The solubility product of PbBr2 is 8.9 10-6. Determine the molar solubility in pure water?

Answer 1

The equilibrium will be
PbBr2 <> Pb2+ + 2Br-

Let x = moles/L PbBr2 that dissolve in water.

When x moles/l PbBr2 dissolve then, according to the reaction we would get x moles/L Pb2+ and 2x moles/L Br-

Substitute these in Ksp expression

8.9 10-6 = (x)(2x)^2 = 4x^3

x= 0.0130

to get grams per 1 L

MM PbBr2 = 376 g/mol

0.0130 mol/L(367 g/mol) (1L)= 0.477 g