Question

During the titration of 25.00 mL of 0.130 M weak base NH3 (Kb = 1.8 x 10^-5) with 0.100 M HCl, calculate the pH of the solution at the following volumes of acid added:

a) 10.00 mL

b) 32.50 mL

c) 50.00 mL

Answer 1

millimoles of NH3 = 25 x 0.13 = 3.25

pKb = - log Kb = - log [1.8 x 10^-5]

pKb = 4.74

a) millimoles of HCl added = 10 x 0.1 = 1.0

3.25 - 1.0 = 2.25 millimoles base left

1.0 millimoles salt formed

[salt] = 1.0 / 35 = 0.0286 M

[base] = 2.25 / 35 = 0.0643 M

pOH = pKb + log [salt] /[base]

pOH = 4.74 + log [0.0286] / [0.0643]

pOH = 4.39

b) millimoles of HCl added = 32.5 x 0.1 = 3.25

means equivalence point reached all base becomes salt

[salt] = 3.25 / 57.5 = 0.0565 M

pOH = 1/2 [pKw + pKb + log C]

pOH = 1/2 [14 + 4.74 + log 0.0565]

pOH = 8.75

pH = 14 - 8.75

pH = 5.25

c) millimoles of HCl added = 50 x 0.1 = 5.0

5.0 - 3.25 = 1.75 millimoles HCl left

[HCl] = 1.75 / 75 = 0.023 M

as HCl is strong acid

[HCl] = [H⁺]

[H⁺] = 0.023 M

pH = -log [H⁺]

pH = - log [0.023]

pH =1.64