Answer
$ K_{sp} (Ag_2S{O_3}) = (1.49 \times 10^{-14})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Ag_2S{O_3}(s) \lt -- \gt 2Ag^{+}(aq) + 1S{O_3}^{2-}(aq)$
$ K_{sp} = [Ag^{+}]^ 2[S{O_3}^{2-}]^ 1$
2. Determine the ions concentrations:
$[Ag^{+}] = [Ag_2S{O_3}] * 2 = [1.55 \times 10^{-5}] * 2 = 3.1 \times 10^{-5}$
$[S{O_3}^{2-}] = [Ag_2S{O_3}] * 1 = 1.55 \times 10^{-5}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (3.1 \times 10^{-5})^ 2 \times (1.55 \times 10^{-5})^ 1$
$ K_{sp} = (9.61 \times 10^{-10}) \times (1.55 \times 10^{-5})$
$ K_{sp} = (1.49 \times 10^{-14})$
