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
Asolution contains a weak monoprotic acid HA and
its sodium salt NaA both at 0.1 M concentration.
Show that [OH�] Kw/Ka.
1 answer