Question

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 \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\left(K_{\mathrm{sp}}=1.8 \times 10^{-18}\right) ?\)

Short answer

The minimum concentration of silver nitrate needed to cause precipitation of solid silver phosphate is \(5.648 \times 10^{-5} \mathrm{M}\).

Step-by-step solution

Write the balanced chemical equation

First, write down the balanced chemical equation for the reaction between sodium phosphate and silver nitrate, which involves the formation of silver phosphate. \(\mathrm{3AgNO}_3 + \mathrm{Na}_3\mathrm{PO}_4 \longrightarrow \mathrm{Ag}_3\mathrm{PO}_4 \downarrow + 3\mathrm{NaNO}_3\) Silver nitrate reacts with sodium phosphate to produce solid silver phosphate and sodium nitrate.

Set up the solubility product expression for silver phosphate

To find the minimum concentration of silver nitrate needed for precipitation, we need to set up the solubility product (Ksp) expression for silver phosphate: \[K_{\mathrm{sp}} = [\mathrm{Ag^+}]^3 [\mathrm{PO}_{4}^{3-}]\] The given solubility product \(K_{\mathrm{sp}}\) for silver phosphate is \(1.8 \times 10^{-18}\).

Use the concentration of sodium phosphate to find the phosphate ion concentration

The concentration of sodium phosphate is given as \(1.0 \times 10^{-5} \mathrm{M}\). In the reaction, one mole of sodium phosphate dissociates to one mole of phosphate ions (\(\mathrm{PO}_{4}^{3-}\)). Therefore, the concentration of phosphate ions in the solution is also \(1.0 \times 10^{-5} \mathrm{M}\).

Find the concentration of silver ions needed for precipitation

Now, we can use the solubility product expression and the concentration of phosphate ions to find the minimum concentration of silver ions needed for precipitation. Since \([\mathrm{PO^{3-}_4}] = 1.0 \times 10^{-5} \mathrm{M}\), we can substitute this value into the Ksp expression and solve for [\(\mathrm{Ag^+}\)]: \[1.8 \times 10^{-18} = [\mathrm{Ag^+}]^3 (1.0 \times 10^{-5})\] Now, solve for [\(\mathrm{Ag^+}\)]: \[[\mathrm{Ag^+}]^3 = \frac{1.8 \times 10^{-18}}{1.0 \times 10^{-5}}\] \[[\mathrm{Ag^+}]^3 = 1.8 \times 10^{-13}\] \[[\mathrm{Ag^+}] = \sqrt[3]{1.8 \times 10^{-13}}\] \[[\mathrm{Ag^+}] = 5.648 \times 10^{-5} \mathrm{M}\]

Determine the minimum concentration of silver nitrate needed for precipitation

We have now found the minimum concentration of silver ions needed for precipitation. Since each silver ion comes from one molecule of silver nitrate, the minimum concentration of silver nitrate needed to cause precipitation of solid silver phosphate is: \[[\mathrm{AgNO}_3]_{\min} = [\mathrm{Ag^+}] = 5.648 \times 10^{-5} \mathrm{M}\] Thus, the minimum concentration of silver nitrate needed to cause precipitation of solid silver phosphate is \(5.648 \times 10^{-5} \mathrm{M}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in which 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. This balance is essential when dealing with solubility products, as it determines at what point a substance will dissolve or precipitate out of solution.

In the context of precipitation, when a solution is saturated with ions that can form a precipitate, it is said to be at equilibrium with the solid form of that compound. The equilibrium favours the solid form when the ion product exceeds the solubility product constant (Ksp), and it leans towards the dissolved form when the ion product is less than Ksp. The Ksp serves as a quantitative measure of a compound's solubility and enables chemists to predict whether a precipitate will form under given conditions.

Precipitation Reactions

Precipitation reactions occur when ions in solution combine to form an insoluble compound, which settles out of the solution as a solid precipitate. To predict and control these reactions, scientists rely on the concept of the solubility product constant (Ksp). The Ksp is specific to each compound at a given temperature and is used to determine the saturation point of a solution.

By knowing the Ksp, chemists can also calculate the minimum ion concentration needed to initiate precipitation. For instance, if a compound's Ksp is known, and the concentration of one ion is given, we can compute the required concentration of the other ion to reach the threshold where precipitation starts, as demonstrated in the original exercise provided.

It's important to recognize that while the Ksp can give us valuable information about whether a precipitate will form, it does not provide a rate for the reaction; it only indicates the point at which the reaction will proceed towards forming a solid precipitate.

Ion Concentration Calculation

Calculating ion concentrations is a vital skill in chemistry, particularly when working with reactions that involve soluble salts and ionic compounds. In the original problem, we're tasked with finding out under what conditions a precipitate will form, which requires us to calculate the concentration of ions in solution.

Using the solubility product constant (Ksp) and the molarity of one reactant, you can find the minimum concentration of another ion necessary for precipitation. The process involves setting up an algebraic expression where the product of the ionic concentrations raised to the power of their stoichiometric coefficients equals the Ksp. This process not only applies to simple precipitation problems but is also crucial in areas like environmental chemistry, where ion concentration dictates the viability of habitats, and in medicine, where the bioavailability of ions can influence health outcomes.