Question

solubility product of m(oh)2 is 10-14 what should be the concentration of m2+ in 0.1m solution of nh4oh if nh4oh gets 10% ionised ?​

Answer 1

Given info : Solubility product of M(OH)₂ is 10¯¹⁴.

To find : what should be the concentration of M²⁺ ion in 0.1 M solution of NH₄OH if it gets 10 % ionised ?

solution :

solubility product of M(OH)₂ is 10¯¹⁴

dissociation reaction of M(OH)₂ is ..

M(OH)₂ ⇔ M²⁺ + 2OH¯

so, Ksp = [M²⁺][OH¯]² = 10¯¹⁴ ...(1)

again dissociation reaction of NH₄OH is..

NH₄OH ⇔NH₄⁺ + OH¯

C(1 - α) Cα Cα

so, concentration of OH¯ = [OH¯] = Cα = 0.1 × 10/100 = 0.1 × 0.1 = 10¯²

now from equation (1) we get,

10¯¹⁴ = [M²⁺] (10¯²)²

⇒ [M²⁺] = 10¯¹⁰ M

Therefore the concentration of M²⁺ ions is 10¯¹⁰ M