Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.130 M LiOH(aq), with 0.130 M HCl(aq).

a) before addition of any HCl

(b) after addition of 13.5 mL of HCl

c) after addition of 21.5 mL of HCl

d) after the addition of 35.0 mL of HCl

e) after the addition of 42.5 mL of HCl

(f) after the addition of 50.0 mL of HCl

Answer 1

1) initially only LiOH present

[OH^-] = [LiOH] = 0.130 M

pOH = - log [OH^-] = - log [0.130]

pOH = 0.89

pH = 14 - 0.89

pH = 13.11

b) millimoles of LiOH = 35 x 0.13 = 4.55

millimoles of HCl added = 13.5 x 0.13 = 1.755

4.55 - 1.755 = 2.795 millimoles LiOH left

[LiOH] = [OH^-] = 2.795 / 35 + 13.5 = 0.0576 M

pOH = - log [0.0576]

pOH = 1.24

pH = 14 - 1.24

pH = 12.76

c) millimoles of HCl added = 21.5 x 0.13 = 2.795

4.55 - 2.795 = 1.755 millimoles LiOH left

[LiOH] = [OH^-] = 1.755 / 35 + 21.5 = 0.0312

pOH = - log [0.0312]

pOH = 1.50

pH = 14 - 1.50

pH = 12.5

d) millimoles HCl added = 35 x 0.13 = 4.55

all LiOH becomes salt

both are strong acid and strong base so at equivalence point

pH = 7.0

e) millimoles of HCl added = 42.5 x 0.13 = 5.525

5.525 - 4.55 = 0.975 millimoles HCl left

[HCl] = [H⁺] = 0.975 / 35 + 42.5 = 0.0126 M

pH = - log [H⁺]

pH = - log [0.0126]

pH = 1.90

f) millimoles of HCl added = 50 x 0.13 = 6.5

6.5 - 4.55 = 1.95 millimoles HCl left

[HCl] = [H⁺] = 1.95 / 35 + 50 = 0.023 M

pH = - log [0.023]

pH = 1.64