Question

A saturated solution of PbI2 has a concentration of 1.65 × 10-3 mol/dm3

.

i. Write the reversible reaction for PbI2 breaking into its constituent ions.

ii. Write the equilibrium expression (Ksp).

iii. Calculate the concentration of each ion, Pb2+ and I-

.

iv. Calculate the Ksp.

Answer 1

(i) PbI₂ = Pb²⁺ + 2I^-

(ii) K_sp = [Pb²⁺]*[I^-]²

(iii) [Pb²⁺] = c(PbI₂) = 1.65*10^-3 mol/L

[I^-] = 2*c(PbI₂) = 2*1.65*10^-3 mol/L = 3.3*10^-3 mol/L

(iv) K_sp = [Pb²⁺]*[I^-]² = (1.65*10^-3)*(3.3*10^-3)² = 1.7968*10^-8