Question

a) A solution contains 0.34 M of a weak acid HA (K_a = 2.0 x 10^-7) and 0.17 M NaA. What is the pH after 0.05 M of HCl is added to this solution (assume no volume change)

b) The pH of a solution containing 0.1 M of a weak acid HA is 6. Calculate K_a for this acid.

Note the acids are unrelated for the two problem parts.

Answer 1

a)

after 0.05 M HCl added

[HA] = 0.34 + 0.05 = 0.39 M

[NaA] = 0.17 - 0.05 = 0.12 M

pH = pKa + log [NaA] / [HA]

pKa = - log Ka = - log 2.0 x 10^-7 = 6.70

pH = 6.70 + log [0.12]/[0.39]

pH = 6.20

b)

HA <---------> H⁺ + A^-

initially

[HA] = 0.1 M

pH = 6.0

[H⁺] = 10^-pH = 10^-6.0 = 1.0 x 10^-6 M

at equilibrium

[H⁺] = [A^-] = 1.0 x 10^-6 M

[HA] = 0.10 - 0.000001 = 0.099999

Ka = [H⁺][A^-]/[HA]

Ka = [1.0 x 10^-6][1.0 x 10^-6] / [0.099999]

Ka = 1.0 x 10^-11