Question

A saturated solution of silver benzoate has a $pH$ of $8.63$. $K_a$ for benzoic acid is $6.5\times {10}^{-5}$. $K_{sp}$ of silver benzoate will be

• A. $1.4\times {10}^{-2}$ ${mol}^2$ ${litre}^{-2}$
• B. $1.9\times {10}^{-2}$ ${mol}^2$ ${litre}^{-2}$
• C. $2.4\times {10}^{-2}$ ${mol}^2$ ${litre}^{-2}$
• D. none of these

Answer 1

Answer: (A)

$1.4\times {10}^{-2}$ ${mol}^2$ ${litre}^{-2}$

We know that,
$pH+pOH=14$
$pOH=14-pH$=$14-8.63$=$5.37$
$\left[O{H}^{-}\right]=Antilog(-pOH)$=$Antilog(-5.37)$=$4.27\times {10}^{-6}mol/d{m}^3$
SIlver Benzoate is a Salt of Weak Acid and Strong Base.
Hence,
$K_h$= $\frac{K_w}{K_a}$ = $\frac{{10}^{-14}}{6.5\ast {10}^{-5}}$ = $1.54\ast {10}^{-10}$
$K_h$=$h^{2}C$
So, $h$=$\sqrt{\frac{K_h}{C}}$
Also,
$\left[O{H}^{-}\right]$=$nhC$=$n\sqrt{\frac{K_h}{C}}C$=$n\sqrt{K_{h}C}$
Hence,
C=$\frac{\left(\right[O{H}^{-}]{)}^2}{K_h}$=$\frac{(4.27\ast {10}^{-6}{)}^2}{(1.54\ast {10}^{-10})}$=$0.1178$
For Silver Benzoate,
$K_{sp}$=$(x\times C{)}^x\times (y\times C{)}^y$=$C^2$=${0.1178}^2$=$1.38\times {10}^{-2}mo{l}^{2}litr{e}^{-2}$
(Since, Silver benzoate is an $AB$ type of Salt, $x=y=1$)