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6-2.Equilibrium-II (Ionic Equilibrium)
hard
The solubility of $S{b_2}{S_3}$ in water is $1.0 \times {10^{ - 5}}$ $mol / litre$ at $298 \,K$. What will be its solubility product
A
$108 \times {10^{ - 25}}$
B
$1.0 \times {10^{ - 25}}$
C
$144 \times {10^{ - 25}}$
D
$126 \times {10^{ - 24}}$
Solution
(a) $S{b_2}{S_3}\; \to \;\mathop {2S{b^{ + 2}}}\limits_{2x\,} + \mathop {3{S^{2 – }}}\limits_{3x\,\,\,\,} $; ${K_{sp}} = {(2x)^2}.{(3x)^3}$
${K_{sp}} = 108{x^5}$ ; ${K_{sp}} = 108 \times {(1 \times {10^{ – 5}})^5}$ $ = 108 \times {10^{ – 25}}$.
Standard 11
Chemistry
