Question

Calculate the concentration of OH^– in a 0.250M solution of the sodium salt of a weak acid (pK_a = 6.00).

A−(aq)+H2​O(l) -> HA(aq)+OH−(aq)

Answer 1

use:

pKa = -log Ka

6.0 = -log Ka

Ka = 1*10^-6

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1*10^-6

Kb = 1*10^-8

A- dissociates as

A- + H2O -----> HA + OH-

0.25 0 0

0.25-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1*10^-8)*0.25) = 5*10^-5

since c is much greater than x, our assumption is correct

so, x = 5*10^-5 M

So,

[OH-] = x = 5*10^-5 M

Answer: 5.00*10^-5 M