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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Equilibrium
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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) $8 \times 10^{-15}$

Can you answer this question?

1 Answer

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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 $L^{-1}$


answered Nov 29, 2013 by mosymeow_1

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