Acid/Base equilibrium

Question

A 0.828 M solution of the salt NaA has a pH of 8.67. Calculate the Ka value for the acid HA. Record your answer in scientific notation to 3 sig figs.

Answer 1

NaA dissociates into Na⁺ and A^- and then A^- is hydrolyzed in water as follows:

A^- + H₂O ==> HA + OH^- ... here A^- is acting as the conjugate base of HA

From pH = 8.67 we can find pOH and then [OH-]: pOH = 14-8.67=5.33 and [OH-] = 4.68x10^-6 M

Kb = [HA][OH]/[A-] = (4.6x10^-6)2/0.828 = 2.56x10^-11

KaKb = Kw = 1x10^-14

Ka(2.56x10^-11) = 1x10^-14

Ka = 1x10^-14/2.56x10^-11

Ka = 3.91x10^-4