Question

Calculate the $O{H}^{-}$ concentration and the $H_{2}P{O}_4$ concentration of a solution prepared by dissolving $0.1$ mol of $N{a}_{3}P{O}_4$ in sufficient water to make $1$L of solution. $K_1=7.1\times {10}^{-3},K_2=6.3\times {10}^{-8},K_3=4.5\times {10}^{-13}$.

Answer 1

${Na}_3{PO}_4$ is the salt produced at the 3rd step of neutralisation of $H_3{PO}_4$. The salt undergoes hydrolysis.

${Na}_3{PO}_4\rightleftharpoons 3{Na}^{+}+{PO}_4^{3-}$

${PO}_4^{3-}+H_{2}O\rightleftharpoons {HPO}_4^{2-}+{OH}^{(-)}$

$K_b=\frac{K_w}{K_{{\alpha }_3}}=\frac{1\times {10}^{-14}}{4.5\times {10}^{-13}}=0.022$

$K_b=\frac{\left[{HPO}_4^{2-}\right]\left[{OH}^{(-)}\right]}{\left[{PO}_4^{3-}\right]}$

$\Rightarrow 0.022=\frac{x.x}{0.1-x}\Rightarrow x^2+0.022x-0.0022=0$

$\therefore$ $x=\frac{-0.022+\sqrt{{\left(0.022\right)}^2+4\times 1\times 0.0022}}{2}=0.037$

$\therefore$ Concentration of ${OH}^{(-)}=$ Concentration of ${HPO}_4^{2-}$

$=0.037M$