A solution contains \(1.0 \times 10^{-5} M \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What concentrations of \(\mathrm{A} \mathrm{g} \mathrm{NO}_{3}\) will cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\) \(\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

For this precipitation reaction, we have: \[\mathrm{3AgNO}_{3}(aq) + \mathrm{Na}_{3}\mathrm{PO}_{4}(aq) \longrightarrow \mathrm{Ag}_{3}\mathrm{PO}_{4}(s) + 3\mathrm{NaNO}_{3}(aq)\]

According to the balanced equation, when \(\mathrm{Ag}_{3}\mathrm{PO}_{4}\) dissolves, it forms 3 moles of \(\mathrm{Ag}^{+}\) ions and 1 mole of \(\mathrm{PO}_{4}^{3-}\) ion. Using these stoichiometric coefficients, we can write the solubility product expression as: \[K_{sp}=[\mathrm{Ag^{+}}]^{3}[{\mathrm{PO}_{4}^{3-}}]\]

The given concentration of \(\mathrm{Na}_{3}\mathrm{PO}_{4}\) in the solution is \(1.0 \times 10^{-5} M\), which produces an equal concentration of \(\mathrm{PO}_{4}^{3-}\) ions, as there is a 1:1 ratio between them in the balanced equation: \[ [\mathrm{PO}_{4}^{3-}] = 1.0 \times 10^{-5} M \]

The reaction quotient (\(Q\)) is a ratio of the concentrations of products and reactants raised to the power of their stoichiometric coefficients in a balanced chemical equation. In this case, it is: \[Q=[\mathrm{Ag^{+}}]^{3}[{\mathrm{PO}_{4}^{3-}}]\] To find the concentration of \(\mathrm{AgNO}_{3}\) that causes precipitation, we have to determine the concentration of \(\mathrm{Ag}^{+}\) ions at which \(Q \geq K_{sp}\). When \(Q\) equals or becomes greater than \(K_{sp}\), the system exceeds its solubility limit and precipitation starts to happen.

Use the given value of \(K_{sp}\) for \(\mathrm{Ag}_{3}\mathrm{PO}_{4}\) and the concentration of \(\mathrm{PO}_{4}^{3-}\) ions determined in Step 3 to solve for the concentration of \(\mathrm{Ag^{+}}\) ions: \[1.8 \times 10^{-18} = [\mathrm{Ag^{+}}]^{3} \times (1.0 \times 10^{-5})\] \[1.8 \times 10^{-13} = [\mathrm{Ag^{+}}]^{3}\] Now, solve for \([\mathrm{Ag^{+}}]\): \[[\mathrm{Ag^{+}}] = (1.8 \times 10^{-13})^{\frac{1}{3}}\] \[[\mathrm{Ag^{+}}] \approx 5.63 \times 10^{-5} M\] Since all the \(\mathrm{Ag}^{+}\) ions are present in the form of \(\mathrm{AgNO}_{3}\) (1:1 stoichiometry), the concentration of \(\mathrm{AgNO}_{3}\) that will cause precipitation is: \[[\mathrm{AgNO}_{3}] = [\mathrm{Ag^{+}}] \approx 5.63 \times 10^{-5} M\]