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 \(\mathrm{mol} / \mathrm{L}\) ) of \(\mathrm{Cu}(\mathrm{OH})_{2}\) in \(5.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\) In \(5.0 \mathrm{M} \mathrm{NH}_{3}\) the concentration of \(\mathrm{OH}^{-}\) is \(0.0095 \mathrm{M} .\)

Short answer

The solubility of Cu(OH)_2 in 5.0 M NH3 solution is approximately \(1.55 \times 10^{-4} \, mol \, L^{-1}\).

Step-by-step solution

Calculate the equilibrium constant K_eq for the reaction

To calculate the equilibrium constant (K_eq) for the given reaction, we need to consider the relationships between K_sp, the overall formation constant (K_f), and the concentrations of the species involved in the reaction. K_eq is the ratio of K_f to K_sp, expressed algebraically as follows: \(K_{eq} = \frac{K_f}{K_{sp}}\) Plug in the given values of K_sp and K_f: \(K_{eq} = \frac{1.0 \times 10^{13}}{1.6 \times 10^{-19}}\) Calculate K_eq: \(K_{eq} = 6.25 \times 10^{31}\)

Calculate the solubility of Cu(OH)_2 in 5.0 M NH3

To find the solubility of Cu(OH)_2 in 5.0 M NH3, we need to consider the reaction's stoichiometry and the given equilibrium constant K_eq. We are looking for [Cu(OH)_2]. Let's denote the solubility of Cu(OH)_2 as S. According to the reaction stoichiometry, the concentrations of the reactants and products at equilibrium will be: - [Cu(OH)_2] = S - [NH3] = 5.0 - 4S (since 4 moles of NH3 reacts with 1 mole of Cu(OH)_2) - [Cu(NH3)_4^(2+)] = S - [OH-] = 0.0095 + 2S (initially 0.0095, and 2 moles of OH- are produced from 1 mole of Cu(OH)_2 at equilibrium) Now, we can construct the equilibrium expression for the given reaction using the calculated K_eq: \(K_{eq} = \frac{[Cu(NH3)_{4}^{2+}] * [OH^-]^2}{[Cu(OH)_2]*[NH3]^4}\) Plug in the values for K_eq and the equilibrium concentrations of the reactants and products: \(6.25 \times 10^{31} = \frac{S * (0.0095 + 2S)^2}{ S(5.0 - 4S)^4}\) Solve this non-linear equation for S, the solubility of Cu(OH)_2: \(S \approx 1.55 \times 10^{-4} \, mol \, L^{-1}\) Hence, the solubility of Cu(OH)_2 in 5.0 M NH3 solution is approximately \(1.55 \times 10^{-4} \, mol \, L^{-1}\).

Key concepts

Solubility

Solubility refers to how much of a substance can dissolve in a given solution. For example, Cu(OH)_2 dissolves slightly in water. The measure of solubility tells us about how many moles per liter (mol/L) of Cu(OH)_2 can dissolve before it reaches saturation. Higher solubility means more of the compound can be dissolved. In many chemical problems, we must calculate solubility using equilibrium expressions, so understanding the basics of solubility is key.

In this exercise, solubility involves finding out how much Cu(OH)_2 can dissolve in a 5.0 M NH3 solution. By setting an equilibrium expression for the chemical reaction, we can determine the maximum concentration of Cu(OH)_2 that dissolves, given the specific conditions.

K_f (Formation Constant)

The formation constant, denoted as K_f, is crucial for understanding how stable a complex ion is when it forms in a solution. It tells us how likely it is for the complex ion to form compared to breaking back into its components.

In the reaction Cu(OH)_2 + 4 NH3 ⇌ Cu(NH3)_4^{2+} + 2 OH^-, the stability of the complex ion Cu(NH3)_4^{2+} is represented by K_f, which is given as 1.0 × 10^{13}. This high value suggests that once Cu(NH3)_4^{2+} forms, it’s not likely to break back down into Cu(OH)_2 and NH3 easily. This indicates a strong tendency for complex formation.

K_sp (Solubility Product)

The solubility product, K_sp, quantifies the solubility of compounds that slightly dissolve in water. It's specific to sparingly soluble salts like Cu(OH)_2, and it helps us understand the extent to which these salts can dissolve.

K_sp is determined by the product of the molar concentrations of the ions involved in the dissolution, each raised to the power of their coefficients in the balanced equation. For Cu(OH)_2, K_sp is given as 1.6 × 10^{-19}, meaning that Cu(OH)_2 isn't very soluble under standard conditions.

In this problem, we're using K_sp to find out how much Cu(OH)_2 can dissolve in a solution influenced by NH3 acting as a ligand, thereby also considering the complex ion's formation.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction matches the rate of the backward reaction in a chemical system. At equilibrium, the concentrations of reactants and products remain constant over time.

In this exercise, for the reaction involving Cu(OH)_2 and NH3, equilibrium means that the forward reaction where Cu(NH3)_4^{2+} and OH^- form, is balanced by the backward reaction. The equilibrium constant, K_eq, relates the concentrations of all species in equilibrium. Calculating K_eq, therefore, involves using both K_sp and K_f.

This balance (equilibrium) helps us predict how much of each substance is present when the system has settled and doesn’t change anymore.

Cu(OH)_2

Copper(II) hydroxide, Cu(OH)_2, is a light blue solid known for being not very soluble in water. It’s important in various reactions and industrial processes.

In this exercise, understanding Cu(OH)_2 is key as it sets the stage for forming Cu(NH3)_4^{2+} in the presence of ammonia. Despite its low solubility, its ability to react with strong ligands like NH3 underlines its importance in forming complex ions.

The study of Cu(OH)_2 in chemical equilibria requires knowledge of both its solubility and its behavior in complex reactions involving ligands.