Answer : The molar solubility of [tex]AlPO_4[/tex] is, [tex]9.92\times 10^{-11}M[/tex]
Explanation :
The balanced equilibrium reaction will be,
[tex]AlPO_4\rightleftharpoons Al^{3+}+PO_4^{3-}[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Al^{3+}][PO_4^{3-}][/tex]
Let the solubility will be, 's'
[tex]K_{sp}=(s)\times (s)[/tex]
[tex]K_{sp}=s^2[/tex]
Now put the value of [tex]K_[sp}[/tex] in this expression, we get the molar solubility of [tex]AlPO_4[/tex].
[tex]9.84\times 10^{-21}=(s)^2[/tex]
[tex]s=9.92\times 10^{-11}M[/tex]
Therefore, the molar solubility of [tex]AlPO_4[/tex] is, [tex]9.92\times 10^{-11}M[/tex]
