PART A. 10 M Acetic Acid (5. 00 mL of 1. 0 M Acetic Acid diluted to 50 PART A 0.10 M Acetic Acid (5.00 mL of 1.0 M Acetic Acid diluted to 50.0 mL) EX6-1 (of 14)

1 Theoretical pH of 0.10 M Acetic Acid HC2H3O2 (aq) + H2O (l) ⇆ H3O+ (aq) + C2H3O2- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 - x + x + x 0.10 - x x x Ka = [H3O+][C2H3O2-] ____________________ [HC2H3O2] 1.8 x 10-5 = x2 ___________ (0.10 – x) 1.34 x 10-3 = x 2.87 = pH EX6-2 (of 14)

2 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl) 1.0 mol HC2H3O2 x 0.00500 L _____________________ L = 0.00500 mol HC2H3O2 1.0 mol HCl x 0.00100 L _______________ L = 0.00100 mol HCl = 0.00100 mol H3O+ The strong acid does not have a base to react with The pH comes from the molarity of unreacted strong acid in the solution 0.00100 mol H3O+ _______________________ 0.05100 L = 0.0196 M H3O+ pH = -log(0.0196 M) = 1.71 EX6-3 (of 14)

the weak acid neutralized the weak conjugate base 3 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH) 1.0 mol HC2H3O2 x 0.00500 L _____________________ L = 0.00500 mol HC2H3O2 1.0 mol NaOH x 0.00100 L __________________ L = 0.00100 mol NaOH = 0.00100 M OH- Strong bases react completely with acids OH- (aq) + HC2H3O2 (aq) → H2O (l) + C2H3O2- (aq) the weak acid neutralized by the strong base the weak conjugate base Is produced EX6-4 (of 14)

3 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH) OH- (aq) + HC2H3O2 (aq) → H2O (l) + C2H3O2- (aq) Initial mol’s Change in mol’s Final mol’s 0.00100 0.00500 – 0.00100 – 0.00100 + 0.00100 0.00400 0.00100 pH = pKa + log nC2H3O2- ___________ nHC2H3O2 = 4.745 + log (0.00100 mol) _________________ (0.00400 mol) = 4.14 EX6-5 (of 14)

PART B. 1 M Sodium Acetate (5. 00 mL of 1 PART B 0.1 M Sodium Acetate (5.00 mL of 1.0 M Sodium Acetate Diluted to 50.0 mL) EX6-6 (of 14)

4 Theoretical pH of 0.10 M Sodium Acetate 0.10 M NaC2H3O2  0.10 M Na+ and 0.10 M C2H3O2- weak base KaKb = Kw Kb = Kw ____ Ka = 1.00 x 10-14 ______________ 1.8 x 10-5 = 5.56 x 10-10 EX6-7 (of 14)

4 Theoretical pH of 0.10 M Sodium Acetate C2H3O2- (aq) + H2O (l) ⇆ HC2H3O2 (aq) + OH- (aq) Initial M’s Change in M’s Equilibrium M’s 0.10 - x + x + x 0.10 - x x x Kb = [HC2H3O2] [OH-] ____________________ [C2H3O2-] 5.56 x 10-10 = x2 ___________ (0.10 – x) 7.46 x 10-6 = x 5.13 = pOH 8.87 = pH EX6-7 (of 14)

the weak conjugate base neutralized by the strong acid 5 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl) 1.0 mol C2H3O2- x 0.00500 L ____________________ L = 0.00500 mol C2H3O2- 1.0 mol HCl x 0.00100 L _______________ L = 0.00100 mol HCl = 0.00100 mol H3O+ Strong acids react completely with bases H3O+ (aq) + C2H3O2 - (aq) → H2O (l) + HC2H3O2 (aq) the weak conjugate base neutralized by the strong acid the weak acid is produced EX6-8 (of 14)

5 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl) H3O+ (aq) + C2H3O2 - (aq) → H2O (l) + HC2H3O2 (aq) Initial mol’s Change in mol’s Final mol’s 0.00100 0.00500 – 0.00100 – 0.00100 + 0.00100 0.00400 0.00100 pH = pKa + log nC2H3O2- ___________ nHC2H3O2 = 4.745 + log (0.00400 mol) __________________ (0.00100 mol) = 5.35 EX6-9 (of 14)

6 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH) 1.0 mol C2H3O2- x 0.00500 L ____________________ L = 0.00500 mol C2H3O2- 1.0 mol NaOH x 0.00100 L __________________ L = 0.00100 mol NaOH = 0.00100 mol OH- The strong base does not have an acid to react with The pH comes from the amount of unreacted strong base in the solution pOH = -log(0.0196 M) = 1.71 pH = 12.29 EX6-10 (of 14)

PART C 0.1 M Acetic Acid / 0.1 M Sodium Acetate (5.00 mL 1.0 M Acetic Acid and 5.00 mL 1.0 M Sodium Acetate diluted to 50.0 mL) EX6-11 (of 14)

7 Theoretical pH of 0.10 M Acetic Acid / 0.10 M Sodium Acetate pH = pKa + log [C2H3O2-] _____________ [HC2H3O2] = 4.745 + log (0.10 M) ___________ (0.10 M) = 4.75 EX6-12 (of 14)

8 Theoretical pH with the Addition of HCl (1.00 mL of 1.0 M HCl) 9 Theoretical pH with the Addition of NaOH (1.00 mL of 1.0 M NaOH) EX6-13 (of 14)

EX6-14 (of 14)