SATHEE: Equilibrium - Result Question 97

Equilibrium - Result Question 97

####101. Which of the following is most soluble? [1994]

(a) $B i_{2} S_{3} (K_{s p} = 1 \times 10^{- 17})$

(b) $M n S (K_{s p} = 7 \times 10^{- 16})$

(d) $A g_{2} S (K_{s p} = 6 \times 10^{- 51})$ .

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Solution:

(a) For $B i_{2} S_{3} \cdot K_{s p} = (2 s)^{2}$ . (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 $s = \sqrt{K_{s p}} = \sqrt{7^{'} 10^{- 16}}$

for $C u S s = \sqrt{K_{s p}} = \sqrt{8 \times 10^{- 37}}$

For $A g_{2} S K_{s p} = 2 s^{2} . s = 4 s^{3}$

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 $B i_{2} S_{3}$ will be in the order of $5^{th}$ root of $10^{- 17}$ , thus, it is most soluble.

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Equilibrium - Result Question 97

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