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vladimir2022 [97]
3 years ago
8

Calculate the molar solubility of Ni(OH)2 in water. Use 2.0 * 10^-15 as the solubility product constant of Ni(OH)2.

Chemistry
2 answers:
finlep [7]3 years ago
8 0
Ni(OH)₂(s) ⇄ Ni²⁺(aq) + 2OH⁻(aq)

Ksp=2.0*10⁻¹⁵

Ksp=[Ni²⁺][OH⁻]²

c=[Ni²⁺]=[OH⁻]/2

Ksp=c×(2c)²=4c³

c=∛(Ksp/4)

c=∛(2.0×10⁻¹⁵/4)=0.01995 mol/L ≈ 0.02 mol/l
mrs_skeptik [129]3 years ago
5 0

Answer:

The molar solubility of Ni(OH)_2 in water 7.93\times10^{-6} mol/L.

Explanation:

Ni(OH)_2\rightleftharpoons Ni^{2+}+2OH^-

                        S        2S

Solubility of Nickel hydroxide =K_{sp}=2.0\times 10^{-15}

K_{sp}=S\times 2S^{2}=4S^3

2.0\times 10^{-15}=4S^3

S=7.93\times10^{-6} mol/L

The molar solubility of Ni(OH)_2 in water 7.93\times10^{-6} mol/L.

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