Question

a. Using the \(K_{\mathrm{sp}}\) value for \(\mathrm{Cu}(\mathrm{OH})_{2}\left(1.6 \times 10^{-19}\right)\) and the overall formation constant for \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\left(1.0 \times 10^{13}\right)\) calculate the value for the equilibrium constant for the following reaction: $$\mathrm{Cu}(\mathrm{OH})_{2}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)$$ b. Use the value of the equilibrium constant you calculated in part a to calculate the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 M\).

Short answer

The equilibrium constant for the given reaction is \(1.6 \times 10^{-32}\), and the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\) is approximately \(4.52 \times 10^{-5} M\).

Step-by-step solution

Calculate the equilibrium constant (K) for the given reaction

To calculate the equilibrium constant for the given reaction, we need to use the given \(K_{\mathrm{sp}}\) value and the overall formation constant. The equation for the equilibrium constant is as follows: \[K = \frac{K_{\mathrm{sp}}}{K_{\mathrm{formation}}}\] In this case, we have: \[K = \frac{1.6 \times 10^{-19}}{1.0 \times 10^{13}}\] Now, we can calculate the equilibrium constant K: \[K = 1.6 \times 10^{-32}\]

Find the solubility (in mol/L) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\)

To find the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\), we can use the following expression involving the given concentrations and the equilibrium constant that we found in step 1: \[K = \frac{\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}[OH^{-}]^2}{\left[\mathrm{NH}_{3}\right]^4}\] From the problem, we are given that \(\mathrm{OH}^{-}\) has a concentration of \(0.0095M\), and \(\mathrm{NH}_{3}\) has a concentration of \(5.0M\). Using these values, we can solve for the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\). We will represent this concentration as \(x\): \[1.6 \times 10^{-32} = \frac{x \cdot (0.0095)^2}{(5.0)^4}\] Solving for \(x\), we obtain the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\): \[x = \frac{1.6 \times 10^{-32} \cdot (5.0)^4}{(0.0095)^2}\] \[x \approx 9.03 \times 10^{-5} M\] Now we can find the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\) using the relationship between the concentration of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) and \(\mathrm{Cu}(\mathrm{OH})_{2}\): \[s = \frac{1}{2} [\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}]\] \[s = \frac{1}{2} (9.03 \times 10^{-5} M)\] \[s \approx 4.52 \times 10^{-5} M\] So, the solubility of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 M \mathrm{NH}_{3}\) is approximately \(4.52 \times 10^{-5} M\).

Key concepts

Solubility Product

The solubility product, often represented as \(K_{sp}\), is a critical concept in chemistry that helps us understand how much of a compound can dissolve in solution before reaching saturation. Specifically, \(K_{sp}\) is the equilibrium constant for the dissociation of a solid substance into its constituent ions in a solution. For any slightly soluble salt, such as \(\mathrm{Cu(OH)}_{2}\), the extent to which it dissolves in water can be quantitatively described using its solubility product.For example, if \(\mathrm{Cu(OH)}_{2}\) dissolves in water, it dissociates as follows:\[\mathrm{Cu(OH)}_{2}(s) \rightleftharpoons \mathrm{Cu^{2+}}(aq) + 2\mathrm{OH^{-}}(aq)\]The \(K_{sp}\) for this reaction would be:\[K_{sp} = [\mathrm{Cu^{2+}}][\mathrm{OH^{-}}]^2\]This mathematical representation allows us to calculate how much of the solid will dissolve and react with ions in solution. The lower the \(K_{sp}\) value, the less soluble the compound is. In the exercise, the \(K_{sp}\) value given for \(\mathrm{Cu(OH)}_{2}\) is \(1.6 \times 10^{-19}\), which indicates low solubility.

Formation Constant

The formation constant, denoted by \(K_{\text{formation}}\), is an equilibrium constant for the formation of a complex ion from a metal ion and ligands. This is essential in coordination chemistry, where transition metals often form complex ions.For the complex ion \(\mathrm{Cu(NH}_3)_4^{2+} \), it involves the copper ion and ammonia acting as ligands. The formation reaction can be described as:\[\mathrm{Cu^{2+}}(aq) + 4\mathrm{NH}_3(aq) \rightleftharpoons \mathrm{Cu(NH}_3)_4^{2+}(aq)\]The equilibrium constant for this reaction, \(K_{\text{formation}}\), reflects how strongly the copper ion binds with ammonia to form this stable complex. In the problem, the \(K_{\text{formation}}\) provided is a high value \((1.0 \times 10^{13})\), indicating a strong tendency for \(\mathrm{Cu^{2+}}\) to form \(\mathrm{Cu(NH}_3)_4^{2+}\). This high constant is crucial for calculating the overall equilibrium constant for reactions where such complexes are formed.

Solubility Calculation

Solubility calculations help chemists determine how much of a compound can dissolve in a given solvent under certain conditions. These calculations are particularly useful when dealing with salts in various environments.In the given exercise, the solubility of \(\mathrm{Cu(OH)}_{2}\) in an ammonia solution was calculated using the equilibrium constant derived from the \(K_{sp}\) and the formation constant. By rearranging the equilibrium expression:\[K = \frac{K_{sp}}{K_{\text{formation}}}\]We find that \(K = 1.6 \times 10^{-32}\). This helps us understand how \(\mathrm{Cu(OH)}_{2}\) behaves in the presence of 5.0 M \(\mathrm{NH}_3\). By inserting the concentrations of \(\mathrm{NH}_3\) and \(\mathrm{OH}^-\) into the equilibrium expression, we can solve for \([\mathrm{Cu(NH}_3)_4^{2+}]\), represented by \(x\):\[1.6 \times 10^{-32} = \frac{x \cdot (0.0095)^2}{(5.0)^4}\]The solution to this equation provided the concentration \([\mathrm{Cu(NH1}_3)_4^{2+}] \approx 9.03 \times 10^{-5} M\). Then, we calculated the solubility of \(\mathrm{Cu(OH)}_{2}\) from this concentration as approximately \(4.52 \times 10^{-5} M\), considering the stoichiometry of the dissolution process.

Chemical Equilibrium

Chemical equilibrium refers to the balanced state of a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of reactants and products.When dealing with systems involving multiple equilibria, such as when both solubility and complex formation are in play, understanding how these equilibria interact is crucial. For the reaction presented in the exercise:\[\mathrm{Cu(OH)}_{2}(s) + 4 \mathrm{NH}_3(aq) \rightleftharpoons \mathrm{Cu(NH}_3)_4^{2+}(aq) + 2 \mathrm{OH}^-(aq)\]The equilibrium constant \(K\) calculated serves to describe how these species are balanced in solution. Given that the dissolution of \(\mathrm{Cu(OH)}_{2}\) and the formation of \(\mathrm{Cu(NH}_3)_4^{2+}\) are concurrent, comprehending their individual constants and using them to find the overall \(K\) helps in predicting the solubility of chemicals under different conditions.Moving to equilibrium concepts helps us understand that changes in the concentration of a reactant or product will shift the reaction in the direction that rebalances these concentrations, which is key in many practical applications of chemistry, like predicting product yields and dissolution rates in various solvent systems.