Chemistry - Ionic Equilibrium Question with Solution | TestHub

At the 2nd equivalence point (50 mL of $\mathrm{NaOH}$) the solution is essentially ${\mathrm{HA}}^{2-}$: $\left[{\mathrm{H}}^{+}\right]=\sqrt{K_2{K}_3}=\sqrt{10^{-19}}=3.2\times 10^{-10}\mathrm{M}$.

$\frac{\left[{\mathrm{H}}_3\mathrm{A}\right]}{\left[{\mathrm{A}}^{3-}\right]}=\frac{{\left[{\mathrm{H}}^{+}\right]}^3}{K_1{K}_2{K}_3}=\frac{3.2\times 10^{-29}}{2\times 10^{-22}}=1.6\times 10^{-7}\approx 10^{-7}$.

Answer: (A).