# Question 6303e

Apr 13, 2017

At 25 °C, 229 mg of ${\text{Mg(OH)}}_{2}$ will remain undissolved.

#### Explanation:

This appears to be a solubility product problem.

We can set up an ICE table.

$\textcolor{w h i t e}{m m m m m m} \text{Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)}$
$\text{I/mol·L"^"-1": } \textcolor{w h i t e}{m m m m m m m m m m m} 0 \textcolor{w h i t e}{m m m m m} 0$
$\text{C/mol·L"^"-1": color(white)(mmmmmmmmmm)"+"xcolor(white)(mmmm)"+2} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m m m m m m m m m m l l} x \textcolor{w h i t e}{m m m m m} 2 x$

At 25 °C,

K_text(sp) = ["Mg"^"2+"]["OH"^"-"]^2 = 5.61×10^"-12"

x(2x)^2 = 4x^3 = 5.61×10^"-12"

x^3 = (5.61×10^"-12")/4 = 1.40 × 10^"-12"

x= 1.12 × 10^"-4"

∴ The solubility of ${\text{Mg(OH)}}_{2}$ at 25 °C is 1.12 × 10^"-4" color(white)(l)"mol/L"#.

$\text{Solubility" = (1.12 × 10^"-4" color(red)(cancel(color(black)("mol"))))/"1 L" × "58 320 mg"/(1 color(red)(cancel(color(black)("mol")))) = "6.53 mg/L}$

The solubility in 500 mL is 2.56 mg.

If you add 232 mg of ${\text{Mg(OH)}}_{2}$ to 500 mL of water, the amount remaining undissolved will be

$\text{232 mg - 2.56 mg = 229 mg}$