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2 October, 22:57

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

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  1. 2 October, 23:12
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    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
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