## Chemistry (7th Edition)

$K_{sp} (Pb(N_3)_2) = (2.5 \times 10^{-9})$
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 $K_{sp}$: $K_{sp} = (8.5 \times 10^{-4})^ 1 \times (1.7 \times 10^{-3})^ 2$ $K_{sp} = (8.5 \times 10^{-4}) \times (2.9 \times 10^{-6})$ $K_{sp} = (2.5 \times 10^{-9})$