Calculate the pH of a solution of 0.80 M of KC7H5O2 (Ka for HC7H5O2 = 6.4×10-5).

Answer 1

KC₇H₅O₂ is the potassium salt of HC₇H₅O₂ It is the salt formed from reaction of a weak acid (HC7H5O2) and a strong base (KOH). A solution of this salt should have a basic pH (>7).

Eliminating the spectator (K⁺), we have the conjugate base which is C₇H₅O₂^-. To find the pH of this solution, we must look at the hydrolysis of this species:

C₇H₅O₂^- + H₂O ==> HC₇H₅O₂ + OH^- (note that C₇H₅O₂^- is acting as a base)

Kb = [HC₇H₅O₂2][OH^-] / [C₇H₅O₂^-] and Kb = 1x10-¹⁴ / 6.4x10^-5 = 1.56x10^-10

1.56x10^-10 = (x)(x) / 0.80 - x (assume x is small relative to 0.80 M and ignore it in denominator)

1.56x10_-10 = x² / 0.80

x² = 1.25x10^-10

x = [OH-] = 1.12x10^-5 M (note: above assumption was valid)

pOH = -log 1.12x10^-5

pOH = 4.95

pH = 14 - pOH

pH = 9.05