Chemistry - Ionic Equilibrium Question with Solution | TestHub

pH at I half equivalence point $={\mathrm{pK}}_{a_3}=12$

$$\begin{array}{rl}& \mathrm{pH}\text{ at I equivalence point }=\frac{{\mathrm{pK}}_{a_2}+{\mathrm{pK}}_{a_3}}{2}=10\\ & \Rightarrow {\mathrm{pK}}_{a_2}=8\\ & \mathrm{pH}\text{ at II equivalence point }=\frac{{\mathrm{pK}}_{a_1}+{\mathrm{pK}}_{a_2}}{2}=6\\ & \Rightarrow {\mathrm{pK}}_{a_1}=4\\ & \text{ at III equivalence point }\\ & \mathrm{N}{\mathrm{a}}_3\mathrm{P}{\mathrm{O}}_4+3\mathrm{HCl}\to 3\mathrm{NaCl}+{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4\\ & \begin{array}{l}100300\end{array}\\ & \times \\ & \mathrm{C}=\frac{100}{400}=\frac{1}{4}\mathrm{M}\\ & \mathrm{K}{\mathrm{a}}_1=10^{-4}=\frac{({\mathrm{H}}^{+}{)}^2}{\mathrm{C}}\Rightarrow ({\mathrm{H}}^{+})=\sqrt{\mathrm{k}{\mathrm{a}}_1\mathrm{C}}\\ & =\frac{10^{-2}}{2}\\ & =5\times 10^{-3}\\ & \mathrm{pH}=3-\log 5\end{array}$$