Question

Will a precipitate of Mg(OH)₂ be formed in a 0.002 M solution of Mg(NO₃)₂, if the pH of solution is adjusted to 9 ? K_sp of Mg(OH)₂ = 8.9 × 10^–12.

Answer 1

pH = 9

∴ [H⁺ ] = 10^–9 M 

or [OH^– ] = 10^–5 M 

Now if Mg(NO₃)₂ is present in a solution of [OH^–] = 10^–5 M, then, 

Product of ionic conc. = [Mg⁺²] [OH^– ]² = [0.002] [[10^–5]² 

 = 2 × 10^–13 lesser than K_sp of Mg(OH)₂ i.e, 8.9 × 10^–12 

∴ Mg(OH)₂ will not precipitate.