If solubility product of HgS{O₄} is 6.4 × {10^{ - 5}} , then its solubility is

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6-2.Equilibrium-II (Ionic Equilibrium)

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If solubility product of $HgS{O_4}$ is $6.4 \times {10^{ - 5}}$, then its solubility is

A

$8 \times {10^{ - 3}} \,mole/litre$

B

$6.4 \times {10^{ - 5}}\,mole/litre$

C

$6.4 \times {10^{ - 3}}\,mole/litre$

D

$2.8 \times {10^{ - 6}}\,mole/litre$

Solution

(a) $HgS{O_4}\; \to \;\mathop {H{g^{ + 2 }}}\limits_{x\,\,\,\,\,} + \mathop {SO_4^{ – – }}\limits_x $

${K_{sp}} = {x^2}$; $x = \sqrt {{K_{sp}}} $; $x = \sqrt {6.4 \times {{10}^{ – 5}}} $

$x = 8 \times {10^{ – 3}}$ $mole/litre.$

Standard 11

Chemistry

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