Question #aa610

1 Answer
Jan 8, 2017

Answer:

WARNING! Long answer!

Explanation:

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"#