Question

Calculate the pH of a 0.10 M solution of sodium phosphate.

Short answer

pH of the given solution comes out to be 10.66.

Step-by-step solution

 Calculation of Kafor Hydrolysis 

Species present are Na⁺, PO₄^3- and H₂O. Na+ does not undergo hydrolysis.

Reaction involved:

${\text{PO}}_{\text{4}}{{}^{\text{3-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}HPO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}$

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{O+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}H}}_{\text{3}}{{\text{O}}^{\text{+}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ {\text{\hspace{1em}K}}_{\text{w}}{\text{=1×10}}^{\text{-14}}\end{array}$

K_b can be calculated as

${\text{K}}_{\text{b}}=\frac{{\text{[K}}_{\text{w}}\text{]}}{{\text{[K}}_a\text{]}}=\frac{1\times {10}^{-14}}{4.8\times {10}^{-13}}=2.08\times {10}^{-2}$

Calculation of pH  forGiven Salt

$\begin{array}{l}{\text{PO}}_{\text{4}}{{}^{\text{3-}}}_{\text{(aq)}}{\text{+H}}_{\text{2}}\text{O}\rightleftarrows {\text{\hspace{0.17em}HPO}}_{\text{4}}{{}^{\text{2-}}}_{\text{(aq)}}{{\text{+\hspace{0.17em}OH}}^{\text{-}}}_{\text{(aq)}}\\ 0.100\\ 0.1-\text{x\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}x\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}x}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{b}}=\frac{\text{[x][x]}}{\text{[0.1-x]}}=2.08\times {10}^{-2}\\ \text{x}={\text{[OH}}^{-}\text{]}=4.56\times {10}^{-4}\end{array}$

${\text{[H}}^{\text{+}}]=\frac{{\text{K}}_{\text{w}}}{{\text{[OH}}^{\text{-}}\text{]}}=\frac{1\times {10}^{-14}}{4.56\times {10}^{-4}}=2.2\times {10}^{-11}$

${\text{pH=-log[H}}^{\text{+}}]=-\log (2.2\times {10}^{-11})=10.66$