Problem 16-53

You need write the chemical reaction of the precipitation

ions presents: Mg²⁺ , NO₃⁻ , Na⁺ , OH⁻

Nitrate and Na⁺ ions are known to make solubles compounds (without exceptions)
Then, the only possible reaction is:

Mg²⁺(aq) + 2OH⁻(aq) → Mg(OH)₂(s)

The K_sp for Mg(OH)₂ is:   K_sp = [Mg²⁺][OH⁻]² = 8.9×10⁻¹²

If the actual concentrations of species in solution are too high, precipitation will occur.
The value Q is the value we are looking for:

Q > K_sp = precipitation
Q = K_sp = at equilibrium (saturation)
Q < K_sp = no precipitation

[Mg²⁺]_o = (0.100 L)(4.0×10⁻⁴ M)/(0.200 L) = 2.0×10⁻⁴ M
[OH⁻]_o = (0.100 L)(2.0×10⁻⁴ M)/(0.200 L) = 1.0×10⁻⁴ M
(don't forget the dilution present when two volumes are mixed)

Q = [Mg²⁺]_o[OH⁻]_o² = (2.0×10⁻⁴)(1.0×10⁻⁴)² = 2.0×10⁻¹²

Since Q < K_sp , no precipitate forms.