Chemistry - Ionic Equilibrium Question with Solution | TestHub

There are $2\mathrm{mmol}$ of ${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4$ and $4\mathrm{mmol}$ of $\mathrm{NaOH}$. Two equivalents of base convert the acid to $\mathrm{N}{\mathrm{a}}_2\mathrm{HP}{\mathrm{O}}_4$:

$${\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_4+2NaOH⟶\mathrm{N}{\mathrm{a}}_2\mathrm{HP}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}.$$

The predominant $\mathrm{HP}{\mathrm{O}}_4^{2-}$ ion is amphiprotic, giving the usual approximation

$$\mathrm{pH}\approx \frac{p{K}_{a2}+p{K}_{a3}}{2}=\frac{7.2+12.3}{2}=\mathrm{9.75.}$$