Answer: The solubility product of [tex]Ag_3PO_4[/tex] is [tex]27s^4[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
[tex]Ag_3PO_4\leftrightharpoons 3Ag^++PO_4^{3-}[/tex]
We are given:
Solubility of [tex]Ag_3PO_4[/tex] = S mol/L
By stoichiometry of the reaction:
1 mole of [tex]Ag_3PO_4[/tex] gives 3 moles of [tex]Ag^{+}[/tex] and 1 mole of [tex]PO_4^{3-}[/tex].
When the solubility of [tex]Ag_3PO_4[/tex] is S moles/liter, then the solubility of [tex]Ag^{+}[/tex] will be 3S moles\liter and solubility of [tex]PO_4^{3-}[/tex] will be S moles/liter.
Expression for the equilibrium constant of [tex]Ag_2CrO_4[/tex] will be:
[tex]K_{sp}=[Ag^+]^3[PO_4^{3-}][/tex]
[tex]K_{sp}=[3s]^3[s]=27s^4[/tex]
The solubility product of [tex]Ag_3PO_4[/tex] is [tex]27s^4[/tex]
