Dear Student,
we know ,
pH = − log10[ H3O+]
pH = 2.32
∴ 2.32 = − log10[ H3O+]
[ H3O+] = 10-2.32 = 0.004786 =4.786 * 10-3
pH + pOH = 14
pOH = 14 - pH == 14 - 2.32
∴ pOH = 11.68
now again,
pOH = - log10[OH-]
11.68 = -log10[OH-]
∴ [OH]- = 10-11.68 =2.1 *10-12
Concentration of [ H3O+] = 4.786 * 10-3 M and [OH]- = 2.1 *10-12 M