A weak tribasic acid, `H_(3)A`, in its aquous solution ionises in the following step:
If the ionisation constants of the steps are `K_(1),K_(2) and K_(3)` respectively, then determine the overall ionisation constant of `H_(3)A`.
`H_(2)A^(-)(aq)+H_(2)O(l)hArrH_(3)O^(+)(aq)+HA^(2-)(aq).`

Explanation

Step 1: `K_(1)=([H_(3)O^(+)][H_(2)A^(-)])/([H_(3)A])`
Step II: `K_(2)=([H_(3)O^(+)][HA^(2-)])/([H_(2)A^(-)])` step III: `K_(3)=([H_(3)O^(+)][A^(3-)])/([HA^(2-)])`
The equation for the overall ionisation of `H_(3)A` is
`H_(3)A(aq)+3H_(2)O(l)hArr A^(3-)(aq)+3H_(3)O^(+)(aq)`
If the overall ionisation constant of `H_(3)A` is K, then
`K=([A^(3-)][H_(3)O^(+)]^(3))/([H_(3)A])`
Multiplying ionisation constants of step (I), (II) and (III),
we have, `K_(1)xxK_(2) xxK_(3)=([A^(3-)][H_(3)O^(+)]^(3))/([H_(3)A])`
Therefore, `K=K_(1)xxK_(2)xxK_(3)`.