Question

Calculate the solubility of PbI₂ in water at 25°C which is 90% dissociated. K_sp (PbI₂) = 1.39 x 10^–8 at 25°C.

(a) 1.68 x 10^–3 mole/litre 

(b) 1.68 x 10^–4 mole/litre

(c) 1.32 x 10^–3 mole/litre 

(d) 2 x 10^–3 mole/litre

Answer 1

Correct option: (a) 1.68 x 10^–3 mole/litre

Explanation: 

Let the solubility of PbI₂ in water be S mole per litre.

PbI₂ ⇌ Pb²⁺ + 2I^–

Since PbI₂ is 90% dissociated