Answer
$ K_{sp} (Ag_2C{O_3}) = (8.39 \times 10^{-12})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Ag_2CO_3(s) \lt -- \gt 2Ag^{+}(aq) + 1{CO_3}^{2-}(aq)$
$ K_{sp} = [Ag^{+}]^ 2[{CO_3}^{2-}]^ 1$
2. Determine the ion concentrations:
** Considering "x", the molar solubility:
$[Ag^{+}] = 2.56 \times 10^{-4}$ $= [Ag_2CO_3] * 2$
$[Ag_2C{O_3}^-] = \frac{2.56 \times 10^{-4}}{2} = 1.28 \times 10^{-4}$
$[{CO_3}^{2-}] = [Ag_2CO_3] * 1 = 1.28 \times 10^{-4}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (2.56 \times 10^{-4})^ 2 \times (1.28 \times 10^{-4})^ 1$
$ K_{sp} = (6.554 \times 10^{-8}) \times (1.28 \times 10^{-4})$
$ K_{sp} = (8.388 \times 10^{-12})$
