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Question

Chemistry Question on Equilibrium

Find out the solubility of Ni(OH)2Ni(OH)_2 in 0.1 M NaOH. Given that the ionic product of Ni(OH)2Ni(OH)_2 is 2×10−152 \times 10^{-15}

A

2×10−13M2 \times 10^{-13} M

B

2×10−8M2 \times 10^{-8} M

C

1×10−13M1 \times 10^{-13} M

D

1×108M1 \times 10^{8} M

Answer

2×10−13M2 \times 10^{-13} M

Explanation

Solution

Ni(OH)2<=>Ni2+(S)+2OH−(2S)Ni(OH)_2 {<=>} \underset{(S)}{Ni^{2+}} + \underset{(2S)}{2OH^-} S=S = molar conc. of Ni(OH)2Ni(OH)_2

NaOH−>Na++OH−(0.1 MNaOH {->} Na^+ +\underset{(0.1\,M}{OH^-}
KSP=[Ni2+][OH−]2K_{SP} = [Ni^{2+}][OH^-]^2
=(S)(2S+0.1)2= (S)(2S + 0.1)^2
=(S)(4S2+0.01+0.4S)=(S)(4S^2 + 0.01 + 0.4S)
=4S3+0.01S+0.4S2= 4S^3 + 0.01S + 0.4 S^2
Neglecting higher power of SS,
2×10−15=0.01S2 \times 10^{-15} = 0.01S
S=2×10−13MS = 2 \times 10^{-13}M