Respuesta :
Answer : The value of [tex]K_{sp}[/tex] of [tex]Pb_3(PO_4)_2[/tex] is, [tex]3.32\times 10^{-209}[/tex]
Explanation :
The equilibrium reaction will be,
[tex]Pb_3(PO_4)_2\rightleftharpoons 3Pb^{2+}+2PO_4^{2-}[/tex]
The expression of solubility product, [tex]K_{sp}[/tex] will be,
[tex]k_{sp}=[Pb^{2+}]^3[PO_4^{2-}]^2[/tex]
Let the molar solubility be 's'.
[tex]k_{sp}=(3s)^3\times (2s)^2[/tex]
[tex]k_{sp}=108\times s^5[/tex]
Now put the value of 's' in this expression, we get :
[tex]k_{sp}=108\times (7.9\times 10^{-43})^5=3.32\times 10^{-209}[/tex]
Therefore, the value of [tex]K_{sp}[/tex] of [tex]Pb_3(PO_4)_2[/tex] is, [tex]3.32\times 10^{-209}[/tex]
