Chemistry
rparker77
2016-04-09 04:53:21
Calculate the molar solubility of Ni(OH)2 in water. Use 2.0 * 10^-15 as the solubility product constant of Ni(OH)2.
ANSWERS
mdsalman
2016-04-09 11:48:29

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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