4 I−(aq)  HgI42−(aq) I− is the limiting reagent [Hg2+] [I−] [HgI42−] initial 3.75E-4 6.25E-4 change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4) equilibrium 2.19E-4 x 1.56E-4 Tro: Chemistry: A Molecular Approach, 2/e 173

Hg2+(aq) + 4 I−(aq)  HgI42−(aq)
Practice – What is [HgI42−] when 125 mL of M KI is reacted with 75 mL of M HgCl2? 4 KI(aq) + HgCl2(aq)  2 KCl(aq) + K2HgI4(aq) Hg2+(aq) + 4 I−(aq)  HgI42−(aq) substitute in and solve for x confirm the “x is small” approximation [Hg2+] [I−] [HgI42−] initial 3.75E-4 6.25E-4 change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4) equilibrium 2.19E-4 x 1.56E-4 2 x 10−8 << 1.6 x 10−4, so the approximation is valid [HgI42−] = 1.6 x 10−4 Tro: Chemistry: A Molecular Approach, 2/e 174

The Effect of Complex Ion Formation on Solubility
The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands AgCl(s)  Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10−10 Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq) Kf = 1.7 x 107 Adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+ Tro: Chemistry: A Molecular Approach, 2/e

Solubility of Amphoteric Metal Hydroxides
Many metal hydroxides are insoluble All metal hydroxides become more soluble in acidic solution shifting the equilibrium to the right by removing OH− Some metal hydroxides also become more soluble in basic solution acting as a Lewis base forming a complex ion Substances that behave as both an acid and base are said to be amphoteric Some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+ Tro: Chemistry: A Molecular Approach, 2/e

Al3+ Al3+ is hydrated in water to form an acidic solution
Al(H2O)63+(aq) + H2O(l)  Al(H2O)5(OH)2+(aq) + H3O+(aq) Addition of OH− drives the equilibrium to the right and continues to remove H from the molecules Al(H2O)5(OH)2+(aq) + OH−(aq)  Al(H2O)4(OH)2+(aq) + H2O (l) Al(H2O)4(OH)2+(aq) + OH−(aq)  Al(H2O)3(OH)3(s) + H2O (l) Tro: Chemistry: A Molecular Approach, 2/e

Chapter 16 Aqueous Ionic Equilibrium

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Chapter 16 Aqueous Ionic Equilibrium

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