Chemistry › Ionic Equilibria › Salt Hydrolysis
The rapid change of $\mathrm{pH}$ near the stoichiometric point of an acid base titration is the basis of…
The rapid change of $\mathrm{pH}$ near the stoichiometric point of an acid base titration is the basis of indicator detection, $\mathrm{pH}$ of the solution is related to ratio of the concentration of the conjugate acid (Hln) and base $\left(\mathrm{In}^{-}\right)$forms of the indicator by the expression:
$\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pK}_{\ln }-\mathrm{pH}$ $\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{pK}_{\mathrm{ln}}-\mathrm{pH}$ $\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{pH}-\mathrm{pK}_{\mathrm{In}}$ $\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\mathrm{In}}$ Solution
Let us consider the formation of a salt of a weak acid and a strong base.
$\begin{aligned}
\mathrm{In}^{-}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{HIn}+\mathrm{OH}^{-} \\
\mathrm{K}_h & =\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]}
\end{aligned}$
$\begin{aligned}
& \mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-} \\
& \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{ln}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]
\end{aligned}$
From (ii) and (iii),
$\begin{aligned}
& \frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}}=\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]} \\
& {\left[\mathrm{OH}^{-}\right] }=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}} \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \log \left[\mathrm{OH}^{-}\right]=\operatorname{logK}_{\mathrm{w}}-\log _{\mathrm{ln}}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}
\end{aligned}$
$\begin{aligned}
& -\mathrm{pOH}=-\mathrm{pK}_w+\mathrm{pK}_{\mathrm{ln}}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \mathrm{pK}_w-\mathrm{pOH}=\mathrm{pK}_{\ln }+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \text {or, } \quad \mathrm{pH}=\mathrm{pK}_{\mathrm{ln}}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\ln } \\
&
\end{aligned}$
Asked in: NEET 2004
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