Asolution contains a weak monoprotic acid HA and

Question

its sodium salt NaA both at 0.1 M concentration.
Show that [OH�] Kw/Ka.

Answer 1

Start with an equation

HA <-> H+ + A-

at start
0.1.....0....0

at equilibrium

0.1-x...x....x+0.1

Ka=[H+][A-]/HA
=x.(0.1+x)/(0.1-x)

if x is small with respect 0.1 then this expression becomes

Ka=x.(0.1)/(0.1)
or Ka=x

Kw=[H+](OH-]
=x[OH-]

so

[OH-]=Kw/x

as x=Ka then

[OH-]=Kw/Ka