The rapid change of pH near the stoichiometric point of an acid base titration is the basis of indicator detection, pH of the solution is related to…

Let us consider the formation of a salt of a weak acid and a strong base. $$\begin{aligned} \mathrm{In}^{-}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{HIn}+\mathrm{OH}^{-} \\ \mathrm{K}_h & =\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]} \end{aligned}$$In−+H2​OKh​​⇌HIn+OH−=[In−][HIn][OH−]​​ $$\begin{aligned} & \mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-} \\ & \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{ln}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\ & \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \end{aligned}$$​HIn⇌H++In−H2​O⇌H++OH−Kln​=[HIn][H+][In−]​Kw​=[H+][OH−]​ From (ii) and (iii), $$\begin{aligned} & \frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}}=\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]} \\ & {\left[\mathrm{OH}^{-}\right] }=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}} \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\ & \log \left[\mathrm{OH}^{-}\right]=\operatorname{logK}_{\mathrm{w}}-\log _{\mathrm{ln}}+\log \\ & \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \end{aligned}$$​KIn​Kw​​=[In−][HIn][OH−]​[OH−]=KIn​Kw​​[HIn][In−]​log[OH−]=logKw​−logln​+log[HIn][In−]​​ $$\begin{aligned} & -\mathrm{pOH}=-\mathrm{pK}_w+\mathrm{pK}_{\mathrm{ln}}+\log \\ & \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\ & \mathrm{pK}_w-\mathrm{pOH}=\mathrm{pK}_{\ln }+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\ & \text { or, } \quad \mathrm{pH}=\mathrm{pK}_{\mathrm{ln}}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\ & \log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\ln } \\ & \end{aligned}$$​−pOH=−pKw​+pKln​+log[HIn][In−]​pKw​−pOH=pKln​+log[HIn][In−]​ or, pH=pKln​+log[HIn][In−]​log[HIn][In−]​=pH−pKln​​