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Q.

Calculate the molar solubility of $Ni {(OH)}_{2}$ in 0.10M NaOH. The ionic product of $Ni {(OH)}_{2}$ is $2.0 \times 10^{15}$

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a

$S = 1 \times 10^{- 15} M$

b

$S = 4.0 \times 10^{- 13} M$

c

$S = 6 \times 10^{- 12} M$

d

$S = 2.0 \times 10^{- 13} M$

answer is A.

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Detailed Solution

$Ni {(OH)}_{2} ⇌ {Ni}^{+ 2} + 2 {OH}^{-}$

$K_{Sp} = [{Ni}^{+ 2}] {[{OH}^{-}]}^{2}$

$K_{SP} = S \times {(2 S + 0.1)}^{2} = S \times 10^{- 2} = 2 \times 10^{- 15}$

$S = 2 \times 10^{- 13}$

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Calculate the molar solubility of Ni(OH)2 in 0.10M NaOH. The ionic product of Ni(OH)2 is 2.0×1015