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

Step-by-step 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 * 10^(-6)`

Let the solubility of given ions be "s".

Since, KBr on dissociation will given bromine ions.

Hence,
`K_(sp) = [Pb^(2+)] * ([Br^(-)])^(2)`

`6.60 * 10^(-6)` =
`s * (2s)^(2)`

=
`1.18 * 10^(-2)` M

Therefore, solubility of
`[PbBr_(2)]` is
`1.18 * 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) * (2s + 0.5)^(2)`

`6.60 * 10^(-6)` =
`4s^(3) + 0.25s + 2s^(2)`

s =
`2.63 * 10^(-5)`

Thus, we can conclude that molar solubility of
`[PbBr_(2)]` in 0.500 m KBr solution is
`2.63 * 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.