Question

Which of the following salts is most soluble?

• A. $B i_{2} S_{3}\left(K_{s p}=1 \times 10^{-17}\right)$
• B. $M n S\left(K_{s p}=7 \times 10^{-16}\right)$
• C. $C u S\left(K_{s p}=8 \times 10^{-37}\right)$
• D. $A g_{2} S\left(K_{s p}=6 \times 10^{-51}\right)$

Answer 1

The Correct Option is A

$B i_{2} S_{3} 2 \underset{2 s}{2} i^{3+}+3 \underset{3 s}{ }^{2-}$ $K_{s p}=(2 s)^{2}(3 s)^{3}=108 s^{5}$ $108 s^{5}=1 \times 10^{-17}$ $s=\left(\frac{1 \times 10^{-17}}{108}\right)^{1 / 5}=1.56 \times 10^{-4}$ $M n S \underset{s}{M n^{2+}}+\underset{s}{S^{2-}}$ $K_{s p}=s^{2} s=\sqrt{K_{s p}}=\sqrt{7 \times 10^{-16}}$ $=2.64 \times 10^{-8} CuSC\underset{s}{u ^{2+}}+\underset{s}{S^{2-}}$ $K_{s p}=s^{2} 8 \times 10^{-37}=s^{2}$ $s=\sqrt{8 \times 10^{-37}}=0.89 \times 10^{-18}$ $A g_{2} S 2 \underset{2 s}{ Ag ^{+}}+\underset{s}{S^{2-}} K_{s p}=4 s^{3}$ $4 s^{3}=6 \times 10^{-51} s=\sqrt[3]{\frac{6 \times 10^{-51}}{4}} $ $=\sqrt[3]{1.5 \times 10^{-51}}=1.14 \times 10^{-17}$ The solubility of $Bi _{2} S _{3}$ is maximum. Hence, it is the most soluble.