Question

The solubility of Mg(OH), is 3.352x10gL' at 290°С. Find out its
solubility product at this temperature. (Mg = 24 u)​

Answer 1

Answer:

Mg(OH)2  ionizes completely in the solution as : Mg(OH)₂→Mg₂+2OH−

∴ [Mg²⁺]=[Mg(OH)₂]and[OH⁻]=2×[Mg(OH)₂]

But Molar mass of Mg(OH)2=58gmol−1

∴ [Mg(OH)₂]=Strength in g / litreMolar mass= 8.352×10−358= 1.44×10⁻⁴ moles/litre

∴[Mg²⁺]=1.44×10⁻⁴ moles/litre and [OH⁻] = 2 × 1.44×10⁻⁴ = 2.88 × 10⁻⁴ moles/litre

∴Ksp for Mg(OH)₂= [Mg²⁺][OH⁻]2 = (1.44×10⁻⁴)×(2.88×10⁻⁴)2 = 1.194×10⁻¹¹

Explanation: