A solution contains \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4} .\) What is the minimum concentration of \(\mathrm{AgNO}_{3}\) that would cause precipitation of solid \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

Chemical equilibrium represents a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. In the context of solubility products, when a sparingly soluble salt like \(\mathrm{Ag_3PO_4}\) is in a saturated solution, it is at equilibrium with its ions \(\mathrm{Ag^+}\) and \(\mathrm{PO_4^{3-}}\).

The equilibrium can be expressed in terms of the solubility product constant, \(K_{sp}\), a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. The \(K_{sp}\) value gives us insight into the extent to which the compound can dissolve under a set of conditions. A key thing to understand is that if you add more of an ion that forms part of the sparingly soluble compound, you can push the equilibrium towards the solid, causing precipitation. This happens when the ionic product of \(\mathrm{Ag^+}\) and \(\mathrm{PO_4^{3-}}\) concentrations exceeds the \(K_{sp}\) value, leading to a shift away from equilibrium caused by the common ion effect.

A precipitation reaction occurs when two soluble salts react in a solution to form an insoluble solid, known as the precipitate. The formation of a precipitate can be predicted using the solubility product constant, \(K_{sp}\), and the concentrations of the ions in solution.

When the product of the concentrations of the ions in solution, raised to their respective stoichiometric coefficients, exceeds the \(K_{sp}\) value, precipitation will occur. This is known as the solubility product principle. In our exercise, the addition of \(\mathrm{AgNO_3}\) to a solution containing \(\mathrm{Na_3PO_4}\) potentially leads to the formation of silver phosphate, \(\mathrm{Ag_3PO_4}\), if the concentration of \(\mathrm{Ag^+}\) ions reaches a critical level. To find this threshold, you would calculate the concentration of \(\mathrm{Ag^+}\) ions needed to exceed the \(K_{sp}\), factoring in the initial concentration of \(\mathrm{PO_4^{3-}}\) ions from the sodium phosphate in the solution.

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction. Understanding stoichiometry is essential for solving problems involving the solubility product constant.

In the provided exercise, the balanced chemical equation indicates that three moles of \(\mathrm{AgNO_3}\) react with one mole of \(\mathrm{Na_3PO_4}\) to produce one mole of \(\mathrm{Ag_3PO_4}\) and three moles of \(\mathrm{NaNO_3}\). This 3:1 ratio is crucial when calculating how the concentration of \(\mathrm{Ag^+}\) changes as it reacts with \(\mathrm{PO_4^{3-}}\). Stoichiometry enables us to set up the relationship between the initial concentrations and the change in concentration (\(x\)) to solve for the minimum \(\mathrm{Ag^+}\) concentration needed to reach the point where \(\mathrm{Ag_3PO_4}\) begins to precipitate.