Answer: The solubility of [tex]Cd_3(PO_4)_2[/tex] in water is [tex]1.18\times 10^{-7}mol/L[/tex]
Explanation:
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:
[tex]Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}[/tex]
3s 2s
The expression for solubility constant for this reaction will be:
[tex]K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2[/tex]
We are given:
[tex]K_{sp}=2.5\times 10^{-33}[/tex]
Putting values in above equation, we get:
[tex]2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L[/tex]
Hence, the solubility of [tex]Cd_3(PO_4)_2[/tex] in water is [tex]1.18\times 10^{-7}mol/L[/tex]
