Problem 43 - Chemistry: Principles and Reactions

Question

Write an overall net ionic equation and calculate \(K\) for the reaction where \(\mathrm{CuCl}\left(K_{s p}=1.9 \times 10^{-7}\right)\) is dissolved by \(\mathrm{NaCN}\) to form \(\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]^{-}\) \(\left(K_{f}=1.0 \times 10^{16}\right)\).

Step-by-Step Solution

Answer

Question: Calculate the overall equilibrium constant for the reaction between \(CuCl\) and \(NaCN\), given that the solubility product constant \(K_{sp}\) for \(CuCl\) is \(1.9 \times 10^{-7}\) and the formation constant \(K_{f}\) for the complex ion \(\left[Cu(CN)_{2}\right]^{-}\) is \(1.0 \times 10^{16}\). Answer: The overall equilibrium constant for the reaction is \(5.26 \times 10^{22}\).

Step 1: Write the balanced reaction equation

To write the balanced equation for the reaction between \(\mathrm{CuCl}\) and \(\mathrm{NaCN}\), we know that \(\mathrm{CuCl}\) will react with \(\mathrm{CN}^{-}\) from the \(\mathrm{NaCN}\) to form \(\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]^{-}\) and \(\mathrm{NaCl}\). The balanced equation will be: \(\mathrm{CuCl + 2 NaCN \rightarrow Na[Cu(CN)_{2}] + NaCl}\)

Step 2: Write the ionic equation

Now, we will break down the ions in the reaction to write the ionic equation. Keep in mind that not all species will dissociate into ions. In this case, only the ionic compounds will dissociate. \(\mathrm{CuCl (s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)}\) \(\mathrm{2 NaCN (aq) \rightarrow 2 Na^{+}(aq) + 2 CN^{-}(aq)}\) \(\mathrm{Na[Cu(CN)_{2}] (aq) \rightarrow Na^{+}(aq) + [Cu(CN)_{2}]^{-}(aq)}\) \(\mathrm{NaCl (aq) \rightarrow Na^{+}(aq) + Cl^{-}(aq)}\) So, the complete ionic equation is: \(\mathrm{Cu^{+}(aq) + Cl^{-}(aq) + 2 Na^{+}(aq) + 2 CN^{-}(aq) \rightarrow Na^{+}(aq) + [Cu(CN)_{2}]^{-}(aq) + Na^{+}(aq) + Cl^{-}(aq)}\)

Step 3: Write the net ionic equation

The net ionic equation is obtained by eliminating the spectator ions, which are those ions that are not involved in the reaction. In this case, the spectator ions are \(\mathrm{Na^{+}}\) and \(\mathrm{Cl^{-}}\). The net ionic equation will be: \(\mathrm{Cu^{+}(aq) + 2 CN^{-}(aq) \rightarrow [Cu(CN)_{2}]^{-}(aq)}\)

Step 4: Calculate the overall equilibrium constant \(K\)

To find the overall equilibrium constant \(K\), we use the solubility product constant \(K_{sp}\) for \(\mathrm{CuCl}\) and the formation constant \(K_{f}\) for the complex ion \(\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]^{-}\) as follows: \(K = \frac{K_{f}}{K_{sp}}\) Given: \(K_{sp}(\mathrm{CuCl}) = 1.9 \times 10^{-7}\) \(K_{f}(\left[\mathrm{Cu}(\mathrm{CN})_{2}\right]^{-}) = 1.0 \times 10^{16}\) Calculate \(K\): \(K = \frac{1.0 \times 10^{16}}{1.9 \times 10^{-7}}\) \(K = 5.26 \times 10^{22}\) The overall equilibrium constant for the reaction is \(K = 5.26 \times 10^{22}\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products over time. It is a dynamic balance, where the reactions are still occurring, but there are no observable changes as the amounts of reactants and products remain constant. This concept is crucial when dealing with reactions such as the one between copper(I) chloride and sodium cyanide, where the formation of a complex ion can be represented by an equilibrium expression.

In equilibrium, the law of mass action applies, allowing for the use of an equilibrium constant to quantify the ratio of the concentration of products over reactants, raised to the power of their stoichiometric coefficients. For the reaction in our exercise, the net ionic equation signifies the change happening in the aqueous solution and helps us focus on the ions directly involved in forming the complex ion, thereby aiding in the determination of the equilibrium constant.

Solubility Product Constant

The solubility product constant (\(K_{sp}\)) is a special type of equilibrium constant that measures the solubility of a sparingly soluble ionic compound. It is the product of the concentrations of the ions each raised to the power of their respective coefficients in the balanced equation. In the context of our problem, the solubility product constant pertains to the dissociation of solid copper(I) chloride (\texttt{CuCl}) into its ions in an aqueous solution.

Importance in Net Ionic Reactions

The value of the solubility product constant provides information on the degree to which \texttt{CuCl} dissolves and can thus be predictive of whether a precipitate will form under certain conditions. If a reaction's quotient exceeds the solubility product, a precipitate will likely form, indicating an imbalance that will shift towards equilibrium. Conversely, if the reaction quotient is less, the system has not reached saturation, and no precipitate is expected to form. Knowing the value of the solubility product constant is fundamental for scientists and chemists when predicting the outcome of reactions involving ionic compounds.

Formation Constant

The formation constant (\(K_f\)) is an equilibrium constant for the formation of a complex ion from its constituent ions in solution. It is a particular case of the equilibrium constant that applies to complex formation reactions, which are vital in the fields of coordination chemistry and analytical chemistry.

Complex Ions in Solution

A high formation constant, like in the reaction given in the exercise where copper cyanide complexes are formed, indicates a very stable complex ion in solution. This implies that once the complex ion is formed, it will not easily dissociate back into its constituent ions under normal conditions. With the formation constant, chemists can infer how tightly bound the ligands (in this case, cyanide ions) are to the metal ion (copper ion). When put to practical use, these constants can help in designing drugs, extracting metals from ores, or even purifying substances.