Question

Given that Ka for the weak acid HA is 3.46 × 10–8, calculate K for the reaction of HA with OH–.

HA(aq) + OH−(aq) ↔ A−(aq) + H2O(l)

Answer 1

HA + H20 ---> H30+ + A-

Ka = [H30+] [A-] / [HA]

now

consider the auto ionization of water

H20 + H20 ---> H30+ + OH-

Kw = [H30+] [OH-]

so

[H30+] = Kw / [OH-]

substitute this in the expression for Ka

Ka = [Kw / [OH-] [A-] / [HA]

Ka = Kw x [A-] / [OH-] [HA]

[A-] / [HA] [OH-] = Ka / Kw

now

consider the given reaction

HA + OH- ---> A- + H20 (l)

the expression for K is given by

K = [A-] / [HA] [OH-]

so

K = Ka / Kw

given

Ka = 3.46 x 10-8

we know that

Kw = 10-14

so

K = 3.46 x 10-8 / 10-14

K = 3.46 x 10^6

so

the value of K is 3.46 x 10^6