Question

\(K_{\text {eq }}=3.9 \times 10^{-11}\) for the dissolution of calcium fluoride in water: \(\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2 \mathrm{~F}^{-}(a q)\) (a) What is another name for \(K_{\mathrm{eq}}\) for this reaction? (b) If the equilibrium calcium ion concentration in a saturated aqueous solution of calcium fluoride is \(3.3 \times 10^{-4} \mathrm{M}\), what is the equilibrium fluoride ion concentration? (c) Which is larger, the rate constant for the forward reaction or the rate constant for the reverse reaction? (d) Which is larger, \(E_{a}\) for the forward reaction or \(E_{\mathrm{a}}\) for the reverse reaction? (e) Which is larger, the rate of the forward reaction or the rate of the reverse reaction? (f) For lithium carbonate, \(K_{\mathrm{sp}}=0.0011\). Write the balanced chemical equation and the equilibrium expression for the dissolution of \(\mathrm{Li}_{2} \mathrm{CO}_{3}\) in water. (g) Which is more soluble in water, calcium fluoride or lithium carbonate?

Short answer

The short version of the answer is as follows: (a) The other name for the equilibrium constant (\(K_{\text{eq}}\)) for this dissolution process is the solubility product constant (\(K_{\text{sp}}\)). (b) The equilibrium fluoride ion concentration in a saturated aqueous solution of calcium fluoride is \(3.435 \times 10^{-4} \mathrm{M}\). (c) The rate constant for the forward reaction is smaller than the rate constant for the reverse reaction. (d) The activation energy of the forward reaction (\(E_a\)) is larger than the activation energy of the reverse reaction. (e) At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. (f) The balanced equation for the dissolution of lithium carbonate is \(\mathrm{Li}_{2}\mathrm{CO}_{3}(s) \rightleftarrows 2\mathrm{Li}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\) and the corresponding equilibrium expression is \(K_{\text{sp}} = [\mathrm{Li}^+]^2 [\mathrm{CO}_3^{2-}]\). (g) Lithium carbonate is more soluble in water than calcium fluoride, as its solubility product constant is higher (\(K_{\text{sp}}=0.0011\)).

Step-by-step solution

Answer:

The other name for the equilibrium constant (\(K_{\text{eq}}\)) for this dissolution process is called the solubility product constant, represented as \(K_{\text{sp}}\). #b) Equilibrium fluoride ion concentration#

Write the balanced equation

The balanced equation for the dissolution of calcium fluoride is given as: \[\mathrm{CaF}_{2}(s) \rightleftarrows \mathrm{Ca}^{2+}(a q)+2\mathrm{F}^{-}(a q)\]

Write the equilibrium expression

According to the balanced equation, the equilibrium expression is: \[K_{\text{sp}} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2\]

Find the fluoride ion concentration

The equilibrium concentrations of calcium ions are given (\(3.3 \times 10^{-4} \mathrm{M}\)). Substitute the given values and \(K_{\text {sp }}=3.9 \times 10^{-11}\) into the equilibrium expression and solve for the fluoride ion concentration. \[3.9 \times 10^{-11} = (3.3 \times 10^{-4})[\mathrm{F}^-]^2\] \[[\mathrm{F}^-]^2 = \frac{3.9 \times 10^{-11}}{3.3 \times 10^{-4}}\] \[[\mathrm{F}^-]^2 = 1.18 \times 10^{-7}\] \[[\mathrm{F}^-] =\sqrt{1.18 \times 10^{-7}}\] \[[\mathrm{F}^-] = 3.435\times 10^{-4} \mathrm{M}\]

Answer:

The equilibrium fluoride ion concentration in a saturated aqueous solution of calcium fluoride is \(3.435 \times 10^{-4} \mathrm{M}\). #c) Comparing forward and reverse rate constants#

Answer:

Because \(K_{\text{sp}}=3.9 \times 10^{-11}\), which is very small, it indicates that the dissolution of calcium fluoride is not very favorable. The forward reaction is not very fast, and the reverse reaction is dominant. So, the rate constant for the forward reaction should be smaller than the rate constant for the reverse reaction. #d) Comparing activation energies of forward and reverse reactions#

Answer:

Since the rate constant for the forward reaction is smaller than the rate constant for the reverse reaction, according to the Arrhenius equation, the activation energy of the forward reaction (\(E_a\)) should be larger than the activation energy of the reverse reaction. #e) Comparing rates of forward and reverse reactions#

Answer:

At equilibrium, the rates of the forward reaction and reverse reaction are equal. Therefore, the rate of the forward reaction is equal to the rate of the reverse reaction. #f) Lithium carbonate dissolution equation and equilibrium expression#

Write the balanced equation

The balanced equation for the dissolution of lithium carbonate is given as: \[\mathrm{Li}_{2}\mathrm{CO}_{3}(s) \rightleftarrows 2\mathrm{Li}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)\]

Write the equilibrium expression

According to the balanced equation, the equilibrium expression is: \[K_{\text{sp}} = [\mathrm{Li}^+]^2 [\mathrm{CO}_3^{2-}]\] #g) Comparing solubility of calcium fluoride and lithium carbonate#

Answer:

To determine which compound is more soluble in water, we can compare their solubility product constants. For calcium fluoride, \(K_{\text{sp}}=3.9 \times 10^{-11}\), whereas for lithium carbonate, \(K_{\text{sp}}=0.0011\). Since the solubility product constant is higher for lithium carbonate, it is more soluble in water than calcium fluoride.

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as K_eq, is a number that describes the ratio of the concentration of products to the concentration of reactants for a reversible chemical reaction at equilibrium. In the context of a solubility reaction, the equilibrium constant is called the solubility product constant, represented by K_sp. This value gives us insight into the extent of a compound's dissolution in water.

For example, in the dissolution of calcium fluoride (CaF₂), the solubility product constant K_sp is extremely low at 3.9 x 10^-11, indicating a reaction that does not proceed extensively towards products, thus CaF₂ is not highly soluble in water. Solubility product constants are crucial for predicting the solubility and precipitation of compounds in solution and are extensively used in areas such as chemistry, environmental science, and engineering.

Equilibrium Concentration

Equilibrium concentration refers to the concentration of each substance in a chemical system at equilibrium. In a saturated solution, it signifies the maximum amount of solute that can be dissolved in the solvent at a given temperature. The equilibrium concentrations of ions in solution can be used in conjunction with the solubility product constant to calculate unknown concentrations.

In our exercise, the relationship between calcium ions and fluoride ions at equilibrium can be expressed mathematically. Given the equilibrium concentration of calcium ions is 3.3 x 10^-4 M, we determine the fluoride ion concentration to be 3.435 x 10^-4 M by manipulating the K_sp expression. This process illustrates the interdependent nature of equilibrium concentrations in soluble product reactions.

Activation Energy

Activation energy, denoted by E_a, is the minimum amount of energy required to initiate a chemical reaction. It is a barrier that reactants must overcome for the reaction to proceed. Higher activation energy means that fewer molecules have enough energy to react, generally resulting in a slower reaction rate.

In the case of calcium fluoride, since the forward reaction has a smaller rate constant, it indicates that it has a higher activation energy compared to the reverse reaction. This is because a smaller rate constant generally correlates with a higher E_a due to the Arrhenius equation. Therefore, the dissolution of calcium fluoride requires more energy to proceed compared to its reformation from ions in solution.

Chemical Dissolution

Chemical dissolution is the process whereby a solid substance (solute) becomes incorporated into a liquid (solvent) to form a solution. The process can be described by a balanced chemical equation indicating the solid and its resultant ions in solution.

For instance, the dissolution of calcium fluoride in water is represented by the equation: CaF₂(s) ⇌ Ca²⁺(aq) + 2F^-(aq). In this process, the solid lattice of calcium fluoride breaks down, and individual ions are solvated by water molecules. The equilibrium of this process is dynamic, with the forward reaction (dissolution) and reverse reaction (precipitation) happening concurrently. Chemical dissolution is influenced by factors like temperature, presence of other ions, and the inherent solubility of the solute.

Rate Constant

The rate constant, symbolized as k, quantifies the speed of a chemical reaction. For any given chemical reaction at a constant temperature, the rate constant connects the reaction rate with the concentrations of the reactants. It is specific to each reaction and can be influenced by factors such as temperature, solvent, and catalysts.

In our problem, it was deduced that calcium fluoride's low K_sp value signals a smaller rate constant for the forward reaction compared to the reverse. This inference stems from the established principle that a small equilibrium constant indicates the reactants are favored over the products, thereby signifying a lower rate of formation of products. Conversely, it implies a rapid reverse reaction, where the rate constant for the formation of the solid from its ions is larger.