Answer
$ K_{sp} (Pd(SCN)_2)= (4.38 \times 10^{-23})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Pd(SCN)_2(s) \lt -- \gt 1Pd^{2+}(aq) + 2SCN^-(aq)$
$ K_{sp} = [Pd^{2+}]^ 1[SCN^-]^ 2$
2. Determine the ions concentrations:
$[Pd^{2+}] = [Pd(SCN)_2] * 1 = [2.22 \times 10^{-8}] * 1 = 2.22 \times 10^{-8}$
$[SCN^-] = [Pd(SCN)_2] * 2 = 4.44 \times 10^{-8}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (2.22 \times 10^{-8})^ 1 \times (4.44 \times 10^{-8})^ 2$
$ K_{sp} = (2.22 \times 10^{-8}) \times (1.97 \times 10^{-15})$
$ K_{sp} = (4.38 \times 10^{-23})$
