Question

Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100

Answer 1

Answer: The solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$

Explanation:

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

$Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}$

                      3s       2s

The expression for solubility constant for this reaction will be:

$K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2$

We are given:

$K_{sp}=2.5\times 10^{-33}$

Putting values in above equation, we get:

$2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L$

Hence, the solubility of $Cd_3(PO_4)_2$ in water is $1.18\times 10^{-7}mol/L$