At 25oC the solubility of lead(II) bromide is 2.70 x 10-2 mol/L. Calculate the value of Ksp at this temperature.

Question

At 25 ^oC the solubility of lead(II) bromide is 2.70 x 10^-2 mol/L. Calculate the value of K_sp at this temperature.

The answer is 7.9e-5

please show the steps

Answer 1

PbBr₂(s) <===> Pb²⁺(aq) + 2Br^-(aq)

Ksp = [Pb²⁺][Br^-]²

[Pb²⁺] = 2.70x10^-2

[Br^-] = 2 x 2.70x10^-2 = 5.40x10^-2

Ksp = (2.70x10^-2)(5.40x10^-2)²

Ksp = 7.87x10^-5 = 7.9x10^-5