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6-2.Equilibrium-II (Ionic Equilibrium)
medium
If ${K_{sp}}$ for $HgS{O_4}$ is $6.4 \times {10^{ - 5}},$ then solubility of the salt is
A
$8 \times {10^{ - 3}}$
B
$8 \times {10^{ - 6}}$
C
$6.4 \times {10^{ - 5}}$
D
$6.4 \times {10^{ - 3}}$
Solution
(a) $HgS{O_4}$ of ${K_{sp}} = {S^2}$
$S = \sqrt {{K_{sp}}} $ ; $S = \sqrt {6.4 \times {{10}^{ – 5}}} $; $S = 8 \times {10^{ – 3}}\,m/l$.
Standard 11
Chemistry