The solubility of #"Ni(OH)"_2# is #1.52 × 10^"-5" color(white)(l)"mol/L"# or #"1.41 mg/L"#.
In #"0.100 mol/L NiSO"_4#, the solubility is #1.87 × 10^"-7" "mol/L" = "17.3 µg/L"#.
Solubility in water
I think your value of #K_"sp"# is in error. A better value would be #1.40 × 10^"-14"#.
The solubility equilibrium is
#color(white)(mmmmmm)"Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-"(aq)#; #K_text(sp) = 1.40 × 10^"-14"#
#"E/mol·L"^"-1":color(white)(mmmmmmmmm)xcolor(white)(mmmmm)2x#
Since #x color(white)(l)"mol of Ni(OH)"_2"(s)"# gives #xcolor(white)(l) "mol of Ni"^"2+"#, the solubility of #"Ni(OH)"_2# is #xcolor(white)(l) "mol/L"#.
The solubility constant expression is
#K_"sp" = ["Ni"^"2+"]["OH"^"-"]^2#
#["Ni"^"2+"]["OH"^"-"]^2 = x × (2x)^2 = 4x^3 = 1.40 × 10^"-14"#
#x^3 = (1.40 × 10^"-14")/4 = 3.50 × 10^"-15"#
#x = root(3)( 3.50 × 10^"-15") = 1.52 × 10^"-5"#
∴ The solubility of #"Ni(OH)"_2# is #1.52 × 10^"-5" color(white)(l)"mol/L"#
#M_r = 92.71#, so
#"Solubility" = (1.52 × 10^"-5" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.41 × 10^"-3" color(white)(l)"g/L" = "1.41 mg/L"#
The solubility listed here is 0.001 27 g/L (close enough!).
Solubility in 0.100 mol/L #"NiSO"_4#
The solubility of the #"Ni(OH)"_2# will be much less because of the common ion effect of the #"Ni"^"2+"# ion.
The solubility equilibrium is now
#color(white)(mmmmmm)"Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-"(aq)#; #K_text(sp) = 1.40 × 10^"-14"#
#"E/mol·L"^"-1":color(white)(mmmmmmmll)0.100 + xcolor(white)(mmmll)2x#
#K_"sp" =["Ni"^"2+"]["OH"^"-"]^2 = (0.100 + x)(2x)^2 = 1.40 × 10^"-14"#
We know that #x < 1.52 × 10^"-5"#, so #x ≪ 0.100#.
∴ #0.100(2x)^2 = 0.400x^2 = 1.40 × 10^"-14"#
#x^2 = (1.40 × 10^"-14")/0.400 = 3.50 × 10^"-14"#
#x =sqrt(3.50 × 10^"-14") = 1.87 × 10^"-7"#
∴ The solubility of #"Ni(OH)"_2# is #1.87 × 10^"-7"color(white)(l) "mol/L"#
Also,
#"Solubility" = (1.87 × 10^"-7" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.73 × 10^"-5" "g/L" = "17.3 µg/L"#
