If the solubility of $M_{3} N_{2}$ is $^{'} S^{'} g m o l L^{- 1}$ , its solubility product is:

2 S^{3}

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8 S^{4}

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108 S^{5}

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27 S^{3}

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Solution

The correct option is C $108 S^{5}$
$M_{3} N_{2} ⇌ 3 M^{2 +} + 2 N^{3 -}$
Let 'S' be solubility of $M_{3} N_{2}$ so
$K_{s p} = [3 S]^{3} [2 S]^{2} = 108 S^{5}$

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If the solubility of M3N2 is ′S′g mol L−1, its solubility product is:

The correct option is C 108S5M3N2⇌3M2++2N3−Let 'S' be solubility of M3N2 soKsp=[3S]3[2S]2=108S5

If the solubility of $M_{3} N_{2}$ is $^{'} S^{'} g m o l L^{- 1}$ , its solubility product is:

The correct option is C $108 S^{5}$
$M_{3} N_{2} ⇌ 3 M^{2 +} + 2 N^{3 -}$
Let 'S' be solubility of $M_{3} N_{2}$ so
$K_{s p} = [3 S]^{3} [2 S]^{2} = 108 S^{5}$