Applications of Aqueous Equilibria McMurry & Fay ch. 15

Vocabulary: BUFFER SOLUTION an aqueous solution containing both an acid and its conjugate base

Calculate the pH in a solution prepared by dissolving 0 Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in 0.500 L of 0.40 M NH3 (Kb = 1.8 x 10-5). I C E [NH4+][OH-] [NH3] Kb = = 1.8 x 10-5

[H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA] A shortcut [H+][A-] [HA] Ka = [H+][A-] [HA] - log Ka = - log [A-] [HA] - log Ka = - log [H+] – log = pKa = pH

Henderson-Hasselbalch Equation pH = pKa + log

Uses for Henderson-Hasselbalch Equation Calculate pH of buffer solution Prepare buffer solution of given pH

Calculating pH of buffer solution Calculate the pH in a solution prepared by dissolving 0.10 mol of solid NH4Cl in 0.500 L of 0.40 M NH3 (Kb = 1.8 x 10-5). pH = pKa + log [A-] [HA]

Calculating pH of buffer solution A buffer solution is created with 0.10 M HF (Ka = 6.8 x 10-4) and 0.15 M NaF. What is the pH? pH = pKa + log [A-] [HA]

Buffer solutions resist changes in pH: adding acid H2CO3 H+ + HCO3-

Buffer solutions resist changes in pH: adding base H2CO3 + OH- H2O + HCO3-

To calculate change in pH: Choose appropriate reaction: HA ⇌ A- + HA if adding acid BH+ + OH- ⇌ B + H2O Make an ICE table that shows number of moles of each species Calculate change using # moles of base or acid added Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

How much would the pH change if 100 mL of 1.0 HCl was added to water? 1.0 L of buffer solution has 0.30 M HNO2 (Ka=7.1 x 10-4) and 0.30 M NaNO2. 100 mL of 1.0 M HCl are added to the solution. How much would the pH change if 100 mL of 1.0 HCl was added to water? How much does the pH change when the HCl is added to the buffer solution? Choose appropriate reaction: HA ⇌ A- + HA if adding acid BH+ + OH- ⇌ B + H2O Make an ICE table that shows number of moles of each species Calculate change using # moles of base or acid added Use final # of moles & Henderson-Hasselbalch eqn to calculate final pH

Preparing buffer solution of given pH 1. Pick acid with pKa within 1 pH unit of desired pH 2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base 3. Multiply ratio by a factor appropriate for necessary buffer strength

Pick acid with pKa near desired pH HF 3.17 CH3CO2H 4.76 Ethylenediamine 6.85 HOCl 7.53 Tris* 8.07 NH4+ 9.25 *Tris(hydroxymethyl)aminomethane

Make a buffer with pH 7.8 [A-] pH = pKa + log [HA] 1. Pick acid with pKa within 1 pH unit of desired pH HF 3.17 CH3CO2H 4.76 Ethylenediamine 6.85 HOCl 7.53 Tris* 8.07 NH4+ 9.25 2. Use Henderson-Hasselbalch equation to calculate ratio between acid & conjugate base 3. Multiply ratio by a factor appropriate for necessary buffer strength pH = pKa + log [A-] [HA]

Acid-Base Titrations

pH changes during Titration: Strong Acid-Strong Base As you add base to an acid solution: Some acid is neutralized; # of moles of H+ decreases Volume of solution increases Molarity = so [H+] will decrease and pH will increase Moles Liters

Calculating pH changes during Strong Acid-Strong Base Titration: before equivalence point Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of H+ remaining: # moles acid at start - # moles base added Calculate volume of sample: Initial volume + volume of base added Calculate [H+] (moles H+/volume) and pH

20. 00 mL of 0. 400 M HCl is titrated with 0 20.00 mL of 0.400 M HCl is titrated with 0.200 M NaOH Calculate pH when you have added: (a) 0 mL of acid (b) 5 mL of acid (c) 7.5 mL of acid (d) 10 mL of acid (e) 15 mL of acid (f) 19 mL of acid (g) 20 mL of acid Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of H+ remaining: # moles acid at start - # moles base added Calculate volume of sample: Initial volume + volume of base added Calculate [H+] (moles H+/volume) and pH

What happens after the equivalence point? All acid has been used up; OH- dominates the pH Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate initial moles of acid : # moles H+ = (Molarity acid) * (volume acid) Calculate moles of OH-: # moles of base – initial moles of acid Calculate volume of sample: Initial volume + volume of NaOH added Calculate [OH-] (moles OH-/volume), pOH, and pH

Equivalence point # moles acid = # moles base

pH changes during Titration: Weak Acid-Strong Base What’s different from strong acid titrations Acid doesn’t completely dissociate so pH depends on Ka Adding base produces the conjugate base of the weak acid Acid + conjugate base both present = buffer solution Use Henderson Hasselbalch to calculate pH

pH changes during Titration: Weak Acid-Strong Base (before equivalence point) Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of conjugate base formed: # moles A- = # moles base added Calculate moles of acid remaining: # moles HA left = initial # moles acid - # moles base added Use Henderson Hasselbalch to calculate pH

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH What is [H+] at the start? Ka = [H+][A-] [HA]

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH Where do we expect our equivalence point to be?

Example: titration of 10 mL 0.500 M benzoic acid (pKa = 4.202) with 0.200 M NaOH What is pH after you have added: 5 mL 10 mL 12.5 mL 15 mL 20 mL 22.5 mL 24.5 mL Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of conjugate base formed: # moles A- = # moles base added pH = pKa + log [A-] [HA] Calculate moles of acid remaining: # moles HA left = initial # moles acid - # moles base added Use Henderson Hasselbalch to calculate pH

After the equivalence point: No acid left A- + H2O HA + OH-

equivalence point # moles acid = # moles base Half equivalence point [A-] = [HA]

Half-equivalence point Point at which [A-] = [HA] Henderson-Hasselbalch equation: [A-] [HA] pKa = pH – log

At half-equivalence point, pH = pKa

pH changes during Titration: Weak Acid-Strong Base (at equivalence point) All acid has been turned into conjugate base Solution of basic salt: calculate pH as done in ch. 14

pH changes during Titration: Weak Acid-Strong Base (after equivalence point) Most OH- comes from excess NaOH Calculate moles of acid you start with: # moles acid = (Molarity acid) * (volume acid) Calculate moles of base added: # moles base = (Molarity base) * (volume base) Calculate moles of OH- remaining: # moles base added - # moles acid Calculate volume of sample: Initial volume + volume of base added Use [H+] = 10-14/[OH-] to calculate pH

Indicators Dyes that are weak acids or bases Color of acid form different from conjugate base HIn (aq)  H+ (aq) + In– (aq) Acid form base form (one color) (another color)

Choosing an indicator Color change begins when two forms are equal: Choose indicator that has pKa = pH of equiv. point!

Polyprotic acids

Carbonate chemistry CO2 + H2O H2CO3 (carbonic acid) H2CO3 H+ + HCO3- (bicarbonate) pKa = 6.35 HCO3- H+ + CO32- (carbonate) pKa = 10.33

2 Equilibrium constants for diprotic acids: H2A H+ + HA- HA- H+ + A2- [H+][HA-] [H2A] [H+][A2-] [HA-] Ka1 = Ka2 = Each contributes to [H+]

Calculating pH [H+]equilibrium = [H+]1stIonization + [H+]2ndIonization

Calculating pH [H+]equilibrium = [H+]1stIonization + [H+]2ndIonization [H+]1stIonization << [H+]2ndIonization

Simplification: [H+]equilibrium = [H+]1stIonization In other words, treat the acid as a monoprotic acid!

Simplification # 2 H2A H+ + HA- HA- H+ + A2- [H+]equilibrium = [H+]1stIonization H2A H+ + HA- [HA-]equilibrium = [HA-]1stIonization HA- H+ + A2-

Simplification #2 means HA- H+ + A2- [H+][A2-] [HA-] Ka2 = [H+][A2-] [H+] Ka2 = [A2-] = Ka2

Calculating equilibrium concentrations for polyprotic acids: 1. Assume [A2-]≈ Ka2 2. Calculate remaining concentrations by treating as monoprotic acid

2. Calculate remaining concentrations by treating as monoprotic acid Calculate concentrations of solute species in 0.30 M oxalic acid (Ka1 = 5.9 x 10-2; Ka2 = 6.4 x 10-5) 1. Assume [A2-]≈ Ka2 2. Calculate remaining concentrations by treating as monoprotic acid

Solubility: An Equilibrium Point of View

Vocabulary SATURATED SOLUTION A solution containing as much solute as can possibly dissolve UNSATURATED SOLUTION A solution containing less solute than can possibly dissolve SUPERSATURATED SOLUTION A solution containing more solute than is supposed to be able to dissolve

Solubility—Equilibrium Point of View NaCl (s) Na+ (aq) + Cl- (aq)

Solubility—Equilibrium Point of View NaCl (s) Na+ (aq) + Cl- (aq) Special equilibrium constant, Ksp: Ksp= [Na+][Cl-]

Slightly soluble salts: How much can actually dissolve? CaCO3 (s) Ca2+ (aq) + CO32- (aq) Ksp = 3.4 x 10-9

Vocabulary: Molar solubility The number of moles of salt required to create one liter of a saturated solution

Calculating Ksp from solubility information Write a balanced equation for ion dissociation. Convert grams per liter to molarity if necessary Use stoichiometry to convert molarity of salt to molarity of ions Use stoichiometry to convert molarity of salt to ion concentrations Use ion concentrations to calculate Ksp

The molar solubility of Ca(OH)2 is 1.06 x 10-2 M. What is the Ksp? Write a balanced equation for ion dissociation. Convert grams per liter to molarity if necessary Use stoichiometry to convert molarity of salt to molarity of ions Use stoichiometry to convert molarity of salt to ion concentrations Use ion concentrations to calculate Ksp

The solubility of PbI2 is 4.34 g/L. What is the Ksp? Write a balanced equation for ion dissociation. Convert grams per liter to molarity if necessary Use stoichiometry to convert molarity of salt to molarity of ions Use stoichiometry to convert molarity of salt to ion concentrations Use ion concentrations to calculate Ksp

Calculating solubility from Ksp Write a balanced equation for ion dissociation. Use Ksp to calculate ion concentrations Use stoichiometry to convert ion concentrations to molarity of salt Convert molarity to grams per liter if necessary

The Ksp for Ag2CO3 is 8. 5 x 10-12. What is the molar solubility The Ksp for Ag2CO3 is 8.5 x 10-12. What is the molar solubility? What is the solubility in grams per liter? Write a balanced equation for ion dissociation. Use Ksp to calculate ion concentrations Use stoichiometry to convert ion concentrations to molarity of salt Convert molarity to grams per liter if necessary

The Ksp for Cu3PO4 is 1. 4 x 10-37. What is the molar solubility The Ksp for Cu3PO4 is 1.4 x 10-37. What is the molar solubility? What is the solubility in grams per liter? Write a balanced equation for ion dissociation. Use Ksp to calculate ion concentrations Use stoichiometry to convert ion concentrations to molarity of salt Convert molarity to grams per liter if necessary

CaCO3 (s) Ca2+ (aq) + CO32- (aq) Common Ion Effect Application of Le Châtelier’s Principle: CaCO3 (s) Ca2+ (aq) + CO32- (aq)

CaCO3 (s) Ca2+ (aq) + CO32- (aq) Common Ion Effect Application of Le Châtelier’s Principle: If we add more CO32-, equilibrium will shift towards reactants CaCO3 (s) Ca2+ (aq) + CO32- (aq)

CaCO3 (s) Ca2+ (aq) + CO32- (aq) Common Ion Effect Application of Le Châtelier’s Principle: If we add more CO32-, equilibrium will shift towards reactants CaCO3 (s) Ca2+ (aq) + CO32- (aq) When common ions are present in solution, calculate solubilities using ICE tables

Calculating Solubility: Common Ion Effect 1. Write a balanced equation for ion dissociation. 2. Make ICE table. Add initial concentration of common ion 3. Use stoichiometry to calculate concentration changes in terms of x 4. Calculate equilibrium concentrations 5. Use Ksp expression to solve for x 6. Use value of x to calculate solubility

The Ksp of CaCO3 is 3.4 x 10-9. What is the molar solubility of CaCO3 in a 0.20 M Na2CO3 solution? 1. Write a balanced equation for ion dissociation. 2. Make ICE table. Add initial concentration of common ion 3. Use stoichiometry to calculate concentration changes in terms of x 4. Calculate equilibrium concentrations 5. Use Ksp expression to solve for x 6. Use value of x to calculate solubility

The Ksp of PbCrO4 is 1.8 x 10-14. What is the molar solubility of PbCrO4 in a 0.0010 M K2CrO4 solution? 1. Write a balanced equation for ion dissociation. 2. Make ICE table. Add initial concentration of common ion 3. Use stoichiometry to calculate concentration changes in terms of x 4. Calculate equilibrium concentrations 5. Use Ksp expression to solve for x 6. Use value of x to calculate solubility

Will a precipitate form? Only if solution is supersaturated Supersaturation occurs if Q > Ksp

To determine if a precipitate will form… 1. Write the mass action expression for dissolution reaction 2. Use given concentrations to calculate Q 3. Compare Q to Ksp. If Q > Ksp, than precipitate forms. If Q < Ksp, then no precipitate forms.

A 0. 10 M solution of NaCl is mixed with a 0. 10 M solution of AgNO3 A 0.10 M solution of NaCl is mixed with a 0.10 M solution of AgNO3. Will a precipitate form? Ksp for AgCl is 1.8 x 10-10 1. Write the mass action expression for dissolution reaction 2. Use given concentrations to calculate Q 3. Compare Q to Ksp. If Q > Ksp, than precipitate forms. If Q < Ksp, then no precipitate forms. Molar solubility = 1.3 x 10-5

A 0. 0010 M solution of CaCl2 is mixed with a 0 A 0.0010 M solution of CaCl2 is mixed with a 0.0010 M solution of K2SO4. Will a precipitate form? Ksp for CaSO4 is 2.4 x 10-5. 1. Write the mass action expression for dissolution reaction 2. Use given concentrations to calculate Q Ksp = 4.93 x 10-5 3. Compare Q to Ksp. If Q > Ksp, than precipitate forms. If Q < Ksp, then no precipitate forms.