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6-2.Equilibrium-II (Ionic Equilibrium)
medium
The solubility of $AgCl_{(s)}$ with solubility product $1.6 \times 10^{-10}$ in $0.1\, M\, NaCl$ 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.
(NEET-2016)
Solution
$\mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-}$
$\mathrm{Ag}^{+}$ is $\mathrm{S}$
$\mathrm{Cl}^{-}$ is $0.1$
$\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]$
$1.6 \times 10^{-10}=S \times 0.1$
$1.6 \times 10^{-9}=S$
Standard 11
Chemistry
