The nickel(II) ion is commonly dissolved in solution using nickel(II) nitrate hexahydrate, Ni(NO3)2.6H2O. The nickel(II) ion precipitates as Ni(OH)2(s) with a Ksp value of 6 x 10-16. Calculate the molar solubility of nickel(II) in a solution with a pH of 10.00

Answer :

Ni(OH)₂ ⇄ Ni⁺² + 2 OH⁻
Ksp = [Ni⁺²][OH⁻]²  = S (2S)² = 4S³
where S is molar solubility.
at pH = 10 
[H⁺] = 10⁻¹⁰
[H⁺][OH⁻] = 10⁻¹⁴ 
so [OH⁻] = 10⁻⁴ M
Ksp = S [10⁻⁴ + 2S]²
Ksp is very small so the molar solubility of OH⁻ will be very small
so (10⁻⁴ + 2S) is about 10⁻⁴
so Ksp = S x 10⁻⁸
S = [tex] \frac{6 x 10^{-16} }{10^{-8} } [/tex] = 6 x 10⁻⁸ M

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