SATHEE: Equilibrium - Result Question 93

Equilibrium - Result Question 93

####97. The solubility product of a sparingly soluble salt $B A_{2}$ is $4 \times 10^{- 12}$ . The solubility of $B A_{2}$ is

[1999]

(a) $4 \times 10^{- 4}$

(b) $4 \times 10^{- 12}$

(d) $1 \times 10^{- 4}$

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Solution:

(d) $B A_{2} \to \underset{s}{B + 2 A}$

Solubility product $= [s] [2 s]^{2} = 4 s^{3}$

$4 \times 10^{- 12} = 4 s^{3}$ or $s = \sqrt[3]{\frac{4^{'} 10^{- 12}}{4}}$

$∴ s = 10^{- 4}$

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Equilibrium - Result Question 93

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