Question

Estimate the total \(\left[\mathrm{Cl}^{-}\right]\) required in a solution that is initially \(0.10 \mathrm{M} \mathrm{CuSO}_{4}\) to produce a visible yellow color. \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}+4 \mathrm{Cl}^{-} \rightleftharpoons\left[\mathrm{CuCl}_{4}\right]^{2-}+4 \mathrm{H}_{2} \mathrm{O}\) \(K_{f}=4.2 \times 10^{5}\) Assume that \(99 \%\) conversion of \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) to \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) is sufficient for this to happen, and ignore the presence of any mixed aqua- chloro complex ions.

Short answer

Therefore, approximately 0.396M concentration of chloride ions is required to produce a yellow color in the 0.10M \(CuSO_4\) solution.

Step-by-step solution

Write down the given quantities

The initial concentration of copper solution is \(0.10M\). The equilibrium constant for the reaction is \(4.2 \times 10^{5}\). And 99% conversion of \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\) to \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) is considered enough for the color change.

Determine the converted concentration

Calculate 99% of the initial concentration to determine the amount that is converted to \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) complex. 0.99 * 0.10 = 0.099M

Use the equilibrium constant

Since the equilibrium constant \(K_f\) is given and we know the concentration of the product, we can use this information to calculate the concentration of \(\mathrm{Cl}^-\). From the reaction we see that 4 \(\mathrm{Cl}^-\) ions are required for each \(\left[\mathrm{Cu}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right]^{2+}\). Therefore we multiply the concentration of \(\left[\mathrm{CuCl}_{4}\right]^{2-}\) by 4 to get the concentration of \(\mathrm{Cl}^-\). 4 * 0.099 = 0.396M.

Key concepts

Understanding Chemical Equilibria

At its core, chemical equilibria refer to a state where a chemical reaction and its reverse are occurring at the same rate, resulting in no net change in the concentration of reactants and products over time. It's a bit like two equally strong teams playing tug of war – neither side wins because they're both applying the same amount of force. In the context of our exercise, the equilibrium is established between the hydrated copper ion, \(\left[\text{Cu}\left(\text{H}_2 \text{O}\right)_4\right]^{2+}\), and the chloro complex, \(\left[\text{CuCl}_4\right]^{2-}\), in water.

The position of the equilibrium in a chemical reaction is quantitative and can be expressed by the equilibrium constant, \(K\). Chemical equilibria are dynamic – while the concentrations of each species remain constant, the molecules continue to react. An amazing aspect of equilibria is that they can be shifted by changing conditions such as concentration, temperature, or pressure, based on Le Châtelier's Principle. This concept is a cornerstone in understanding chemical reactions in various fields like biochemistry, environmental science, and engineering.

In learning environments, leveraging visual aids such as graphs showing the rates of forward and reverse reactions or interactive simulations can help students visualize this dynamic balance and better grasp the concept of chemical equilibria.

Equilibrium Constant Calculation

Now let's dive into the equilibrium constant calculation, a crucial aspect in predicting the extent of a reaction. The equilibrium constant \(K\) provides a numeric measure of the position of the equilibrium. It's calculated based on the concentrations of the products and the reactants raised to the power of their stoichiometric coefficients in the balanced chemical equation. Generally, \(K\) is expressed as \([Products]/[Reactants]\), where concentration is denoted in molarity (M).

In the case of our exercise, we're given the equilibrium constant \(K_f = 4.2 \times 10^{5}\) for the formation of the copper chloride complex. If \(K_f\) is very large, as it is here, the reaction strongly favors the products at equilibrium - meaning almost all reactants convert into products. This knowledge is vital, as it informs us that to achieve the visible yellow color, a significant fraction of the initial copper must be converted to the complex. Understanding how to calculate and interpret \(K\) enables students to predict how a system will react to changes and is a fundamental concept in chemistry.

Copper Chloride Complexes

Copper chloride complexes, such as the \(\left[\text{CuCl}_4\right]^{2-}\) mentioned in our problem, are fascinating examples of transition metal complexes where copper pairs with chloride ions to form a compound. These complexes are known for their distinct colors and are useful in many applications, including art for producing vibrant pigments and in industry for their catalytic properties.

The color change to a visible yellow occurs when copper ions in solution bind with chloride ions to form the \(\left[\text{CuCl}_4\right]^{2-}\) complex. This reaction is reversible, and the ratio between the concentrations of the different species involved is determined by the equilibrium constant \(K_f\). These complexes also illustrate the principle of coordination chemistry, where the number of ligands (in this case, chloride ions) attached to a central metal ion (copper) defines the complex's properties. In educational settings, discussing copper chloride complexes can make an appealing practical demonstration of coordination chemistry and equilibrium concepts.

Ionic Concentration Calculations

Ionic concentration calculations play a pivotal role in understanding chemical reactions in solution. They are the backbone of realizing how much of each ion is present in a solution and how those ions interact to form various compounds. In our textbook solution, we had to calculate the total chloride ion concentration required for copper sulfate to convert into the copper chloride complex, leading to a color change.

The calculation emphasized that for every one mole of \(\left[\text{Cu}\left(\text{H}_2 \text{O}\right)_4\right]^{2+}\), four moles of chloride ions are needed to form \(\left[\text{CuCl}_4\right]^{2-}\). The careful accounting for these stoichiometric ratios is essential for accuracy in laboratory work and industry processes. To improve understanding, teachers often use hands-on activities where students mix actual solutions and measure concentrations, giving them practical experience with the abstract concept of molarity. This practice reinforces the connection between theoretical calculations and real-world applications, aiding the student in grasping ionic concentration calculations thoroughly.