Question

The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution

Answer 1

Explanation:

The given equation is as follows.

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

                          s       2s

It is given that,

       $K_{sp} = [Pb^{2+}][Br^{-}]^{2} = 6.60 \times 10^{-6}$

Let the solubility of given ions be "s".

Since, KBr on dissociation will given bromine ions.

Hence,  $K_{sp} = [Pb^{2+}] \times ([Br^{-}])^{2}$

        $6.60 \times 10^{-6}$  = $s \times (2s)^{2}$

                        = $1.18 \times 10^{-2}$ M

Therefore, solubility of $[PbBr_{2}]$ is $1.18 \times 10^{-2}$ M  in KBr.

Now, we will calculate the molar solubility of $PbBr_{2}$ in 0.5 M KBr solution as follows.

           $K_{sp} = (s) \times (2s + 0.5)^{2}$

  $6.60 \times 10^{-6}$  = $4s^{3} + 0.25s + 2s^{2}$

                         s = $2.63 \times 10^{-5}$

Thus, we can conclude that molar solubility of $[PbBr_{2}]$ in 0.500 m KBr solution is $2.63 \times 10^{-5}$.

Answer 2

The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.