Question

A solution is \(0.10 \mathrm{M}\) in free \(\mathrm{NH}_{3}, 0.10 \mathrm{M}\) in \(\mathrm{NH}_{4} \mathrm{Cl}\), and \(0.015 \mathrm{M}\) in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+} .\) Will \(\mathrm{Cu}(\mathrm{OH})_{2}(\mathrm{s})\) precipitate from this solution? \(K_{\mathrm{sp}}\) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) is \(2.2 \times 10^{-20}\).

Short answer

After calculating the values in step 2, if \(Q_{sp} > K_{sp}\) (i.e. \(Q_{sp} > 2.2 \times 10^{-20}\)), \(\mathrm{Cu}(\mathrm{OH})_{2}\) will precipitate from the solution.

Step-by-step solution

Expression for Ion Product

The first step is to write the expression for the Ion Product \(Q_{sp}\) of the reaction. For the salt \(\mathrm{Cu}(\mathrm{OH})_{2}\), \(Q_{sp} = [\mathrm{Cu^{2+}}] [\mathrm{OH^-}]^2\)

Calculation of \(Q_{sp}\)

Now calculate \(Q_{sp}\). As the solution is 0.015 M in \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}\), this will be the concentration of \(\mathrm{Cu^{2+}}\). The \(\mathrm{OH^-}\) ions will come from the dissociation of \(\mathrm{NH}_{3}\). The dissociation of \(\mathrm{NH}_{3}\) in water is \(\mathrm{NH}_{3} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{NH}_{4^{+}} + \mathrm{OH^{-}}\). Since the solution is 0.10 M in \(\mathrm{NH}_{3}\) and in \(\mathrm{NH}_{4^{+}}\), it can be assumed that equilbrium is achieved and thus the concentration of \(\mathrm{OH^{-}}\) should also be 0.10 M. Giving, \(Q_{sp} = (0.015) \times (0.10)^2\)

Compare \(Q_{sp}\) and \(K_{sp}\)

The final step is to compare \(Q_{sp}\) and \(K_{sp}\) to check whether the precipitation of \(\mathrm{Cu}(\mathrm{OH})_{2}\) will occur or not. If \(Q_{sp} > K_{sp}\), \(\mathrm{Cu}(\mathrm{OH})_{2}\) will precipitate out, and if \(Q_{sp} < K_{sp}\), it won't.

Key concepts

Solubility Product Constant

The solubility product constant, often denoted as K_{sp}, is a numerical representation of a compound’s solubility in water. Specifically, it is the equilibrium constant for a solid substance dissolving in an aqueous solution to form a saturated solution.

This constant is highly useful in predicting the solubility of an ionic compound under a given set of conditions. The expression depends on the ions that the solid produces upon dissolution. For instance, for a compound like Cu(OH)_{2}, which dissociates into Cu^{2+} and 2OH^- ions, the K_{sp} is written as K_{sp} = [Cu^{2+}] [OH^-]^2.

However, it's important to note that the K_{sp} is constant for a given substance only at a constant temperature. Furthermore, it doesn’t provide any information on the rate of dissolution or precipitation; it only indicates the extent to which a compound can dissolve.

Ion Product

The ion product, denoted as Q_{sp}, refers to the product of the concentrations of the ions of a solute at any given moment in solution, raised to the power of their coefficients in the balanced dissolution equation. This is an essential concept when assessing the likelihood of precipitation in a solution.

The Q_{sp} is similar to the K_{sp} expression except that it applies to any moment in time, not just at equilibrium. If Q_{sp} > K_{sp}, the solution is supersaturated, and precipitation is likely. Conversely, if Q_{sp} < K_{sp}, the solution is unsaturated with respect to the solute, and no precipitation will occur. When Q_{sp} = K_{sp}, the solution is at equilibrium, and the system is in a state where the rate of dissolution equals the rate of precipitation.

Precipitation Reactions

Precipitation reactions occur when a solid, known as a precipitate, forms in a solution during a chemical reaction. These reactions typically involve two soluble salts that, when mixed, yield an insoluble compound that falls out of the solution.

For example, when solutions containing Cu^{2+} and OH^- ions are mixed, Cu(OH)_{2} may form as a precipitate if the concentrations of these ions surpass the solubility limit of Cu(OH)_{2}.
The concept of precipitation is crucial in areas like waste water treatment, where it helps in removing harmful ions from water, and in the pharmaceutical industry, where it assists in the isolation and purification of compounds.

Dissociation of Weak Bases

Weak bases, such as NH_3 (ammonia), do not completely dissociate in water. Instead, they establish an equilibrium with their ions and the water molecules. The dissociation of NH_3 in water is represented by the following equilibrium equation: NH_3 + H_2O leftrightarrow NH_4^+ + OH^-.

Knowing the concentration of the weak base and its K_b (the base dissociation constant), one can calculate the extent of dissociation and thus the concentration of OH^- ions in solution. This is particularly important when determining the pH of the solution or when figuring out the solubility of other substances, like in the example of Cu(OH)_{2} precipitation.