This appears to be a solubility product problem.
We can set up an ICE table.
color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^"2+""(aq)" + "2OH"^"-""(aq)"mmmmmmMg(OH)2(s)⇌Mg2+(aq)+2OH-(aq)
"I/mol·L"^"-1": "color(white)(mmmmmmmmmmm)0color(white)(mmmmm)0I/mol⋅L-1:mmmmmmmmmmm0mmmmm0
"C/mol·L"^"-1": color(white)(mmmmmmmmmm)"+"xcolor(white)(mmmm)"+2"xC/mol⋅L-1:mmmmmmmmmm+xmmmm+2x
"E/mol·L"^"-1": color(white)(mmmmmmmmmmll)xcolor(white)(mmmmm)2xE/mol⋅L-1:mmmmmmmmmmllxmmmmm2x
At 25 °C,
K_text(sp) = ["Mg"^"2+"]["OH"^"-"]^2 = 5.61×10^"-12"Ksp=[Mg2+][OH-]2=5.61×10-12
x(2x)^2 = 4x^3 = 5.61×10^"-12"x(2x)2=4x3=5.61×10-12
x^3 = (5.61×10^"-12")/4 = 1.40 × 10^"-12"x3=5.61×10-124=1.40×10-12
x= 1.12 × 10^"-4"x=1.12×10-4
∴ The solubility of "Mg(OH)"_2Mg(OH)2 at 25 °C is 1.12 × 10^"-4" color(white)(l)"mol/L"1.12×10-4lmol/L.
"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 "Mg(OH)"_2 to 500 mL of water, the amount remaining undissolved will be
"232 mg - 2.56 mg = 229 mg"
