A solution contains $1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\( and \)5.0 \times 10^{-7} M$ \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? $\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.$

: First, let's write the balanced chemical equation for the formation of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) solid: \(3 \mathrm{Sr}^{2+}(aq) + 2 \mathrm{PO}_{4}^{3-}(aq) \rightleftharpoons \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) Now we can write the expression for \(K_{sp}\) of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\): \(K_{sp} = [\mathrm{Sr}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\) The given value of \(K_{sp}\) for \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(1.0 \times 10^{-31}\).

: The given initial concentration for the \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) solution is \(1.0 \times 10^{-6} \mathrm{M}\) and for the \(\mathrm{K}_{3} \mathrm{PO}_{4}\) solution is \(5.0 \times 10^{-7} \mathrm{M}\). Since one mole of \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) dissociates completely to produce one mole of \(\mathrm{Sr}^{2+}\) ions, the initial concentration of \(\mathrm{Sr}^{2+}\) ions in the solution is also \(1.0 \times 10^{-6} \mathrm{M}\). Likewise, one mole of \(\mathrm{K}_{3} \mathrm{PO}_{4}\) dissociates completely to produce one mole of \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions, so the initial concentration of \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions is \(5.0 \times 10^{-7} \mathrm{M}\).

: Now, we can calculate the ionic product of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) using the initial concentrations of \(\mathrm{Sr}^{2+}\) and \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions: \(Q_{sp} = [\mathrm{Sr}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2} = (1.0 \times 10^{-6})^{3}(5.0 \times 10^{-7})^{2} = 1.25 \times 10^{-34}\) Now, we need to compare the ionic product (\(Q_{sp}\)) with the solubility product constant (\(K_{sp}\)): If \(Q_{sp} > K_{sp}\), then precipitation will occur. If \(Q_{sp} < K_{sp}\), then no precipitation will occur. In this case: \(Q_{sp} = 1.25 \times 10^{-34} < K_{sp} = 1.0 \times 10^{-31}\) Since \(Q_{sp} < K_{sp}\), no precipitation of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) will occur.