Respuesta :
Explanation:
[tex]PbBr_{2}[/tex] will dissociate into ions as follows.
[tex]PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)[/tex]
Hence, [tex]K_{sp}[/tex] for this reaction will be as follows.
[tex]K_{sp} = [Pb^{2+}][Br^{-}]^{2}[/tex]
We take x as the molar solubility of [tex]PbBr_{2}[/tex] when we dissolve x moles of solution per liter.
Hence, ionic molarities in the saturated solution will be as follows.
[tex][Pb^{2+}][/tex] = [tex][Pb^{2+}]_{o}[/tex] + x
[tex][Br^{-}]^{2}[/tex] = [tex][Br^{-}]_{o}[/tex] + 2x
So, equilibrium solubility expression will be as follows.
[tex]K_{sp}[/tex] = [tex]([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}[/tex]
Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [tex][Br^{-}]_{o}[/tex] = 0.10 and there will be no lead ions. So, [tex][Pb^{2+}]_{o}[/tex] = 0
So, [tex][Br^{-}]_{o} + 2x[/tex] will approximately equals to [tex][Br^{-}]_{o}[/tex].
Hence, [tex]K_{sp} = x[Br^{-}]^{2}_{o}[/tex]
[tex]4.67 \times 10^{-6} = x \times (0.10)^{2}[/tex]
x = [tex]4.67 \times 10^{-4}[/tex] M
Thus, we can conclude that molar solubility of [tex]PbBr_{2}[/tex] is [tex]4.67 \times 10^{-4}[/tex] M.
Answer:
Molar solubility of [Pb]^{2+}][Br^-]^2[/tex] = [tex]4.67\times 10^{-4}\ M[/tex]
Explanation:
[tex]PbBr_2 \leftrightharpoons Pb^{2+} + 2Br^-[/tex]
[tex]Ksp = 4.76 \times 10^{-6}[/tex]
Let the molar solubility of [tex]PbBr_2[/tex] be x
[tex]PbBr_2 \leftrightharpoons Pb^{2+} + 2Br^-[/tex]
At equi. x 2x
[tex][Pb]^{2+}[/tex] = x
[tex][Br]^{-}[/tex] = 2x
In the presence of 0.10 M NaBr
[tex][Br]^{-}[/tex] = 2x + 0.10
[tex]Ksp = [Pb]^{2+}][Br^-]^2[/tex]
[tex]4.67 \times 10^{-6} = x \times (2x + 0.10)^2[/tex]
As [tex]PbBr_2[/tex] is weakly soluble, so 2x<<0.1.
2x can be neglected as compared to 0.1
[tex]4.67 \times 10^{-6} = x \times (0.10)^2[/tex]
[tex]x=\frac{4.67 \times 10^{-6}}{(0.10)^2} = 4.67\times 10^{-4}\ M[/tex]
