Question

Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.

Answer 1

Molar solubility of [Pb]^{2+}][Br^-]^2[/tex] =
`4.67* 10^(-4)\ M`

Step-by-step explanation:

`PbBr_2 \leftrightharpoons Pb^(2+) + 2Br^-`

`Ksp = 4.76 * 10^(-6)`

Let the molar solubility of
`PbBr_2` be x

`PbBr_2 \leftrightharpoons Pb^(2+) + 2Br^-`

At equi. x 2x

`[Pb]^(2+)` = x

`[Br]^(-)` = 2x

In the presence of 0.10 M NaBr

`[Br]^(-)` = 2x + 0.10

`Ksp = [Pb]^(2+)][Br^-]^2`

`4.67 * 10^(-6) = x * (2x + 0.10)^2`

As
`PbBr_2` is weakly soluble, so 2x<<0.1.

2x can be neglected as compared to 0.1

`4.67 * 10^(-6) = x * (0.10)^2`

`x=(4.67 * 10^(-6))/((0.10)^2) = 4.67* 10^(-4)\ M`