Equilibrium Ques 102
102. Which of the following is most soluble?
[1994]
(a) $Bi_2 S_3(K _{s p}=1 \times 10^{-17})$
(b) $MnS(K _{s p}=7 \times 10^{-16})$
(c) $CuS(K _{s p}=8 \times 10^{-37})$
(d) $Ag_2 S(K _{s p}=6 \times 10^{-51})$.
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Answer:
Correct Answer: 102.(a)
Solution:
- (a) For $Bi_2 S_3 \cdot K _{s p}=(2 s)^{2}$. (3s) $)^{3}$
$ =4 s^{2} \cdot 27 s^{3}=108 s^{5} $
or $s=\sqrt[5]{\frac{K _{s p}}{108}}=\sqrt[5]{\frac{1 \times 10^{-17}}{108}}$
For $MnS$. $K _{s p}=s^{2}$
or $s=\sqrt{K _{s p}}=\sqrt{7^{\prime} 10^{-16}}$
for $C u S \quad s=\sqrt{K _{s p}}=\sqrt{8 \times 10^{-37}}$
For $A g_2 S \quad K _{s p}=2 s^{2} . s=4 s^{3}$
or $s=\sqrt[3]{\frac{K _{s p}}{4}}=\sqrt[3]{\frac{6 \times 10^{-51}}{4}}$
thus $B i_2 S_3$ has maximum solubility.
The solubility of $Bi_2 S_3$ will be in the order of $5^{\text{th }}$ root of $10^{-17}$, thus, it is most soluble.
BETA
