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# The value of $K^{sp}$ of $HgCl_2$ at room temperature is $4 \times 10^{-15}$. The concentration of $Cl^-$ ion in its aqueous solution at saturation point is

(a) $1 \times 10^{-5}$

(b) $2 \times 10^{-5}$

(c) $2 \times 10^{-15}$

(d) $8 \times 10^{-15}$

Answer: $1 \times 10^{-5}$ mol $L^{-1}$

$HgCl_2 \rightleftharpoons Hg^{2+} + 2Cl^-$

If solubility is X $mol L^-$ then $[Cl^-] = 2X$

Now, $4X^3 = K_{sp}$ = solubility product

$X = [\frac{K_{sp}}{4}]^{\frac{1}{3}} = [\frac{4 \times 10^{-15}}{4}]^{\frac{1}{3}} = 1 \times 10^{-5}$ mol $L^{-1}$
or $2X = 2 \times 10^{-5}$ mol $L^{-1}$