Question

Copper hydroxide is only very slightly soluble in water. The reaction by which it goes into solution is: $$\mathrm{Cu}(\mathrm{OH}{)}_2(\mathrm{s})\rightleftarrows {\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{OH}}^{-}(\mathrm{aq})$$ a. Formulate the expression for the equilibrium constant, $K_{\mathrm{sp}},$ for the above reaction. b. It is possible to dissolve significant amounts of $\mathrm{Cu}(\mathrm{OH}{)}_2$ in solutions in which the concentration of either ${\mathrm{Cu}}^{2+}$ or ${\mathrm{OH}}^{-}$ is very, very small. Explain, using $K_{\mathrm{sp}},$ why this is the case. c. Explain why $\mathrm{Cu}(\mathrm{OH}{)}_2$ might have very appreciable solubility in $1\mathrm{M}$ $HCl$. (Consider the effect of Reaction 3 on the $\mathrm{Cu}(\mathrm{OH}{)}_2$ solution reaction.)

Short answer

a. The expression for the equilibrium constant, $K_{sp}$, for the reaction is given by: $$K_{sp}=[{\mathrm{Cu}}^{2+}][{\mathrm{OH}}^{-}{]}^2$$ b. If the concentration of either ${\mathrm{Cu}}^{2+}$ or ${\mathrm{OH}}^{-}$ ions is very small, the reaction shifts to the right according to Le Châtelier's principle, leading to an increased solubility of $\mathrm{Cu}(\mathrm{OH}{)}_2$ in the solution. c. $\mathrm{Cu}(\mathrm{OH}{)}_2$ has appreciable solubility in 1 M HCl because the reaction between ${\mathrm{OH}}^{-}$ and ${\mathrm{H}}^{+}$ ions removes ${\mathrm{OH}}^{-}$ ions from the solution, causing the equilibrium to shift to the right and increasing the solubility of $\mathrm{Cu}(\mathrm{OH}{)}_2$.

Step-by-step solution

1. Formulate the expression for the equilibrium constant Ksp

We are given the following reaction for copper hydroxide dissolving in water: $$\mathrm{Cu}(\mathrm{OH}{)}_2(\mathrm{s})\rightleftarrows {\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{OH}}^{-}(\mathrm{aq})$$ To find the expression for the solubility product constant, $K_{sp}$, we just need to multiply the concentrations of the products raised to their respective stoichiometric coefficients, and divide by the concentration of the reactants raised to their respective coefficients. Since the reactant is a solid, we don't include it in the expression. Hence, for this reaction, the equation for $K_{sp}$ will be: $$K_{sp}=[{\mathrm{Cu}}^{2+}][{\mathrm{OH}}^{-}{]}^2$$

2. Explain why significant amounts of Cu(OH)₂ can dissolve in solutions with very small concentrations of Cu²⁺ or OH⁻

According to Le Châtelier's principle, a system at equilibrium will shift to counteract any disturbance applied to it. In this case, if the concentration of either ${\mathrm{Cu}}^{2+}$ or ${\mathrm{OH}}^{-}$ ions is very small, the reaction will shift to the right to re-establish the equilibrium and return to the solubility product constant, $K_{sp}$. This means more copper hydroxide will dissolve, leading to an increased solubility of $\mathrm{Cu}(\mathrm{OH}{)}_2$ in the solution.

3. Explain why Cu(OH)₂ might have very appreciable solubility in 1 M HCl

When $\mathrm{Cu}(\mathrm{OH}{)}_2$ is added to a 1 M HCl solution, the following reaction occurs between ${\mathrm{OH}}^{-}$ and ${\mathrm{H}}^{+}$ ions, which are present due to the dissociation of HCl: $${\mathrm{OH}}^{-}(\mathrm{aq})+{\mathrm{H}}^{+}(\mathrm{aq})\to {\mathrm{H}}_2\mathrm{O}(\mathrm{l})$$ This reaction will remove ${\mathrm{OH}}^{-}$ ions from the solution. According to Le Châtelier's principle, the equilibrium of the copper hydroxide solubility reaction will shift to the right, trying to compensate for the removed ${\mathrm{OH}}^{-}$ ions. As a result, more $\mathrm{Cu}(\mathrm{OH}{)}_2$ will dissolve in the solution, hence increasing its solubility in 1 M HCl.

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is pivotal when examining reactions such as the dissolution of copper hydroxide in water. In general, the equilibrium constant for a reaction expresses the ratio of the concentration of products to reactants at equilibrium. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For insoluble solids dissolving into ions, the equilibrium constant is known as the solubility product constant (Ksp), which only includes the concentrations of the dissolved ions. Let's consider the specific case of copper hydroxide:
For the dissolution reaction $$\mathrm{Cu}(\mathrm{OH}{)}_2(\mathrm{s})\rightleftarrows {\mathrm{Cu}}^{2+}(\mathrm{aq})+2{\mathrm{OH}}^{-}(\mathrm{aq})$$ the equilibrium constant expression (Ksp) is given by $$K_{sp}=[{\mathrm{Cu}}^{2+}][{\mathrm{OH}}^{-}{]}^2$$ Note that the concentration of solid copper hydroxide is excluded because its concentration remains constant over time.

Le Châtelier's Principle

Le Châtelier's principle is a valuable tool in predicting how a change in conditions can affect the equilibrium position of a reaction. It states that if an external stress is applied to a system at equilibrium, the system adjusts itself to minimize the stress and re-establish equilibrium. Let's apply this to the dissolution of copper hydroxide: if the concentration of either ${\mathrm{Cu}}^{2+}$ ions or ${\mathrm{OH}}^{-}$ ions is reduced, for instance by adding another reagent that reacts with one of them, Le Châtelier's principle suggests that the equilibrium will shift to produce more of those ions. This shift increases the solubility of $\mathrm{Cu}(\mathrm{OH}{)}_2$ because more solid dissolves to replace the ions that were removed, aligning with the principle's prediction.

Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is a specific type of equilibrium constant that applies to the solubility of sparingly soluble compounds. It is a direct measure of how much of a solid can dissolve in water to form a saturated solution. A higher Ksp implies greater solubility, and vice versa. For example, copper hydroxide has a low Ksp, indicating that under normal circumstances, it is not very soluble in water. However, certain conditions, such as a reduction in the concentration of ${\mathrm{Cu}}^{2+}$ or ${\mathrm{OH}}^{-}$ ions in the solution, can increase the extent of its dissolution, as the system seeks to restore equilibrium in response to these changes.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical process are equal, leading to no net change in the concentration of reactants and products over time. It is a dynamic state, meaning that the reactions continue to occur, but the overall concentrations remain constant. The concept of chemical equilibrium is particularly relevant when discussing soluble and insoluble substances. Even insoluble substances like copper hydroxide are in a state of dynamic equilibrium in saturated solutions, where solid particles dissolve at the same rate as dissolved particles precipitate out. By applying stresses to the equilibrium, such as the addition of an acid like HCl, we can shift the equilibrium to favor the forward or reverse reaction, altering the solubility of the compound in question.