Question

The $pH$ of $0.1M$ $K_3{PO}_4$ solution is:
The third dissociation constant of $H_3{PO}_4$ is $1.3\times {10}^{-12}$. Assume that the hydrolysis proceeds only in the first step.

• A. $pH=12.44$
• B. $pH=12.04$
• C. $pH=12.64$
• D. $pH=12.94$

Answer 1

The correct option is A $pH=12.44$

${PO}_4^{3+}+H_{2}O\rightleftharpoons HP{O}_4^{2-}+{OH}^{-}$

Since, hydrolysis proceeds only in I step

$\left[{OH}^{-}\right]=c\cdot h=c\sqrt{\frac{K_w}{K_{\left(HP{O}_4^{2-}\right)}\cdot c}}$

$=\sqrt{\frac{K_w\cdot c}{K_{\left(HP{O}_4^{2-}\right)}}}=\sqrt{\frac{K_w\cdot c}{K_{3\left(H_3{PO}_4\right)}}}$

$\because K_{\left(HP{O}_4^{2-}\right)}=K_{3}of{H}_3{PO}_4$

$HP{O}_4^{2-}\rightleftharpoons H^{+}+{PO}_4^{3+}$

$\therefore \left[{OH}^{-}\right]=\sqrt{\frac{{10}^{-14}\times 0.1}{1.3\times {10}^{-12}}}$

$\therefore pOH=1.56,\therefore pH=12.44$