Question

a 40.0 mL sample of 0.0600 M CH3COONa is titrated with 0.0600 M HClO4. Calculate the pH of the solution after the addition of the following volumes of acid. Ka of CH3COOH is 1.8 x 10^-5.

a) 0.0 mL acid added b) 20.0 mL c) 30.0 mL d) 38.0 mL e) 40.0 mL f) 45.0 mL g) 50.0 mL

Answer 1

a)

pKa = 4.74

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (4.74 + log 0.0600)

pH = 8.76

b)

mmoles of HClO4 = 20 x 0.06 = 1.2

mmoles of CH3COO- = 40 x 0.06 = 2.4

CH3COO- +   H+     --------------> CH3COOH

   2.4              1.2                                  0

   1.2                0                                  1.2

here pH = pKa

pH = 4.74

c)

mmoles of H+ = 30 x 0.06 = 1.8

CH3COO- +   H+     --------------> CH3COOH

   2.4              1.8                                  0

   0.6                0                                  1.8

pH = 4.74 + log (0.6 / 1.8)

pH = 4.26

d)

mmoles of H+ = 2.28

pH = 4.74 + log (0.12 / 2.28)

pH = 3.46

e)

mmoles of H+ = 2.4

pH = 1/2 (pKa - log C)

     = 1/2 (4.74 - log 0.03)

pH = 3.13

f)

pH = 2.45

g)

pH = 2.18