Respuesta :
Answer :
The solubility equilibrium reaction will be:
[tex]PbCrO_4\rightleftharpoons Pb^{2+}+CrO_4^{2-}M^2[/tex]
The solubility in grams per liter in water at [tex]25^oC[/tex] is, [tex]1.7\times 10^{-4}g/L[/tex]
Explanation :
The solubility equilibrium reaction will be:
[tex]PbCrO_4\rightleftharpoons Pb^{2+}+CrO_4^{2-}M^2[/tex]
Let the solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Pb^{2+}][CrO_4^{2-}][/tex]
[tex]K_{sp}=(s)\times (s)[/tex]
[tex]K_{sp}=s^2[/tex]
Given:
Solubility constant = [tex]K_{sp}[/tex] = [tex]2.8\times 10^{-13}[/tex]
Now put all the given values in the above expression, we get:
[tex]K_{sp}=s^2[/tex]
[tex]2.8\times 10^{-13}=s^2[/tex]
[tex]s=5.3\times 10^{-7}M=5.3\times 10^{-7}mol/L[/tex]
Now we have to convert the solubility in gram per liter.
Solubility = [tex](5.3\times 10^{-7}mol/L)\times \text{Molar mass of }PbCrO_4[/tex]
[tex]\text{Molar mass of }PbCrO_4=323.2g/mol[/tex]
Solubility = [tex](5.3\times 10^{-7}mol/L)\times 323.2g/mol[/tex]
Solubility = [tex]1.7\times 10^{-4}g/L[/tex]
Therefore, the solubility in grams per liter in water at [tex]25^oC[/tex] is, [tex]1.7\times 10^{-4}g/L[/tex]
