${\mathrm{H}}_3\mathrm{A}$ is a weak triprotic acid $({\mathrm{K}}_{{\mathrm{a}}_1}={10}^{-5},{\mathrm{K}}_{{\mathrm{a}}_2}={10}^{-9},{\mathrm{K}}_{\mathrm{a}}={10}^{-13})$. What is the value of $\mathrm{pX}$ of $0.1\mathrm{M}{\mathrm{H}}_3\mathrm{A}$ (aq)solution? Where  $\mathrm{pX}=-\log \mathrm{X} \& \mathrm{X}=\frac{\left[{\mathrm{A}}^{-3}\right]}{\left[{\mathrm{HA}}^{-2}\right]}$

First we will calculate $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ from ${\mathrm{K}}_{{\mathrm{a}}_1}$

${\mathrm{K}}_{{\mathrm{a}}_1}=\frac{\left[{\mathrm{H}}_2{\mathrm{A}}^{-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{H}}_3\mathrm{A}\right]}$

${10}^{-5}=\frac{\mathrm{Y}\times \mathrm{Y}}{0.1-\mathrm{Y}}$

Since ${\mathrm{K}}_{{\mathrm{a}}_1}$ is small, we can approximate 0.1- Y to 0.1.

${10}^{-5}=\frac{\mathrm{Y}\times \mathrm{Y}}{0.1}$

${10}^{-5}\times 0.1={\mathrm{Y}}^2$

$\mathrm{Y}={10}^{-3}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

Now from ${\mathrm{K}}_{{\mathrm{a}}_3}$, we will calculate X

$\mathrm{X}=\frac{\left[{\mathrm{A}}^{3-}\right]}{\left[{\mathrm{HA}}^{2-}\right]}$

${\mathrm{K}}_{{\mathrm{a}}_3}=\frac{\left[{\mathrm{A}}^{3-}\right]\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}{\left[{\mathrm{HA}}^{2-}\right]}$

${\mathrm{K}}_{{\mathrm{a}}_3}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

${10}^{-13}=\mathrm{X}\times {10}^{-3}$

$\mathrm{X}={10}^{-10}$

$\mathrm{pX}=-\log \mathrm{X}=-\log {10}^{-10}=10$