Answer
$ K_{sp} (Pb(N_3)_2) = (2.5 \times 10^{-9})$
Work Step by Step
1. Write the $K_{sp}$ expression:
$ Pb(N_3)_2(s) \lt -- \gt 1Pb^{2+}(aq) + 2{N_3}^{-}(aq)$
$ K_{sp} = [Pb^{2+}]^ 1[{N_3}^{-}]^ 2$
2. Determine the ion concentrations:
$[Pb^{2+}] = [Pb(N_3)_2] * 1 = [8.5 \times 10^{-4}] * 1 = 8.5 \times 10^{-4}$
$[{N_3}^{-}] = [Pb(N_3)_2] * 2 = 1.7 \times 10^{-3}$
3. Calculate the $K_{sp}$:
$ K_{sp} = (8.5 \times 10^{-4})^ 1 \times (1.7 \times 10^{-3})^ 2$
$ K_{sp} = (8.5 \times 10^{-4}) \times (2.9 \times 10^{-6})$
$ K_{sp} = (2.5 \times 10^{-9})$
