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The solubility (in mol $\mathrm{L}^{-1}$ ) of $\mathrm{AgCl}$ $\left(K_{\mathrm{sp}}=1.0 \times 10^{-10}\right)$ in a $0.1 \mathrm{M} \mathrm{KCl}$ solution will be
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$1.0 \times 10^{-9}$
$1.0 \times 10^{-9}$
Let solubility of $\mathrm{AgCl}=x \mathrm{~mole} / \mathrm{L}$
$$
\begin{aligned}
& \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\
& \text {i.e., } K_{\mathrm{sp}(\mathrm{AgCl})}=x \times x \\
& \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \\
& {\left[\mathrm{Cl}^{-}\right] \text {from } \mathrm{KCl}=0.1 \mathrm{~m}} \\
& \text { Total }\left[\mathrm{Cl}{ }^{-}\right] \text {in solution }=x+0.1 \\
& K_{s p}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=x(x+0.1) \\
& 1.0 \times 10^{-10}=x(x+0.1) \\
& 1.0 \times 10^{-10}=x^2+0.1 x \\
& 1.0 \times 10^{-10}=0.1 x \quad\left(\text { as } x^2< < 1\right) \\
& x=1.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{AgCl} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\
& \text {i.e., } K_{\mathrm{sp}(\mathrm{AgCl})}=x \times x \\
& \mathrm{KCl} \longrightarrow \mathrm{K}^{+}+\mathrm{Cl}^{-} \\
& {\left[\mathrm{Cl}^{-}\right] \text {from } \mathrm{KCl}=0.1 \mathrm{~m}} \\
& \text { Total }\left[\mathrm{Cl}{ }^{-}\right] \text {in solution }=x+0.1 \\
& K_{s p}(\mathrm{AgCl})=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=x(x+0.1) \\
& 1.0 \times 10^{-10}=x(x+0.1) \\
& 1.0 \times 10^{-10}=x^2+0.1 x \\
& 1.0 \times 10^{-10}=0.1 x \quad\left(\text { as } x^2< < 1\right) \\
& x=1.0 \times 10^{-9} \mathrm{~mol} / \mathrm{L}
\end{aligned}
$$
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