SATHEE: Equilibrium - Result Question 92

Equilibrium - Result Question 92

####96. Solubility of a $M_{2} S$ salt is $3.5 \times 10^{- 6}$ , then its solubility product will be

[2001]

(a) $1.7 \times 10^{- 16}$

(b) $1.7 \times 10^{- 6}$

(c) $1.7 \times 10^{- 18}$

(d) $1.7 \times 10^{- 12}$

Show Answer

Solution:

(a) $M_{2} S \to \underset{2 s}{2 M^{+}} + \underset{s}{S^{-}}$

Solubility product $= (2 s)^{2} (s) = 4 s^{3}$

$= 4 (3.5 \times 10^{- 6})^{3} = 1.7 \times 10^{- 16}$

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