Answer to Question #151289 in Inorganic Chemistry for hasnain naqvi

Question

calculate the solubility of Mg(OH) in moles per liter if the solubility product of Mg(OH)2 is 5 x 10^12

Answer 1

Mg(OH)² ------------- Mg²⁺ +2(OH)^-

Solubility product=[Mg²⁺][OH^-]²

Let the conc of Mg²⁺= x

So (OH^-) =2x

So we have;

5×10¹²=(x)(2x)²

5×10¹²=4x³

x³=1.25×10¹²

x=1.08×10⁴

Therefore the Solubility of Mg(OH)2=1.08×10⁴