Question

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

Answer 1

Explanation:

$PbBr_{2}$ will dissociate into ions as follows.

         $PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)$

Hence, $K_{sp}$ for this reaction will be as follows.

                   $K_{sp} = [Pb^{2+}][Br^{-}]^{2}$

We take x as the molar solubility of $PbBr_{2}$ when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               $[Pb^{2+}]$ = $[Pb^{2+}]_{o}$ + x

               $[Br^{-}]^{2}$ = $[Br^{-}]_{o}$ + 2x

So, equilibrium solubility expression will be as follows.

            $K_{sp}$ = $([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}$

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains $[Br^{-}]_{o}$ = 0.10 and there will be no lead ions. So, $[Pb^{2+}]_{o}$ = 0

So, $[Br^{-}]_{o} + 2x$ will approximately equals to $[Br^{-}]_{o}$.

Hence, $K_{sp} = x[Br^{-}]^{2}_{o}$

            $4.67 \times 10^{-6} = x \times (0.10)^{2}$

                        x = $4.67 \times 10^{-4}$ M

Thus, we can conclude that molar solubility of $PbBr_{2}$ is $4.67 \times 10^{-4}$ M.

Answer 2

Answer:

Molar solubility of [Pb]^{2+}][Br^-]^2[/tex] = $4.67\times 10^{-4}\ M$

Explanation:

$PbBr_2 \leftrightharpoons Pb^{2+} + 2Br^-$

$Ksp = 4.76 \times 10^{-6}$

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

$Ksp = [Pb]^{2+}][Br^-]^2$

$4.67 \times 10^{-6} = x \times (2x + 0.10)^2$

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

2x can be neglected as compared to 0.1

$4.67 \times 10^{-6} = x \times (0.10)^2$

$x=\frac{4.67 \times 10^{-6}}{(0.10)^2} = 4.67\times 10^{-4}\ M$