If the ionic product of `Ni(OH)_(2) is 1.9 xx 10^(-15)`, the molar solubility of `Ni(OH)_(2)`, in 1.0 M NAOH is

Options

A. `1.9 xx 10^(-18)`M
B. `1.9 xx 10^(-13)`M
C. `1.9 xx 10^(-15)`M
D. `1.9 xx 10^(-14)`M

Answer

C

Explanation

`Ni(OH)_(2) hArr Ni^(2+) + 2OH^(-)`
`K_(sp) = [Ni^(2+)][OH^(-)]^(2)`
`K_(sp) = a(a+c)^(2)(:' OH^(-)`ion is common for both NiOH and NaOH)
`K_(sp) = a(a^(2)+2ac+c^(2))`
Neglect higher orders of a
`K_(sp) = a.C^(2)`
`:. A = K_(sp)/C^(2) = 1.9 xx 10^(-15)/1^(2) = 1.9 xx 10^(-15)`