Ncert Exercises & Solutions

7.70 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Since pH = 3.19,

$[H_{3} O^{+}] = 6.46 \times 10^{- 4} M$
$C_{6} H_{5} COOH + H_{2} O \leftrightarrow C_{6} H_{5} {COO}^{-} + H_{3} O$
$K_{a} \frac{[C_{6} H_{5} {COO}^{-}] [H_{3} O^{+}]}{[C_{6} H_{5} COOH]}$
$\frac{[C_{6} H_{5} COOH]}{[C_{6} H_{5} {COO}^{-}]} = \frac{[H_{3} O^{+}]}{K_{a}} = \frac{6.46 \times 10^{- 4}}{6.46 \times 10^{- 5}} = 10$

$Let the solubility of C_{6} H_{5} COOAg be x mol / L . Then,$

$[{Ag}^{+}] = x$
$[C_{6} H_{5} {COO}^{-}] + [C_{6} H_{5} {COO}^{-}] = x$
$10 [C_{6} H_{5} {COO}^{-}] = \frac{x}{11}$
$K_{sp} [{Ag}^{+}] [C_{6} H_{5} {COO}^{-}]$
$2.5 \times 10^{- 13} = x (\frac{x}{11})$
$x = 1.66 \times 10^{- 6} mol / L$

$Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 \times 10^{- 6} mol / L .$
$Now, let the solubility of C_{6} H_{5} COOAg be x ’ mol / L .$
$Then, [{Ag}^{+}] = x' M and [{CH}_{3} {COO}^{-}] = x' M .$
$K_{sp} = [{Ag}^{+}] [{CH}_{3} {COO}^{-}]$
$K_{sp} = {(x')}^{2} x' = \sqrt{K_{sp}} = \sqrt{2.5 \times 10^{- 13}} = 5 \times 10^{- 7} mol / L$
$∴ \frac{x}{x'} = \frac{1.66 \times 10^{- 6}}{5 \times 10^{- 7}} = 3.32$
$Hence, C_{6} H_{5} COOAg is approximately 3.317 times more soluble in a low pH solution .$

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