Question

The solubility of pbcl, at 298 k is 2x10mol calculate k, at this temperature.

Answer 1

Given,

solubility of PbCl2 = 2 x 10mol/L

Therefore,

[Pb2+] = 2 x 10-2 mol/L

[Cl-] = 2 x 2x 10-2 mol/L

Solubility product,

=Multiply this both