CH160 General Chemistry II Lecture Presentation Applications of Acid-Base Equilibria Chapter 17 Sections 1-4 11/8/2018 Chapter 17

HA + H2O <=> H3O+ + A- Common Ion Effect Consider the ionization of weak acid HA: HA + H2O <=> H3O+ + A- What affect will adding salt NaA to the solution have on the acid ionization and solution pH? NaA  Na+ + A- 11/8/2018 Chapter 17

Common Ion Effect Consider the ionization of weak acid HA: HA + H2O <=> H3O+ + A- A- has 2 sources: HA and NaA. Adding NaA increases [A-]. 11/8/2018 Chapter 17

HA + H2O <=> H3O+ + A- Common Ion Effect Consider the ionization of weak acid HA: HA + H2O <=> H3O+ + A- A- is a “common ion” 11/8/2018 Chapter 17

Example 1 (1a on the Example Problem Handout) Calculate the percent ionization of the acid and the pH of the solution that contains 0.500M HC2H3O2 (Ka = 1.8 x 10-5) and 0.200 M NaC2H3O2. (ans.: pH = 4.34, 0.0090%) 11/8/2018 Chapter 17

Calculations Using Ka Basic Steps for Weak Acid Calculations Using Ka Write balanced chemical equation and the expression for Ka Look up value for Ka For each chemical species involved in the equilibrium (except H2O), write: Initial concentration Equilibrium concentration Let the change in the [H3O+] be the variable “x” Substitute the equilibrium concentrations into Ka and solve for x using either quadratic approach simplified approach Calculate pH, equilibrium concentrations, % ionization, etc., as specified in the problem. 11/8/2018 Chapter 17

(problem 1a) (problem 8a) + CH3COONa 0.5 M CH3COOH + 0.200 M CH3COONa pH = 4.3 (problem 1a) 0.5 M CH3COOH (problem 8a) pH = 2.5 + CH3COONa 0.6% 0.009% 11/8/2018 Chapter 17

HA + H2O <=> H3O+ + A- Common Ion Effect What affect does NaA have on weak acid HA ionization?: HA + H2O <=> H3O+ + A- Which way does equilibrium shift? What do the results of problems 8a and 1a tell us? Does this agree with LeChatelier’s principle? 11/8/2018 Chapter 17

Common Ion Effect What affect does NaA have on weak acid HA ionization?: HA + H2O <=> H3O+ + A- In presence of NaA, HA ionization shifts left. “common-ion effect” 11/8/2018 Chapter 17

Buffer Solutions What is a buffer solution? Requirements solution with ability to resist pH changes upon addition of small amounts of either acid or base Requirements must contain an acid to neutralize added OH- ions must contain a base to neutralize added H3O+ ions acidic and basic species in buffer must not neutralize each other. 11/8/2018 Chapter 17

Buffer Action How do buffers work? Consider buffer with weak acid HA and salt NaA: Addition of acid A- + H+  HA 11/8/2018 Chapter 17

Buffer Action How do buffers work? Consider buffer with weak acid HA and salt NaA: Addition of acid (small amount) A- + H+  HA [A-] decreases slightly [HA] increases slightly Added acid is neutralized. 11/8/2018 Chapter 17

Buffer Action How do buffers work? Consider buffer with weak acid HA and salt NaA: Addition of base HA + OH-  H2O + A- 11/8/2018 Chapter 17

Buffer Action How do buffers work? Consider buffer with weak acid HA and salt NaA: Addition of base HA + OH-  H2O + A- [HA] decreases slightly [A-] increases slightly Added base is neutralized. 11/8/2018 Chapter 17

Buffer Action Two important properties of buffer solutions: Buffer capacity Amount of acid or base the buffer can react with before giving a significant pH change (1 pH unit) Determined by how much buffer acid and base are used to make buffer pH Determined by Ka and relative amounts of buffer acid and base present 11/8/2018 Chapter 17

Calculation of Buffer pH Calculating pH for buffer containing both weak acid HA and salt NaA. The major equilibrium is: HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA] 11/8/2018 Chapter 17

Calculation of Buffer pH Calculating pH for buffer containing both weak acid HA and salt NaA. HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA] A- has 2 sources: HA and NaA (This seems familiar! Didn’t we just do this?) 11/8/2018 Chapter 17

Calculation of Buffer pH Calculating pH for buffer containing both weak acid HA and salt NaA. HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA] Neither CHA nor CNaA change much since only a very small amount of HA ionizes. 11/8/2018 Chapter 17

Calculation of Buffer pH Calculating pH for buffer containing both weak acid HA and salt NaA. HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA] Since % ionization is small: CA-  [A-] and CHA  [HA] 11/8/2018 Chapter 17

Calculation of Buffer pH Calculating pH for buffer containing both weak acid HA and salt NaA. HA + H2O <=> H3O+ + A- Ka = [H3O+][A-]/[HA] Since % ionization is small: CA-  [A-] and CHA  [HA] Solving for [H3O+] gives: [H3O+] = KaCHA/CA- pH = -log[H3O+] (This is just a common ion effect problem.) 11/8/2018 Chapter 17

Calculation of Buffer pH We can also take the –log of our equation: [H3O+] = KaCHA/CA- -log[H3O+] = -log(KaCHA/CA-) -log[H3O+] = -logKa - log CHA/CA- pH = pKa - log CHA/CA- 11/8/2018 Chapter 17

Example 2 (2a on Example Problem Handout) Calculate the pH of a buffer solution that contains 0.25 M sodium acetate, NaC2H3O2, and 0.35 M acetic acid, HC2H3O2 (Ka = 1.8 x 10-5). (ans.: pH = 4.60) 11/8/2018 Chapter 17

Calculating pH Changes in Buffers How do we calculate buffer pH after adding acid (H3O+) or base (OH-)? HA/NaA + H3O+ or OH- H3O+ + A-  HA + H2O or OH- + HA  A- + H2O Buffer Neutralization Rxn Calculate new [H3O+] from [H3O+] = KaCHA/CA- Calculate new [HA] & [A-] pH 11/8/2018 Chapter 17

Calculating pH Changes in Buffers How do we calculate buffer pH after adding acid (H3O+) or base (OH-)? HA/NaA + H3O+ or OH- H3O+ + A-  HA + H2O or OH- + HA  A- + H2O Buffer Neutralization Rxn This much is a stoichiometry problem. (Oh, oh! General Chemistry I stuff here.). Calculate new [HA] & [A-] 11/8/2018 Chapter 17

Calculating pH Changes in Buffers How do we calculate buffer pH after adding acid (H3O+) or base (OH-)? This part is an equilibrium calculation. Calculate new [H3O+] from [H3O+] = KaCHA/CA- pH 11/8/2018 Chapter 17

Example 3 (3a on Example Problem Handout) Calculate the pH of the solution formed and the change in pH observed when (a) 0.050 moles of HCl and (b) 0.025 moles of NaOH are added to 500 mL of the buffer in example (2a). (c) Calculate the change in pH that occurs when 0.050 moles HCl are added to 500 mL H2O. (ans.: (a) pH = 4.27, pH = -0.33, (b) pH = 4.74, pH = +0.14, (c) pH = -6) 11/8/2018 Chapter 17

Buffer Preparation What if I need to make a buffer solution of known pH? Select: Buffer system. Often pKa of buffer acid is close to desired pH. Relative amounts of buffer acid and base. Buffer capacity increases with concentrations. Buffer effectiveness best with concentrations on same order of magnitude. 11/8/2018 Chapter 17

Example 4 (4 on Example Problem Handout) Starting with 1.0L of 0.100 M CH3COOH (Ka = 1.8 x 10-5), how many grams of sodium acetate, CH3COONa (FW = 82.034 g/mol), to give a buffer with a pH of 4.40? (Assume no volume change.) (ans.: 3.7 g) 11/8/2018 Chapter 17

Quantitative Acid-Base Chemistry How do we calculate the pH of a solution formed by mixing an acid solution with a base solution? Consider addition of 0.1 M strong base, MOH, to 0.1 M strong acid, HX pH changes can be observed from titration curve pH vs. volume standard 11/8/2018 Chapter 17

Strong Acid-Strong Base Titration HX + NaOH  NaX + H2O pH Rxn is complete = equivalence point. pH = 7.0 since only neutral NaX present. mL NaOH 11/8/2018 Chapter 17

Calculating pH in Acid-Base Reactions How do we calculate pH after adding a strong base to a strong acid? HX + MOH HX + MOH  MX + H2O Strong acid Neutralization Rxn Calculate new [H3O+] from [HX] or [MOH] Calculate [HX] or [MOH] left (ignore neutral MX) pH 11/8/2018 Chapter 17

Calculating pH in Acid-Base Reactions How do we calculate pH after adding a strong base to a strong acid? Considerations: Stoichiometry Limiting reagent At what point in rxn does calculation take place? Initial Pre-equivalence Equivalence Post-equivalence Does dilution occur? If mixing 2 solutions: Vtotal = V1 + V2 11/8/2018 Chapter 17

Strong Acid-Strong Base Titration HX + NaOH  NaX + H2O -NaOH in excess -NaOH/NaX left -[OH-] = CMOH -All HX & MOH consumed -NaX left (neutral) -pH = 7.0 pH -HX in excess -HX/NaX left -[H3O+] = CHX Only HX [H3O+] = CHX mL NaOH 11/8/2018 Chapter 17

Example 5 (5 of Example Problem Handout) Calculate the pH for a solution prepared by mixing 25.00 mL of 0.100 M HCl(aq) with a) 10.00 mL b) 25.00 mL and c) 35.00 mL of 0.100 M NaOH(aq). (ans.: a) 1.37, b) 7.0, c) 12.22) 11/8/2018 Chapter 17

Weak Acid-Strong Base Titration HA + NaOH  NaA + H2O Rxn is complete = equivalence point. pH > 7.0 since only basic NaA present. pH 11/8/2018 Chapter 17 mL NaOH

Weak Acid-Strong Base Titration HA + NaOH  NaA + H2O -NaOH in excess -NaOH/NaA left -[OH-]  CMOH -HA in excess -HA/NaA left = buffer -[H3O+] = Ka(CHA/CNaA) -All HA & MOH consumed -NaA left (weak base) -[OH-]  (KbCNaA)1/2 -pH > 7.0 pH -only HA [H3O+]  (KaCHA)1/2 11/8/2018 Chapter 17

Strong Acid-Weak Base Titration HX + B  HB+ + X- pH Rxn is complete = equivalence point. pH < 7.0 since only acidic HB+ present. 11/8/2018 Chapter 17