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JEE · NEET Previous Year Question

$Solubility of a salt A_{2} B_{3} is 1 \times 10^{- 3} mol dm^{- 3} . What is the value of its solubility product?$

Topic: JEE PYQ

Answer: A. Full step-by-step solution below.

A $1.08 \times 10^{- 13}$
B $8.1 \times 10^{- 15}$
C $2.7 \times 10^{- 15}$
D $2.0 \times 10^{- 13}$

Correct answer: A

Solution

(a) : For A_{2} B_{3}, A_{2} B_{3} ⟶ 2 S 2 S A^{3 +} + 3 B^{2 -} Solubility product = (2 S)^{2} (3 S)^{3} = 108 S^{5} = 108 \times (10^{- 3})^{5} = 1.08 \times 10^{- 13}

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