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Q.

The solubility of a sparingly soluble salt ${AB}_{2}$ , in water is $1.0 x 10^{- 5} M$ . Its solubility product will be

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a

$4 X 10^{- 10} M^{3}$

b

$1.0 X 10^{- 10} M^{3}$

c

$1 X 10^{- 15} M^{3}$

d

$4 X 10^{- 15} M^{3}$

answer is B.

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Detailed Solution

$K_{sp} = [A^{2 +}] {[B^{-}]}^{2} = (1.0 \times 10^{- 5} M) {(2 \times 1.0 \times 10^{- 5} M)}^{2} = 4 \times 10^{- 15} M^{3}$

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