Question

The solubility of AgBr in water is 1.20 x 10-5 mol dm -3

. Calculate the

solubility product of Ag Br.​

Answer 1

Given-

The solubility of AgBr in water is 1.20 x 10⁻⁵ mol dm⁻³

To Find:-

The  solubility product of Ag Br

Solution:-

AgBr → Ag⁺ (aq) + Br⁻(aq)

Ksp = [Ag⁺][Br⁻]

= (1.2 x 10 ⁻⁵)(1.2 x 10⁻⁵)

= 1.44 x 10 ⁻¹⁰

Hence, the solubility product of AgBr is 1.44 x 10⁻¹⁰ mol/dm³