Question

Solid PbI₂ was added to a 0.030 M NaI solution. Calculate the molar concentration of lead ion in this solution. Ksp = 7.9 x 10-9 (at 25ºC) (Please show work)

PbI_2(s) <=> Pb²⁺ + 2I^-

Answer 1

Answer

8.77×10^-6M

Explanation

PbI2(s) <------> Pb²⁺(aq) + 2I^-(aq)

Ksp = [Pb²⁺] [ I^-]² = 7.9×10^-9

Initial concentration

[Pb²⁺] = 0

[I^-] = 0.030

change in concentration

[Pb²⁺ ] = + x

[I^-] = +2x

Equillibrium concentration

[Pb²⁺] = x

[I^-] = 0.030 + 2x

so,

x (0.030 + 2x)² = 7.9 ×10^-9

solving for x

x = 8.77×10^-6

Therefore

molar concentration of Pb²⁺ = 8.77×10^-6M