The solubility of AgCl(s) with solubility product 1.6×10-10 in 0.1M of NaCl solution would be

A
$1.26 \times 10^{- 5} M$
B
$1.6 \times 10^{- 9} M$
C
$1.6 \times 10^{- 11} M$
D
Zero

Solution:

$Let s be the solubility of AgCl in moles per litre .$

${AgCl}_{(aq)} ⇌ {Ag}_{(aq)}^{+} + {Cl}_{(aq)}^{-}$

s s s+0.1

$(∵ 0.1 MNaCl solution also provides 0.1 M {Cl}^{-} ion)$

$\begin{array}{l} K_{sp} = [{Ag}^{+}] [{Cl}^{-}] \\ 1.6 \times 10^{- 10} = s (s + 0.1) \\ 1.6 \times 10^{- 10} = s (0.1) (∵ s < < < < 0.1) \\ s = \frac{1.6 \times 10^{- 10}}{0.1} = 1.6 \times 10^{- 9} M \end{array}$

The solubility of ${AgCl}_{(s)}$ with solubility product $1.6 \times 10^{- 10}$ in $0.1 M of NaCl$ solution would be

Solution:

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