1. PART A M C V C = M D _______ V D = (1.0 M)(5.0 mL) ___________________ (50.0 mL) = 0.10 M HC 2 H 3 O 2

2. 1.34 x 10 -3 = x xx HC 2 H 3 O 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + C 2 H 3 O 2 - (aq) Initial M’s Change in M’s Equilibrium M’s 0.1000 - x+ x 0.10 - x K a = [H 3 O + ][C 2 H 3 O 2 - ] ____________________ [HC 2 H 3 O 2 ] 1.8 x 10 -5 = x 2 ___________ (0.10 – x) 2.87 = pH 1Theoretical pH of 0.10 M Acetic Acid

3. M C V C = M D _______ V D = (0.10 M)(50.0 mL) ______________________ (51.0 mL) = 0.0980 M HC 2 H 3 O 2 = (1.0 M)(1.0 mL) ___________________ (51.0 mL) = 0.0196 M HCl= 0.0196 M H 3 O + The strong acid does not have a base to react with The pH comes from the amount of unreacted strong acid in the solution 2Theoretical pH with the Addition of HCl pH = -log(0.0196 M)= 1.71

4. M C V C = M D _______ V D = (0.10 M)(50.0 mL) ______________________ (51.0 mL) = 0.0980 M HC 2 H 3 O 2 = (1.0 M)(1.0 mL) ___________________ (51.0 mL) = 0.0196 M NaOH= 0.0196 M OH - 3Theoretical pH with the Addition of NaOH Strong bases react completely with acids OH - (aq) + HC 2 H 3 O 2 (aq) → H 2 O (l) + C 2 H 3 O 2 - (aq) the weak acid neutralized by the strong base the weak conjugate base Is produced

5. OH - (aq) + HC 2 H 3 O 2 (aq) → H 2 O (l) + C 2 H 3 O 2 - (aq) Initial M’s Change in M’s Final M’s 0.01960.09800 0 0.0784 0.0196 – 0.0196 + 0.0196 3Theoretical pH with the Addition of NaOH pH = pK a + log [C 2 H 3 O 2 - ] ______________ [HC 2 H 3 O 2 ] = 4.745 + log (0.0196 M) ______________ (0.0784 M) = 4.14

6. PART B 2A-1 (of 24) M C V C = M D _______ V D = (1.0 M)(5.0 mL) ___________________ (50.0 mL) = 0.10 M NaC 2 H 3 O 2 = 0.10 M C 2 H 3 O 2 - weak base K b = K w ____ K a K a K b = K w = 5.56 x 10 -10 = 1.00 x 10 -14 _______________ 1.8 x 10 -5

7. 7.46 x 10 -6 = x xx C 2 H 3 O 2 - (aq) + H 2 O (l) ⇆ HC 2 H 3 O 2 (aq) + OH - (aq) Initial M’s Change in M’s Equilibrium M’s 0.1000 - x+ x 0.10 - x K b = [HC 2 H 3 O 2 ] [OH - ] _____________________ [C 2 H 3 O 2 - ] 5.56 x 10 -10 = x 2 ____________ (0.10 – x) 8.87 = pH 4Theoretical pH of 0.10 M Sodium Acetate 5.13 = pOH

8. M C V C = M D _______ V D = (0.10 M)(50.0 mL) ______________________ (51.0 mL) = 0.0980 M C 2 H 3 O 2 - = (1.0 M)(1.0 mL) ___________________ (51.0 mL) = 0.0196 M HCl= 0.0196 M H 3 O + 5Theoretical pH with the Addition of HCl Strong acids react completely with bases H 3 O + (aq) + C 2 H 3 O 2 - (aq) → H 2 O (l) + HC 2 H 3 O 2 (aq) the weak conjugate base neutralized by the strong acid the weak acid Is produced

9. H 3 O + (aq) + C 2 H 3 O 2 - (aq) → H 2 O (l) + HC 2 H 3 O 2 (aq) Initial M’s Change in M’s Final M’s 0.019600.0980 00.01960.0784 – 0.0196+ 0.0196– 0.0196 5Theoretical pH with the Addition of HCl pH = pK a + log [C 2 H 3 O 2 - ] ______________ [HC 2 H 3 O 2 ] = 4.745 + log (0.0784 M) ______________ (0.0196 M) = 5.35

10. M C V C = M D _______ V D = (0.10 M)(50.0 mL) ______________________ (51.0 mL) = 0.0980 M C 2 H 3 O 2 - = (1.0 M)(1.0 mL) ___________________ (51.0 mL) = 0.0196 M NaOH= 0.0196 M 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 6Theoretical pH with the Addition of NaOH pOH = -log(0.0196 M)= 1.71 pH = 12.29