Answer : The molar solubility of [tex]PbBr_2[/tex] in pure water is, 0.0118 M
Explanation : Given,
[tex]K_{sp}=6.60\times 10^{-6}[/tex]
The solubility equilibrium reaction will be:
[tex]PbBr_2\rightleftharpoons Pb^{2+}+2Br^{-}[/tex]
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Pb^{2+}][Br^{-}]^2[/tex]
[tex]K_{sp}=(s)\times (2s)^2[/tex]
Now put all the given values in the above expression, we get:
[tex]6.60\times 10^{-6}=4s^3[/tex]
[tex]s=0.0118M[/tex]
Therefore, the molar solubility of [tex]PbBr_2[/tex] in pure water is, 0.0118 M
