Calculate the concentration of cadmium ions in a saturated solution of cadmium(II) phosphate, Cd3(PO4)2, Ksp = 2.53 x 10^{-33}. - HSC - SSCE Chemistry - Question 32 - 2024 - Paper 1

The solubility product constant ( K_sp) is defined as:

$K_{sp} = [\text{Cd}^{2+}]^3 [\text{PO}_4^{3-}]^2$

If we let the solubility of Cd3(PO4)2 be 's' mol/L, then:

• [ [\text{Cd}^{2+}] = 3s ]
• [ [\text{PO}_4^{3-}] = 2s ]

Substituting these into the Ksp expression gives:

$K_{sp} = (3s)^3 (2s)^2 = 108s^5$

Substituting the given Ksp value into the expression:

$2.53 \times 10^{-33} = 108s^5$

Solving for 's':

$s^5 = \frac{2.53 \times 10^{-33}}{108} \approx 2.34 \times 10^{-35}$

Taking the fifth root:

$s \approx (2.34 \times 10^{-35})^{1/5} \approx 1.19 \times 10^{-7} \text{ mol L}^{-1}$

Finally, the concentration of cadmium ions is:

$[\text{Cd}^{2+}] = 3s = 3 \times 1.19 \times 10^{-7} \approx 3.56 \times 10^{-7} \text{ mol L}^{-1}$