# Question aa610

Jan 8, 2017

#### Explanation:

The solubility of ${\text{Ni(OH)}}_{2}$ is 1.52 × 10^"-5" color(white)(l)"mol/L" or $\text{1.41 mg/L}$.

In ${\text{0.100 mol/L NiSO}}_{4}$, the solubility is 1.87 × 10^"-7" "mol/L" = "17.3 µg/L".

Solubility in water

I think your value of ${K}_{\text{sp}}$ is in error. A better value would be 1.40 × 10^"-14".

The solubility equilibrium is

$\textcolor{w h i t e}{m m m m m m} \text{Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-} \left(a q\right)$; K_text(sp) = 1.40 × 10^"-14"
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m m m m m m m m} x \textcolor{w h i t e}{m m m m m} 2 x$

Since $x \textcolor{w h i t e}{l} \text{mol of Ni(OH)"_2"(s)}$ gives $x \textcolor{w h i t e}{l} \text{mol of Ni"^"2+}$, the solubility of ${\text{Ni(OH)}}_{2}$ is $x \textcolor{w h i t e}{l} \text{mol/L}$.

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 ${\text{Ni(OH)}}_{2}$ is 1.52 × 10^"-5" color(white)(l)"mol/L"

${M}_{r} = 92.71$, so

$\text{Solubility" = (1.52 × 10^"-5" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.41 × 10^"-3" color(white)(l)"g/L" = "1.41 mg/L}$

The solubility listed here is 0.001 27 g/L (close enough!).

Solubility in 0.100 mol/L ${\text{NiSO}}_{4}$

The solubility of the ${\text{Ni(OH)}}_{2}$ will be much less because of the common ion effect of the $\text{Ni"^"2+}$ ion.

The solubility equilibrium is now

$\textcolor{w h i t e}{m m m m m m} \text{Ni(OH)"_2"(s)" ⇌ "Ni"^"2+"(aq) + "2OH"^"-} \left(a q\right)$; K_text(sp) = 1.40 × 10^"-14"
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m m m m m m l l} 0.100 + x \textcolor{w h i t e}{m m m l l} 2 x$

${K}_{\text{sp" =["Ni"^"2+"]["OH"^"-"]^2 = (0.100 + x)(2x)^2 = 1.40 × 10^"-14}}$

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 ${\text{Ni(OH)}}_{2}$ is 1.87 × 10^"-7"color(white)(l) "mol/L"#

Also,

$\text{Solubility" = (1.87 × 10^"-7" color(red)(cancel(color(black)("mol"))))/"1 L" × "92.71 g"/(1 color(red)(cancel(color(black)("mol")))) = 1.73 × 10^"-5" "g/L" = "17.3 µg/L}$