Question

Calculate the molar solubility of Mg(OH)2 in the following solvents. K_sp = 1.8 x 10¯¹¹

pure water

8.68×10^?2 M MgCl2

3.65×10^?2 M KOH(aq).

Really i just need someone to explain how these things affect solubility numerically

Answer 1

Mg(OH)₂$\rightleftharpoons$ Mg²⁺ + 2OH^-

In pure water,

K_sp = [ Mg²⁺] * [OH^-]²   

= x*(2x)² = 4x³ =  1.8 x 10^-11 => x = 1.65*10^-4mol/lt

In MgCl₂,

	Mg(OH)2	Mg2+	OH-
Initial		8.68*10-2 ( Already MgCl2 has Mg2+ ions)	0
Change		+x	+2x
Equilibrium		8.68*10-2 + x	2x

K_sp = [ Mg²⁺] * [OH^-]²   

= (8.68*10^-2 + x)(2x)² = 1.8*10^-11 as k_sp is very low (8.68*10^-2 + x) = 8.68*10^-2

=> 8.68*10^-2*4x² = 1.8*10^-11

Solving, x = 7.2*10^-6M

In KOH,

	Mg(OH)2	Mg2+	OH-
Initial		0	3.65*10-2 (( Already KOH has OH- ions)
Change		+x	3.65*10-2+2x
Equilibrium		x	3.65*10-2+2x

K_sp = [ Mg²⁺] * [OH^-]²   

= ( x)(3.65*10^-2+2x)² = 1.8*10^-11 as k_sp is very low (3.65*10^-2+2x) = 3.65*10^-2

=> x*(3.65*10^-2)² = 1.8*10^-11

Solving, x = 1.35*10^-8M