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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
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Verified Answer
The correct answer is:
$K_1 \times K_2 \times K_3$
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 $K, K=K_1 \times K_2 \times K_3$
$$
K=K_1 \times K_2 \times K_3
$$
$\mathrm{H}_3 \mathrm{PO}_4 \rightleftharpoons \mathrm{H}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
$\mathrm{H}_2 \mathrm{PO}_4^{-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{HPO}_4^{2-}$
$\mathrm{HPO}_4^{2-} \rightleftharpoons \mathrm{H}^{+}+\mathrm{PO}_4^{3-}$
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 $K, K=K_1 \times K_2 \times K_3$
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