Question

What is the molar solubility of Mg(OH)₂, K_sp is 5.6 x 10^-12 for Mg(OH)₂, in the following aqueous solutions?

A. Calculate the molar solubility in 0.24 M NaOH:

B. Calculate the solubility in 0.24 M MgSO₄?

Answer 1

Mg(OH)2 ----->Mg+2+2OH-

KSp = [Mg+2] [OH-]2 = 5.6*10-12

let x= solubility of Mg(OH)2

then [Mg+2] =x and [OH-]=2x

hence KSp= x*(2x)²= 5.6*10^-12,x=0.000112

when 0.24M NaOH is added, there will be 0.24M OH-

let x= solubility of Mg(OH)2, then [Mg+2]=x and [OH-]= 2x+0.24

KSp= x*(2x+0.24)²= 5.6*10^-12, compared to 2x, 0.24 is larger and hence x compared to 0.24 can be neglected and hence x*0.24*0.24= 5.6*10^-12

x= 9.72*10^-11

whne 0.24M MgSO4 is added, [Mg+2] is added. Now [Mg+2] =x+0.24 and [OH-]=2x

hence (x+0.24)*(2x)²= 5.6*10^-12

(x+0.24)*x²= 1.4*10^-12

when solved using excel, x=2.4*10^-6