Equilibrium, Acids and Bases

Equilibrium, Acids and Bases
A. Dynamic Equilibrium equilibrium theories and principles apply to a variety of phenomena in our world eg) blood gases in scuba diving CO2 in carbonated beverages buffers in our blood reactions are often which means that not only are the reversible… products formed but the reactants can be reformed we use the double arrow to show this relationship eg) A + B ⇌ C + D

the forward and reverse reaction will proceed at different rates…it depends on the concentration of the reactants and products if we start with only the reactants A and B, the forward reaction will initially be the fastest as it is the only reaction possible as the products C and D are formed, the reaction will and the reaction will forward slow down reverse speed up at some point, the rates of forward and reverse reactions become equal

Dynamic Equilibrium equilibrium forward reaction Rate reverse reaction
Time

a system is said to be in a state of when:
dynamic equilibrium 1. the of the forward and reverse reactions are rates equal 2. the of the system, such as temperature, pressure, concentration, pH are observable (macroscopic) properties constant 3. the system is a system at closed constant temperature

B. Classes of Reaction Equilibria
there are three classes of chemical equilibria: favoured (percent rxn ) reactants <50% <50% A + B ⇌ C + D favoured (percent rxn ) products >50% >50% A + B ⇌ C + D

3. to the right (percent rxn ) quantitative >99%
A + B ⇌ C + D or A + B  C + D

C. The Equilibrium Constant
experiments have shown that under a given set of conditions (P and T) a specific quantitative relationship exists between the equilibrium concentrations of the reactants and products one reaction that has been studied intensively is that between H2(g) and I2(g) (simple molecules and takes place in gas phase no solvent necessary!) H2(g) + I2(g) ⇌ 2 HI(g)

when different combinations of H2(g), I2(g), and HI(g) were mixed and the concentrations measured, it was discovered that equilibrium was reached in all cases even though the equilibrium [ ] are , the different end quotient was the same each time (within experimental error)

this led to the empirical generalization known as the
this led to the empirical generalization known as the which says that there is a between the concentrations of the products and the concentrations of the reactants at equilibrium Law of Equilibrium constant ratio

this law can be expressed mathematically:
For the reaction aA + bB ⇌ cC + dD The law is: Kc = [C]c [D]d [A]a [B]b where: Kc = A, B = C, D = a, b, c, d = coefficients equilibrium constant reactants products balancing

is constant for a reaction at a given
is constant for a reaction at a given …if you change the temperature, Kc also changes Kc temperature it is common to ignore the units for Kc and list it only as a numerical value (since depends on the powers of the various [ ] terms) when determining Kc use only the species that are in or gas aqueous ***unless all states are the same, then use them all

the the value of Kc, the greater the tendency for the reaction to favor the
higher forward direction (the products) if Kc is then the reaction is favoured greater than 1, products if Kc is then the reaction is favoured less than 1, reactants Kc indicates the and not the percent reaction rate of the reaction catalysts will not affect the [ ] at equilibrium… they only increase the rate of the rxn

Example 1 Write the equilibrium law for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas. 2 NO(g) + O2(g) ⇌ 2 NO2(g) Kc = [NO2(g)]2 [NO(g)]2[O2(g)]

Example 2 Write the equilibrium law for the following reaction: CaCO3(s) ⇌ CaO(s) + CO2(g) Kc = [CO2(g)] *** do not include solids in Kc

Example 3 Write the equilibrium law for the following reaction: 2 H2O(l) ⇌ 2 H2(g) + O2(g) Kc = [H2(g)]2[O2(g)] *** do not include liquids in Kc

Example 4 Phosphorus pentachloride gas can be decomposed into phosphorus trichloride gas and chlorine gas. a) Write the equilibrium law for this reaction. PCl5(g) ⇌ PCl3(g) + Cl2(g) Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)]

b) If the [PCl5(g)]eq = 4. 3 x 10-4 mol/L, the [PCl3(g) ]eq = 0
b) If the [PCl5(g)]eq = 4.3 x 10-4 mol/L, the [PCl3(g) ]eq = mol/L and the [Cl2(g)]eq = mol/L then calculate Kc. Kc = [PCl3(g) ][Cl2(g)] [PCl5(g)] = (0.014)(0.014) (4.3 x 10-4) = 0.46

Example 5 Find the [SO3(g)] for the following reaction if Kc = 85.0 at 25.0C. 2 SO2(g) O2(g) ⇌ 2 SO3(g) 0.500 mol/L mol/L ??? Kc = [SO3(g) ]2____ [SO2(g)]2[O2(g)] = [SO3(g) ]2 (0.500)2(0.500) [SO3(g) ]2 = [SO3(g) ] = 3.26 mol/L

D. Graphical Analysis a graph of vs can be used to see when equilibrium has been reached…as soon as the concentrations , you can read this time off the graph concentration time don’t change any more

Consider this rxn: 2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Example 1 Consider this rxn: 2 SO2(g) + O2(g) ⇌ 2 SO3(g) Concentration (mol/L) Time (s) O2(g) SO2(g) SO3(g) 10 20 30 75 50 25 At what time does equilibrium get reached and what is the value for Kc?

Kc = [SO3]2 [SO2]2[O2] = (75)2 (50)2(25) = Equilibrium is reached at approximately 20 seconds.

E. Le Châtelier’s Principle
states that when a chemical system at is disturbed by a the system adjusts in a way that Le Châtelier’s principle equilibrium change in property of the system, opposes the change this takes place in a three-stage process 1. initial equilibrium state 2. shifting non-equilibrium state 3. new equilibrium state

a system can be affected by a change in concentration, temperature and or volume (pressure)
1. Concentration Changes an in the [ ] of the products or reactants favours increase the other side of the equation a in the [ ] of the products or reactants favours decrease the same side of the equation

eg) N2(g) H2(g) ⇌ 2 NH3(g) ***Haber-Bosch process ↑ [N2(g)] will shift the equilibrium ↑ [NH3(g)] will shift the equilibrium  [NH3(g)] will shift the equilibrium to the products to the reactants to the products changes in concentration have on the value of no effect Kc

2. Temperature Changes energy is treated like a or reactant product eg) reactants + energy ⇌ products reactants ⇌ products + energy if cooled, the equilibrium shifts so more heat is produced (same side) if heated, the equilibrium shifts away from the heat so it cools down (opposite side)

a change in temperature is the only stress that the value of Kc!!!!!!!
will change if the shift is towards the side, Kc will product increase if the shift is towards the side, Kc will reactant decrease

3. Volume and Pressure Changes
with gases, volume and pressure are related (volume , pressure )  ↑ the concentration of a gas is related to volume (pressure)…volume , concentration ↓ ↑ an caused by a in volume causes a shift towards the side of the equation with increase in [ ] drop fewer moles of gas eg) N2(g) H2(g) ⇌ 2 NH3(g) 4 moles moles will shift to NH3(g)

if the number of moles are the on both sides of the reaction, a change in volume (pressure) has
same no effect changes in volume and pressure have on the value of no effect Kc

4. Colour Changes in many equilibrium systems, the reactants will have a different colour than the products predictions can be made about the equilibrium shift and the resulting change in colour

Example Use the following reaction to predict the equilibrium shift and resulting colour change when the stresses are applied 2 CrO42(aq) H3O+(aq) ⇌ 2 Cr2O72(aq) H2O(l) yellow orange a) a crystal of Na2CrO4(s) is added products, orange b) a crystal of K2Cr2O7(s) is added reactants, yellow c) a few drops of concentrated acid is added products, orange d) water is removed products, orange e) a few crystals of NaOH(s) are added reactants, yellow

all of the changes that can happen to systems in equilibrium can be shown graphically:
Example State what change to the equilibrium takes place at each of the labelled parts of the graph:

Manipulations of An Equilibrium System
N2(g) H2(g) ⇌ 2 NH3(g) + energy Concentration (mol/L) Time (min) NH3(g) N2(g) H2(g) A B C D Equilibrium Time Stress A addition of H2(g) B addition of inert gas, addition of catalyst C decrease in volume D increase in energy

F. ICE Tables we can use a table set-up to calculate the equilibrium concentrations and/or Kc for any system you must be able to calculate all before you can use the equilibrium law equilibrium [ ]

Example 1 Hydrogen iodide gas decomposes into hydrogen gas and iodine gas. If 2.00 mol of HI(g) is place in a 1.00 L container and allowed to come to equilibrium at 35C, the final concentration of H2(g) is mol/L. Find the value for Kc. 2 HI(g) ⇌ H2(g) I2(g) Initial 2.00 mol/L Change –0.214 mol/L x 2/1 mol/L mol/L x 1/1 = –0.428 Equil. 1.572 mol/L 0.214 mol/L mol/L

Kc = [H2(g)][I2(g)] [HI(g)]2 = (0.214)(0.214) (1.572)2 =

CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
Example 2 In a 500 mL stainless steel reaction vessel at 900C, carbon monoxide and water vapour react to produce carbon dioxide and hydrogen. Evidence indicates that this reaction establishes an equilibrium with only partial conversion of reactants to products. Initially, 2.00 mol of each reactant is placed in the vessel. Kc for this reaction is 4.20 at 900C. Calculate the concentration of each substance at equilibrium. CO(g) H2O(g) ⇌ CO2(g) + H2(g) I 2.00/0.5L = 4.00 mol/L 2.00/0.5L = 4.00 mol/L C –x mol/L –x mol/L +x mol/L x 1/1 +x mol/L x 1/1 E 4.00  x mol/L 4.00  x mol/L x mol/L x mol/L

Kc = [CO2(g)][H2(g)] [CO(g)][H2O(g)] 4.20 = (x)(x)
***Note, this is a perfect square so to solve for x simply square root both sides of the equation, then solve = x (4.00  x) 2.05(4.00  x) = x 8.05  2.05x = x 8.05 = 3.05x x = 2.69 mol/L

[CO(g)] = 4.00 – x = [H2O (g)] = 4.00 – x = [CO2(g)] = x = [H2(g)] = x = 4.00 – 2.69 = 1.31 mol/L 4.00 – 2.69 = 1.31 mol/L 2.69 mol/L 2.69 mol/L

Example 3 Gaseous phosphorus pentachloride decomposes into gaseous phosphorus trichloride and chlorine gas at a temperature where Keq = 1.00  103. Suppose 2.00 mol of PCl5(g) in a 2.00 L vessel is allowed to come to equilibrium. Calculate the equilibrium [ ] of each species. PCl5(g) ⇌ PCl3(g) Cl2(g) 2.00mol/2.00L = 1.00 mol/L I 0 mol/L 0 mol/L C –x mol/L x 1/1 +x mol/L +x mol/L x 1/1 x mol/L E 1.00 – x mol/L x mol/L

Kc = [PCl3(g)][Cl2(g)] [PCl5(g)] 1.00  = (x)(x) ( x) ***at this point, you would have to use the quadratic formula to solve for x when the concentrations are greater than the equilibrium constant, we can make an that greatly simplifies our calculations 1000 X approximation if Kc is very small, the equilibrium doesn’t lie very far to the right and  x is a very small number

in this example 1. 00 – x can be assumed to be 1
***in this example 1.00 – x can be assumed to be 1.00 since x is really small, so… 1.00  = (x)(x) (1.00 ) x2 =  10-3 x 1.00 x = ***now you can calculate the [ ]eq for each species …substitute x into the equilibrium values in the ICE table [PCl5(g)]eq = 1.00 mol/L – mol/L = mol/L [PCl3(g)]eq = mol/L = mol/L [Cl2(g)]eq = mol/L = mol/L

Example 4 Gaseous NOCl decomposes to form gaseous NO and Cl2. At 35C the equilibrium constant is 1.6  Calculate the equilibrium [ ] of each species when 1.0 mol of NOCl is placed in a 2.0 L covered flask. 2 NOCl(g) ⇌ 2 NO(g) + Cl2(g) I 0 mol/L 0 mol/L 1.0mol/2.0L = 0.50 mol/L C –x mol/L x 2/1 +x mol/L x 2/1 +x mol/L = –2x mol/L = +2x mol/L E 0.50 – 2x mol/L 2x mol/L x mol/L

[NOCl(g)]eq = 0.50 mol/L – (2)(0.010) mol/L = 0.48 mol/L
Kc = [NO(g)]2[Cl2(g)] [NOCl(g)]2 1.6  = (2x)2(x) ( x)2 ***using approximation, 0.50 – 2x = 0.50 1.6  = (4x 2)(x) (0.50 )2 4x3 = 1.6  10-5 x 0.502 x3 = 4.0  10-6 / 4 x = mol/L [NOCl(g)]eq = 0.50 mol/L – (2)(0.010) mol/L = 0.48 mol/L [NO(g)]eq = 0 + (2)(0.010 mol/L) = mol/L [Cl2(g)]eq = mol/L = mol/L

the equilibrium of water can be written as follows:
G. Ionization of Water the equilibrium of water can be written as follows: H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) the equilibrium law is: Kc = [H3O+(aq) ][ OH-(aq)] the equilibrium constant for water is designated as Kw at 25C, neutral water has [H+(aq)] = [OH-(aq)] = 1.0  10-7 mol/L  Kw = = (on pg 3 of Data Booklet) (1.0  10-7) (1.0  10-7) 1.0  10-14

Kw is constant and therefore can be used to determine the or the
[H3O+(aq) ] [ OH-(aq)] eg) if [H3O+(aq)] = 1.0  104 mol/L then [OH(aq)] = 1.0 1014 = 1.0  104 1.0  1010 mol/L if [H3O+(aq)] = [OH(aq)], then solution is if [H3O+(aq)] > [OH(aq)], then solution is if [H3O+(aq)] < [OH(aq)], then solution is neutral acidic basic

Try These: 1. [H3O+(aq)] = 1.0  109 mol/L [OH(aq)] = 2. [H3O+(aq)] = 1.0  101 mol/L 3. [H3O+(aq)] = [OH(aq)] = 1.0  102 mol/L 4. [H3O+(aq)] = 6.3  109 mol/L 5. [H3O+(aq)] = 8.1  103 mol/L 6. [H3O+(aq)] = [OH(aq)] = 2.8  107 mol/L basic 1.0  105 mol/L acidic 1.0  1013 mol/L 1.0  1012 mol/L basic basic 1.6  106 mol/L acidic 1.2  1012 mol/L 3.6  108 mol/L basic

H. Review of pH and pOH the number of digits following the in the is equal to the number of in the decimal place pH value sig digs [H3O+(aq)] pH =  log[H3O+(aq)] [H3O+(aq)] = 10-pH pOH =  log[OH-(aq)] [OH-(aq)] = 10-pOH pH + pOH = 14

Example 1 Find the pH of a solution where the [H3O+(aq)] = 4.7  mol/L. pH =  log[H3O+(aq)] =  log(4.7  10-11) = 10.33

Example 2 Find the pH of a solution where the [OH-(aq)] = 2.4  10-3 mol/L. pOH =  log[OH(aq)] =  log(2.4  103 ) = 2.619… pH = 14 – pOH = 14 – 2.619… = 11.38

Example 3 Calculate the [H3O+(aq)] if the pH of the solution is 5.25. [H3O+(aq)] = 10-pH = = 5.6  10-6 mol/L

Example 4 Calculate the pH of a solution where 10.3 g of Ca(OH)2(s) is dissolved in 500 mL of water. Ca(OH)2(s)  Ca2+(aq) OH-(aq) m = 10.3 g M = g/mol n = m  M = 10.3 g  g/mol = 0.139…mol v = L n = 0.139…mol  2/1 = 0.278…mol C = n  V = …mol  0.500L = 0.556…mol/L

Example 4 (continued) pOH =  log[OH(aq)] =  log(0.556… ) = … pH = 14 – pOH = 14 – 0.254… =

I. Brønsted-Lowry Definition of Acids & Bases (1923)
this theory looks at the role of the acid or base an acid is a chemical species (anion, cation or molecule) that loses a proton a base is a chemical species that gains a proton like in electrochemistry where e are transferred…now we transfer H+

water does not have to be involved!
HCl(aq) + H2O(l) ⇌ Cl-(aq) + H3O+(aq) H+ NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) water does not have to be involved! H+ NH4Cl(s) HCl(g) + NH3(g) ⇌

a Brønsted-Lowry acid doesn’t necessarily have to produce an acidic solution…it depends on what accepts the proton an acid/base reaction is a chemical reaction in which a is transferred from an to a forming a and a proton (H+) acid base new acid new base this theory explains how some chemical species can be used to neutralize both acids and bases eg) HCO3-(aq) + H3O+(aq) ⇌ H2O(l) + H2CO3(aq) HCO3-(aq) + OH-(aq) ⇌ H2O(l) + CO32-(aq)

a substance that appears to act as a Brønsted-Lowry acid in some rxns and a Brønsted-Lowry base in other rxns is said to be amphiprotic or amphoteric eg) H2O, HCO3, HSO4, HOOCCOO etc

J. Conjugate Acids and Bases
a pair of substances that differ only by a proton is called a …the is on one side of the reaction and the is on the other conjugate acid-base pair acid base in general, the reaction can be shown as follows: HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) acid base conjugate acid conjugate base the an acid, the its conjugate base stronger weaker the an acid, the its conjugate base weaker stronger

K. Strengths of Acids and Bases
two different acids (or bases) can have the same [ ] but have different strengths eg) 1 M CH3COOH(aq) and 1 M HCl(aq) will react in the same way but not to the same degree the stronger the acid, the electricity it conducts, the the pH and the it reacts with other substances more lower faster

1. Strong Acids acids that ionize in water to form H3O+(aq) quantitatively percent rxn = 100% the bigger the Ka (Kc for acids) the more the are favoured products top 6 acids on the table (pg 8-9 in Data Book) have a very large Ka …note the H3O+ is the strongest acid on the chart (leveling effect)…all strong acids react to form H3O+(aq) so it is the strongest

when calculating pH, the so use
[SA] = [H3O+(aq)] pH = -log[H3O+(aq)] Example What is the pH of a mol/L solution of HNO3(aq)? [H3O+(aq)] = [HNO3(aq)] = mol/L pH = -log[H3O+(aq)] = -log(0.500 mol/L) = 0.301

2. Weak Acids a weak acid is one that only partially ionizes in water to form H+(aq) ions most ionize <50% Ka value is small (<1) to calculate pH, you need to use the …you cannot use just the because it is not Ka value [WA] 100% dissociated

the Ka law is an and is devised the same way we did
equilibrium law Kc eg) HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) Ka = [H3O+(aq)][A-(aq)] [HA(aq)] you will be required to figure out the before you can calculate the pH [H3O+(aq)]

you have the and the value but you don’t have the
[WA] Ka [A-(aq)] since the mole ratio for is , they have the same [ ] (this is a !) [H3O+(aq)]:[A-(aq)] 1:1 dissociation Ka = [H3O+(aq)]2 [HA(aq)] now you can solve for x to get the [H3O+(aq)] [H3O+(aq)] = (Ka)([WA])

Example 1 What is the pH of a 0.10 mol/L acetic acid solution? ***check in DB…weak acid!!!!! CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq) [H3O+(aq)] = (Ka)([WA]) = (1.8 x 10-5 mol/L )(0.10 mol/L) = 1.34… 10-3 mol/L pH = -log[H3O+(aq)] = -log(1.34… 10-3 mol/L) = 2.87

Example 2 What is the pH of a 1.0 mol/L acetic acid solution? CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq) [H3O+(aq)] = (Ka)([WA]) = (1.8 x 10-5 mol/L )(1.0 mol/L) = 4.24… 10-3 mol/L pH = -log[H3O+(aq)] = -log(4.24… 10-3 mol/L) = 2.37

Example 3 A 0.25 mol/L solution of carbonic acid has a pH of Calculate Ka. H2CO3(aq) + H2O(l) ⇌ H3O+(aq) + HCO3-(aq) [H3O+(aq)] = 10-pH = = 3.31…  10-4 mol/L Ka = [H3O +(aq)]2 [H2CO3 (aq)] = (3.31…x10-4 mol/L)2 0.25 mol/L = 4.4 x 10-7 mol/L

the can be written as a above the in a chemical reaction:
% reaction (% ionization) % ⇌ 1.3% eg) CH3COOH(aq) H2O(l) ⇌ H3O+(aq) + CH3COO-(aq) Ka = 1.8 x 10-5 the % reaction be calculated using [H3O+] and [WA] % ionization = [H3O+(aq)]  100 [WA(aq)]

Example 1 Calculate the % ionization for a mol/L solution of hydrosulphuric acid if the [H+(aq)] is 5.0  10-4 mol/L. % ionization = [H3O +(aq)]  [WA(aq)] = 5.0  10-4 mol/L  100 0.500 mol/L = 0.10 %

Example 2 The pH of a 0.10 mol/L solution of methanoic acid is Calculate the % ionization. [H3O+] = 10-pH = = … mol/L % ionization = [H3O +(aq)]  [WA(aq)] = ( …mol/L) x (100) 0.10 mol/L = 4.2 %

3. Strong Bases according to Arrhenius, bases are substances that increase the of a solution hydroxide [ ] all are strong bases ionic hydroxides percent rxn = 100% strength depends on … is a stronger base than at the same [ ] because it produces # of hydroxide ions Ba(OH)2(aq) NaOH (aq) 2 OH-(aq)

where x is the number of ions (think about the dissociation equation!)
[OH-(aq)] = x[BH(aq)] hydroxide Example Calculate the pH of a mol/L solution of Ca(OH)2(aq). [OH-(aq)] = x[BH(aq)] = x[Ca(OH)2(aq)] = 2( mol/L) = mol/L pOH = log(0.120) = … pH = 14 – … =

4. Weak Bases do not in water…just like weak acids dissociate completely is the dissociation constant or equilibrium constant for bases Kb B + H2O(l) ⇌ BH+(aq) + OH-(aq) Kb = [BH+(aq)][OH-(aq)] [WB(aq)]

you can calculate Kb using for the
Ka base Ka  Kb = KW = 1.00  1014 eg) Calculate Kb for SO42(aq). Kb = KW Ka = 1.00  1014 1.0  102 = 1.0  1012 mol/L

once you have Kb, you can then solve for [OH(aq)] using the equilibrium law (just like with weak acids!) Kb = [BH+(aq)][OH-(aq)] [WB(aq)] Kb = [OH-(aq)]2 now you can solve for [OH(aq)] [OH-(aq)] = (Kb)([WB])

Example Find the pH of a 15.0 mol/L NH3(aq) solution. NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq) Ka = 5.6  mol/L Kb = Kw Ka = 1.00  10-14 5.6  10-10 = 1.78…  10-5 mol/L

[OH-(aq)] = (Kb)([NH3(aq)])
[OH-(aq)] = 1.63… x 10-2 mol/L pOH = -log[OH-(aq)] = -log(1.63…x 10-2 mol/L) = 1.78… pH = 14 – pOH = 14 – 1.78… =

L. Predicting Acid-Base Equilibria
acids are listed in order of strength on the left side and bases are listed in order of strength on the right side decreasing increasing when predicting reactions, the substance with the will react with the substance that greatest attraction for protons (the strongest base) gives up its proton most easily (strongest acid) we will assume that only is transferred per reaction one proton

to predict the acid-base reaction, follow the following steps:
1. List all species (ions, atoms, molecules) initially present. Note:  strong acids ionize into H3O+ and the anion  weak acids are NOT dissociated  dissociate ionic compounds  don’t forget to include water 2. Identify all possible acids and bases.

3. Identify the and …like redox rxns the and the
strongest acid (SA) strongest base (SB) SA is top left SB is bottom right. 4. To write the reaction, transfer one proton from the acid to the base to predict the conjugate acid and conjugate base.

5. Predict the position of the equilibrium.
Note:  if acid is above base, then >50% (favours products) ⇌  if base is above acid, then <50% (favours reactants) ⇌

Example 1 Predict the acid-base reaction that occurs when sodium hydroxide is mixed with vinegar. List: Na+(aq) OH-(aq) CH3COOH(aq) H2O(l) B A A/B SB SA OH-(aq) + CH3COOH(aq) H2O(l) + CH3COO-(aq) ⇌

Example 2 Predict the acid-base reaction when ammonia is mixed with HCl(aq). List: NH3(aq H3O+(aq) Cl-(aq) H2O(l) B A B A/B SB SA NH3(aq) + H3O+(aq) H2O(l) + NH4+ (aq) ⇌

M. Monoprotic vs. Polyprotic Acids and Bases
an acid capable of donating only one proton is called monoprotic eg) HCl(aq), HNO3(aq), HOCl(aq) etc. if an acid can transfer more than one proton, it is called ( if 2 protons, if 3 protons) polyprotic diprotic triprotic

eg) Label each of the following acids as monoprotic or polyprotic: 1. H2SO4(aq) 2. HOOCCOOH(aq) 3. HCOOH(aq) 4. CH3COOH(aq) 5. H2PO4-(aq) 6. NH4+(aq) polyprotic polyprotic monoprotic monoprotic polyprotic monoprotic

a base capable of accepting only one proton is called a
monoprotic base a base that can accept more than one proton is called a polyprotic base ( or ) diprotic triprotic eg) can accept up to 3 H+ to form and respectively PO43-(aq) HPO42-(aq), H2PO4-(aq), H3PO4(aq)

eg) Label each of the following as monoprotic or polyprotic acids, monoprotic or polyprotic bases: monoprotic acid; monoprotic base 1. HSO4-(aq) 2. H2PO4-(aq) 3. HPO42-(aq) 4. HCO3-(aq) 5. H2O(l) polyprotic acid; monoprotic base monoprotic acid; polyprotic base monoprotic acid; monoprotic base monoprotic acid; monoprotic base

reactions involving polyprotic acids or polyprotic bases substances involve the same principles of reaction prediction only is transferred at a time and always from strongest acid to strongest base one proton

Example 1 Potassium hydroxide is continuously added to oxalic acid until no more reaction occurs. OOCCOO2-(aq) List: K+(aq) OH-(aq) HOOCCOOH(aq) H2O(l) HOOCCOO-(aq) B A A/B A/B B SB SA SA OH-(aq) + HOOCCOOH(aq) ⇌ H2O(l) + HOOCCOO-(aq) OH-(aq) + HOOCCOO-(aq) ⇌ H2O(l) + OOCCOO2-(aq)

Net Reaction: Add all reactions together (only if all quantitative), cancelling out any species that occur in the same quantity on both the reactant and product sides and summing any species that occur more than once on the same side OH-(aq) + HOOCCOOH(aq) H2O(l) HOOCCOO-(aq) OH-(aq) + HOOCCOO-(aq) H2O(l) OOCCOO2-(aq) 2 OH-(aq) + HOOCCOOH(aq) 2 H2O(l) + OOCCOO2-(aq)

Example 2 Sodium hydrogen phosphate is titrated with hydroiodic acid. If we assume all steps are quantitative, give the net reaction. H3O+(aq) List: Na+(aq) HPO42-(aq) I-(aq) H2O(l) H2PO4-(aq) H3PO4(aq) A/B A/B A B A A/B SB SB SA H3O+(aq) + HPO42-(aq) H2O(l) + H2PO4-(aq) H3O+(aq) + H2PO4-(aq) H2O(l) + H3PO4(aq) 2 H3O+(aq) + HPO42-(aq) 2 H2O(l) + H3PO4(aq)

O. Titrations titrations are used to determine the pH of the of acid-base reactions endpoint the information from the titration can be plotted on a graph, buffer regions can be analyzed and stoichiometric calculations can be performed

1. pH Curves a is a graph showing the continuous change of pH during an acid-base reaction pH curve graph of vs volume of titrant added to sample (x-axis) pH (y-axis) the is the point (usually shown by a change in indicator colour) when the reaction has gone to completion endpoint the is the of titrant required for the reaction to go to completion equivalence point volume

they contain a relatively flat region called the
buffer region all pH curves have 4 major features: the initial pH of the curve must be the pH of the sample the co-ordinate of the equivalence point must be correct in terms of pH and volume the “over-titration” must be asymptotic with the pH of the titrant number of equivalence points must match the number of quantitative reactions occurring

titrant selection: if the sample is an acid, titrant should be a such as strong base NaOH(aq) or KOH(aq) if the sample is a base, titrant should be HCl(aq)

you need to be able to interpret pH curves:
1. Strong Monoprotic with Strong Monoprotic pH of at the equivalence point 7 SA titrated with SB SB titrated with SA 14 14 pH EP 7 pH EP 7 volume volume

2. Weak Monoprotic with Strong Monoprotic
if weak acid, then pH of at equivalence point >7 if weak base, then pH of at equivalence point <7 bottom “flat” region is not as flat as with strong acid/strong base WA titrated with SB WB titrated with SA 14 14 pH EP 7 pH 7 EP volume volume

3. Polyprotic with Strong Monoprotic more than 1 equivalence point
WA(poly) titrated with SB WB(poly) titrated with SA 14 14 EP2 EP1 7 pH 7 EP1 EP2 pH volume volume

an indicator is a substance that changes colour when it reacts with an
2. Indicators an indicator is a substance that changes colour when it reacts with an acid or base and are usually weak acids themselves they exist in one of two conjugate forms that are reversible and distinctly different in color HIn(aq) H2O(l) ⇌ In-(aq) + H3O+(aq) acid base conjugate base conjugate acid eg) litmus red blue

recall that to show the equivalence point of an acid-base titration, choose an indicator:
1. whose includes the of the titration colour change range equivalence point 2. that will react …this means the indicator is a weaker acid or base than the sample right after the sample reacts

3. Buffers are chemicals that, when added to water, protect the solution from buffers large pH changes when acids or bases are added to them they are used to and control the rate of pH sensitive reactions (eg. in the blood) calibrate pH meters typical buffers are solutions containing relatively large amounts of such as a and the conjugate pairs weak acid salt of the conjugate base eg) H2CO3 and NaHCO3

can be selected by using …this tells you the pH at which the buffer is most useful
pKa = –logKa eg) Choose a buffer that would be useful for each of the following solutions: 1. pH of 7.0 H2S – HS- pKa = -log(8.9 × 10-8) = 7.1 2. pH of 4.5 CH3COOH – CH3COO- pKa = -log(1.8 × 10-5) = 4.7 3. pH of 10.0 HCO3- – CO32- pKa = -log(4.7 × 10-11) = 10.3

the in the conjugate pair of the buffer protects against any added
acid base the in the conjugate pair of the buffer protects against any added base acid buffers can be by the addition of overwhelmed too much acid or base

a) a small amount of HCl(aq) is added
Example Using an acetic acid – sodium acetate buffer system, show what happens when: a) a small amount of HCl(aq) is added CH3COOH(aq) Na+(aq) CH3COO-(aq) H3O+(aq) Cl-(aq) H2O(l) A – B A B A/B SB SA CH3COO-(aq) + H3O+(aq) ⇌ CH3COOH(aq) + H2O(l)

b) a small amount of NaOH(aq) is added
CH3COOH(aq) Na+(aq) CH3COO-(aq) OH-(aq) H2O(l) A – B B A/B SA SB CH3COOH(aq) + OH-(aq) ⇌ CH3COO-(aq) + H2O(l)

Equilibrium, Acids and Bases

Similar presentations

Presentation on theme: "Equilibrium, Acids and Bases"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Equilibrium, Acids and Bases

Similar presentations

Presentation on theme: "Equilibrium, Acids and Bases"— Presentation transcript:

Similar presentations

About project

Feedback