Problem 94 - Chemistry

Question

Will a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) form if \(1.0 \mathrm{mL}\) of \(1.0 \mathrm{M}\) \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(1.0 \mathrm{L}\) of \(5.0 \mathrm{MNH}_{3} ?\) $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ $$\mathrm{Cd}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cd}^{2+}(a q)+2 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=5.9 \times 10^{-15}$$

Step-by-Step Solution

Answer

Based on the calculations, there are no free \(\mathrm{Cd}^{2+}\) ions remaining in the solution due to complexation with \(\mathrm{NH}_{3}\), and the concentration of \(\mathrm{OH}^-\) ions is very small. The reaction quotient, Q, is close to 0, which is less than \(K_{\mathrm{sp}}=5.9\times10^{-15}\). Therefore, a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will not form under these conditions.

Step 1: Calculate the moles and final concentrations of all species after mixing

First, we need to calculate the moles of each species after mixing. For that, we can use the formula: moles = volume × concentration. - Moles of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\): \((1.0 \times 10^{-3}\,L) \times (1.0\,\mathrm{M}) = 1.0 \times 10^{-3} \, \text{moles}\) - Moles of \(\mathrm{NH}_{3}\): \((1.0\,L) \times (5.0\,\mathrm{M}) = 5.0\,\text{moles}\) Next, we need to calculate the final concentrations of each species in the mixed solution: - Final volume of the solution: \(1.0\,L + 1.0\times10^{-3}\,L = 1.001\,L\) - Concentration of \(\mathrm{Cd}^{2+}\): \(\frac{1.0\times10^{-3}\,\text{moles}}{1.001\,L} \approx 9.99\times10^{-4}\,\mathrm{M}\) - Concentration of \(\mathrm{NH}_{3}\): \(\frac{5.0\,\text{moles}}{1.001\,L} \approx 4.995\,\mathrm{M}\)

Step 2: Calculate the concentration of \(\mathrm{Cd}^{2+}\) remaining in solution after complexation with \(\mathrm{NH}_{3}\)

Now, we will calculate the final concentration of \(\mathrm{Cd}^{2+}\) after reacting with \(\mathrm{NH}_{3}\) to form the complex \(\mathrm{Cd}(\mathrm{NH}_{3})_{4}^{2+}\) using the reaction: $$\mathrm{Cd}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) \quad K=1.0 \times 10^{7}$$ According to the stoichiometry of the reaction, 1 mole of \(\mathrm{Cd}^{2+}\) reacts with 4 moles of \(\mathrm{NH}_{3}\). Thus, all 1.0×10^(-3) moles of \(\mathrm{Cd}^{2+}\) react with \(\frac{1}{4} \times 1.0\times10^{-3}\, \text{moles} = 2.5\times10^{-3}\, \text{moles}\) of \(\mathrm{NH}_{3}\). After the reaction, the concentration of each species becomes: - \(\mathrm{Cd}^{2+}\): \(0\,\mathrm{M}\) - \(\mathrm{NH}_{3}\): \(\frac{5.0\,\text{moles} - 2.5\times10^{-3}\,\text{moles}}{1.001\,L} \approx 4.993\,\mathrm{M}\) There will be no free \(\mathrm{Cd}^{2+}\) ions remaining in the solution since all are complexed with ammonia.

Step 3: Calculate the concentration of \(\mathrm{OH}^-\) ions

As ammonia is a weak base, it can react with water to produce \(\mathrm{OH}^-\) ions. $$\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K_{b} = 1.8 \times 10^{-5}$$ However, since the concentration of \(\mathrm{NH}_{3}\) is very high (about 4.993 M) compared to the \(K_{b}\) value, we can assume that the \(\mathrm{OH}^-\) concentration is very small and negligible.

Step 4: Calculate the reaction quotient and compare to \(K_{sp}\)

Now, we can calculate the reaction quotient, Q, for the solubility equilibrium of \(\mathrm{Cd}(\mathrm{OH})_{2}\) using the concentrations of the ions in the solution: $$Q = [\mathrm{Cd}^{2+}][\mathrm{OH}^{-}]^{2}$$ With no free \(\mathrm{Cd}^{2+}\) ions and an insignificant concentration of \(\mathrm{OH}^-\), Q will be close to 0. Since \(Q < K_{\mathrm{sp}}=5.9\times10^{-15}\), a precipitate of \(\mathrm{Cd}(\mathrm{OH})_{2}\) will not form under these conditions.

Key Concepts

Solubility Product

In chemistry, the solubility product constant, denoted as \( K_{sp} \), is the equilibrium constant for a solid substance dissolving in an aqueous solution. The \( K_{sp} \) value indicates the extent to which a compound can dissolve in water. It is specific to a particular compound at a given temperature. Consider the scenario of a sparingly soluble compound like cadmium hydroxide, \( \mathrm{Cd(OH)_2} \).
When this compound dissolves in water, it breaks down into cadmium ions and hydroxide ions: \( \mathrm{Cd(OH)_2 (s) \rightleftharpoons Cd^{2+} (aq) + 2OH^- (aq)} \).
The solubility product expression for this is \( K_{sp} = [\mathrm{Cd^{2+}}][\mathrm{OH^-}]^2 \). When the concentrations of the ions in a solution multiply to a value equal to or greater than the \( K_{sp} \), the solution becomes saturated, and no more solute can dissolve. If the value of the product is less than \( K_{sp} \), the solution is unsaturated. Calculating and understanding \( K_{sp} \) helps predict whether a precipitate will form when two solutions are mixed.

Complex Ion Formation

Complex ion formation involves the interaction between a metal ion and ligands (molecules or ions that donate electrons to the metal ion), resulting in the formation of a complex ion. This is an important concept in coordination chemistry.
Consider the case of cadmium ions \( \mathrm{Cd^{2+}} \) in the presence of ammonia \( \mathrm{NH_3} \). The cadmium ions can react with ammonia to form a complex ion: \( \mathrm{Cd^{2+} + 4 NH_3 \rightleftharpoons Cd(NH_3)_4^{2+}} \).
The equilibrium constant for this reaction is denoted by \( K_{f} \) and is very large \( (K = 1.0 \times 10^{7}) \), indicating that in the presence of a considerable amount of ammonia, virtually all \( \mathrm{Cd^{2+}} \) ions will form the complex ion. The formation of complex ions often increases the solubility of otherwise insoluble compounds by binding the free metal ions in the solution, reducing their concentration and preventing precipitation.

Precipitation Reaction

Precipitation reactions occur when two aqueous solutions are mixed, resulting in the formation of an insoluble solid called a precipitate. This reaction is a type of double displacement reaction.
For a precipitation to occur, the product of the concentrations of the resulting ions must exceed the solubility product \( K_{sp} \) of the compound. Consider the mixing of solutions containing \( \mathrm{Cd^{2+}} \) and \( \mathrm{OH^-} \), which can form cadmium hydroxide \( \mathrm{Cd(OH)_2} \) as a precipitate if their ion product exceeds the \( K_{sp} \) value of \( 5.9 \times 10^{-15} \).
If the ion concentrations are such that this product is less than \( K_{sp} \), then the solution remains unsaturated, and no precipitate forms. In the case study, complex ion formation with ammonia reduces the concentration of free \( \mathrm{Cd^{2+}} \), thus preventing cadmium hydroxide precipitation by keeping the ion product below the \( K_{sp} \). Understanding these reactions allows chemists to predict and control the formation of precipitates.