Answer : The solubility of [tex]Ag_3PO_4[/tex] in water is, [tex]1.8\times 10^{-5}mol/L[/tex]
Explanation :
The solubility equilibrium reaction will be:
[tex]Ag_3PO_4\rightleftharpoons 3Ag^++PO_4^{3-}[/tex]
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ag^{+}]^3[PO_4^{3-}][/tex]
[tex]K_{sp}=(3s)^3\times (s)[/tex]
[tex]K_{sp}=27s^4[/tex]
Given:
[tex]K_{sp}[/tex] = [tex]2.8\times 10^{-18}[/tex]
Now put all the given values in the above expression, we get:
[tex]K_{sp}=27s^4[/tex]
[tex]2.8\times 10^{-18}=27s^4[/tex]
[tex]s=1.8\times 10^{-5}M=1.8\times 10^{-5}mol/L[/tex]
Therefore, the solubility of [tex]Ag_3PO_4[/tex] in water is, [tex]1.8\times 10^{-5}mol/L[/tex]
