The solubility product of AgCl at 25^oC is 1.8X10^-10 . then solubility of AgCl in pure water

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$2.5 \times 10^{- 6} m o l / L$

$0.34 \times 10^{- 5} m o l / L$

$1.67 \times 10^{- 5} m o l / L$

$1.34 \times 10^{- 5} m o l / L$

answer is A.

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Detailed Solution

$\begin{array}{l} A g C l \to A g^{+} + C l^{-} \\ K_{s p} = S^{2} \Rightarrow 1.8 \times 10^{- 10} = S^{2} \Rightarrow S = 1.34 \times 10^{- 5} \end{array}$

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