Calculate the pH of an aqueous solution of the salt NaA, which was made to a concentration of 0.080 M NaA. The Ka of the weak acid HA is 2.0 x 10-5.

Answer 1

Look at the hydrolysis of the conjugate base, A-.

A- + H₂O ==> HA + OH^-

Recall that KaKb = Kw and so we can find the Kb for A^- as follows:

Kb for A^- = 1x10^-14 / 2.0x10^-5 = 5x10^-10

5x10^-10 = [HA][OH^-] / [A^-] = (x)(x) / 0.080 - x (we can ignore the -x for now to simplify the calculations)

5x10-10 = x2/0.08

x² = 4x10^-10

x = 2x10^-5 M = [OH^-] (note: our assumption above was valid since this is very small compared to 0.08)

pOH = -log 2x10^-5 = 4.7

pH = 14 - pOH

pH = 9.3