Question

Part ADetermine the pH during the titration of 29.2 mL of 0.274 M perchloric acid by 0.352 M potassium hydroxide at the following points:

(1) Before the addition of any potassium hydroxide

(2) After the addition of 11.4 mL of potassium hydroxide

(3) At the equivalence point

(4) After adding 28.8 mL of potassium hydroxide

Part B Calculate the pH and the equilibrium concentrations of HS^- and S^2- in a 0.0590 M hydrosulfuric acid solution, H₂S (aq).

For H₂S, K_a1 = 1.00×10^-7 and K_a2 = 1.00×10^-19

pH =	?
[HS-] =	?M
[S2-] =	? M

Part C When a 27.3 mL sample of a 0.314 M aqueous hydrofluoric acid solution is titrated with a 0.351 M aqueous potassium hydroxide solution,

(1) What is the pH at the midpoint in the titration?

(2) What is the pH at the equivalence point of the titration?

(3) What is the pH after 36.6 mL of potassium hydroxide have been added?

Answer 1

Part A

a) Before the addition of any potassium hydroxide

pH = -log[H+]

[H+] = [HClO4] = 0.274 M (because HClO4 is strong acid)

pH = -log0.274

pH = 0.562

2) After the addition of 11.4 mL of potassium hydroxide

No of mol of HClO4 = M*V = 29.2*0.274 = 8.0 mmol

No of mol of KOH added = 11.4*0.352 = 4.0 mmol

concentration of excess HClO4 = (8-4)/(29.2+11.4)

                              = 0.099 M

pH = -log[H+]

pH = -log0.099

pH = 1

3) At the equivalence point

as HClO4 and KOH is strong acid,strong base at equivalence point the pH = 7 (neutral).

4) After adding 28.8 mL of potassium hydroxide

No of mol of HClO4 = M*V = 29.2*0.274 = 8.0 mmol

No of mol of KOH added = 28.8*0.352 = 10.14 mmol

concentration of excess KOH = (10.14-8)/(29.2+28.8)

                            = 0.037 M

pOH = -log[OH-]

   = -log0.037

   = 1.43

pH = 14-1.43 = 12.57