Question

Calculate the pH of a $0.1M{K}_{3}P{O}_4$ solution at ${25}^{\circ }C$. The third dissociation constant $K_{a_3}$ of orthophosphoric acid is $1.3\times {10}^{-12}$. Assume that hydrolysis proceeds only in the first step.
Take $log\left(4.6\right)=0.66$

• A. 11.28
• B. 10.24
• C. 9.34
• D. 12.34

Answer 1

The correct option is D 12.34

$K_{3}P{O}_4+H_{2}O\rightleftharpoons K_{2}HP{O}_4+KOHInitial:0.100$
or
$P{O}_4^{3-}+H_{2}O\rightleftharpoons HP{O}_4^{2-}+O{H}^{-}Equilibrium:0.1(1-h)0.1h0.1h$
Since $K_h$ is determined by the dissociation constant of weak acid $HP{O}_4^{2-}$ i.e. the third dissociation constant of $H_{3}P{O}_4$
$K_h=\frac{K_w}{K_{a_3}}=\frac{{10}^{-14}}{1.3\times {10}^{-12}}=7.7\times {10}^{-3}$

Finding $\frac{C}{K_h}$

$\frac{C}{K_h}=\frac{0.1}{7.7\times {10}^{-3}}=12.98$
Since $\frac{C}{K_h}<100$ , we cannot take $1-h\approx 1$
$\frac{C{h}^2}{(1-h{)}^2}=K_h\frac{h^2}{(1-h{)}^2}=\frac{7.7\times {10}^{-3}}{0.1}=0.077h^2=0.077(h^2+1-2h)0.923h^2+0.154h-0.077=0$
$h=\frac{-0.154\pm \sqrt{(0.154{)}^2+(4\times 0.923\times 0.077)}}{2\times 0.923}h=0.217$

$\left[O{H}^{-}\right]=Ch=0.1\times 0.217=2.17\times {10}^{-2}\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{2.17\times {10}^{-2}}=4.6\times {10}^{-13}pH=-log\left[H^{+}\right]=-log(4.6\times {10}^{-13})pH=(13-0.66)=12.34$