The successive equilibrium constants for the stepwise dissociation of a tribasic acid are $K_1$, $K_2$ and $K_3$, respectively. The equilibrium constant for the overall dissociation is

• A

  $\left(K_1+K_2+K_3\right)$
• B

  $\sqrt[3]{\left(K_1+K_2+K_3\right)}$
• C

  $\left(K_1 \times K_2 \times K_3\right)^3$
• Correct answer
  D

  $K_1 \times K_2 \times K_3$

Solution

For a tribasic acid like $\mathrm{H}_3 \mathrm{PO}_4$ has three ionisation constant. The overall ionisation constants $(K)$ will be product of ionisation constants of three steps, i.e. $$ K=K_1 \times K_2 \times K_3 $$ $\mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
$\mathrm{H}_2 \mathrm{PO}_4^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_4^{2-}$
$\mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_4^{3-}$
Adding Eqs. (i), (ii) and (iii), we get $$ \begin{gathered} \mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons 3 \mathrm{H}^{+}+\mathrm{PO}_4^{3-} \\ K=\frac{\left[\mathrm{PO}_4^{3-}\right]\left[\mathrm{H}^{+}\right]^3}{\left[\mathrm{H}_3 \mathrm{PO}_4\right]} \end{gathered} $$ Multiply mathematical Eqs. (i), (ii) and (iii), we get $$ \begin{gathered} K_1 \times K_2 \times K_3=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]}{\left[\mathrm{H}_3 \mathrm{PO}_4\right]} \times \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HPO}_4^{2-}\right]}{\left[\mathrm{H}_2 \mathrm{PO}_4^{-}\right]} \\ \times \frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{PO}_4^{3-}\right]}{\left[\mathrm{HPO}_4^{2-}\right]} \\ K_1 K_2 K_3=\frac{\left[\mathrm{H}^{+}\right]^3\left[\mathrm{PO}_4^{3-}\right]}{\left[\mathrm{H}_3 \mathrm{PO}_4\right]} \end{gathered} $$ Overall ionisation constant $K, K=K_1 \times K_2 \times K_3$

Hence, the correct answer is option: 'D'.