Problem 104 - Chemistry The Science in Context

Question

In which of the following solutions will \(\mathrm{CaF}_{2}\) be most soluble? (a) \(0.010 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ;\) (b) \(0.01 M \mathrm{NaF}\) (c) \(0.001 M\) NaF; (d) \(0.10 M \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\)

Step-by-Step Solution

Answer

Answer: CaF₂ will be most soluble in solution (c) with 0.001 M NaF.

Step 1: Write the balanced equation and \(K_{sp}\) expression for the dissolution of \(\mathrm{CaF}_{2}\)

First, we should write the balanced equation representing the dissolution of \(\mathrm{CaF}_{2}\) in water: \(\mathrm{CaF}_{2(s)} \rightleftharpoons \mathrm{Ca}^{2+}(aq) + 2\mathrm{F}^-(aq)\) Now, let's write the expression for the solubility product constant (\(K_{sp}\)) for the dissolution of \(\mathrm{CaF}_{2}\): \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2\)

Step 2: Calculate the initial concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^-\) ions in each solution

For each given solution, we will consider the initial concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^-\) ions as follows: (a) \(0.010\,\mathrm{M}\, \mathrm{Ca}(\mathrm{NO}_{3})_{2}\): \(\mathrm{[Ca}^{2+}]_{i} = 0.010\,\mathrm{M}\) and \(\mathrm{[F}^-]_{i} = 0\,\mathrm{M}\) (b) \(0.01\,\mathrm{M}\,\mathrm{NaF}\): \(\mathrm{[Ca}^{2+}]_{i} = 0\,\mathrm{M}\) and \(\mathrm{[F}^-]_{i} = 0.01\,\mathrm{M}\) (c) \(0.001\,\mathrm{M}\,\mathrm{NaF}\): \(\mathrm{[Ca}^{2+}]_{i} = 0\,\mathrm{M}\) and \(\mathrm{[F}^-]_{i} = 0.001\,\mathrm{M}\) (d) \(0.10\,\mathrm{M}\, \mathrm{Ca}(\mathrm{NO}_{3})_{2}\): \(\mathrm{[Ca}^{2+}]_{i} = 0.10\,\mathrm{M}\) and \(\mathrm{[F}^-]_{i} = 0\,\mathrm{M}\)

Step 3: Apply Le Chatelier's principle and common ion effect to analyze the solubility of \(\mathrm{CaF}_{2}\) in each solution

According to Le Chatelier's principle, the solubility of \(\mathrm{CaF}_{2}\) will be influenced by the presence of common ions, \(\mathrm{Ca}^{2+}\) and \(\mathrm{F}^-\). From the provided solutions: - Solution (a) has a higher concentration of \(\mathrm{Ca}^{2+}\), which indicates a decrease in the solubility of \(\mathrm{CaF}_{2}\) due to the common ion effect. - Solution (b) has a higher concentration of \(\mathrm{F}^-\), which also indicates a decrease in the solubility of \(\mathrm{CaF}_{2}\) due to the common ion effect. - Solution (c) has a lower concentration of \(\mathrm{F}^-\), which would promote the solubility of \(\mathrm{CaF}_{2}\) as compared to solution (b). - Solution (d) has a much higher concentration of \(\mathrm{Ca}^{2+}\), indicating the lowest solubility of \(\mathrm{CaF}_{2}\) among the solutions.

Step 4: Determine in which solution \(\mathrm{CaF}_{2}\) will be most soluble

Comparing the presence of common ions in the given solutions, it can be concluded that solution (c), with \(0.001\,\mathrm{M}\,\mathrm{NaF}\), will have the highest solubility of \(\mathrm{CaF}_{2}\), as it has the lowest concentration of the common ion, \(\mathrm{F}^-\), among the provided solutions with a common ion effect.

Key Concepts

Common Ion Effect

When it comes to understanding the solubility of salts in solutions containing a common ion, the common ion effect is an essential concept. This phenomenon occurs because the solubility of an ionic compound decreases when another compound containing one of the ions increases their concentration in the same solution. To put it simply, if you have a saturated solution of a salt, adding more of one of the ions present in that salt from another compound will push the equilibrium toward the undissolved form, thus reducing its solubility.

For example, in the context of our \texttt{CaF\(_2\)} problem, a solution with \texttt{Ca(NO\(_3\))\(_2\)} would release \texttt{Ca\(^{2+}\)} ions into the solution. Since \texttt{CaF\(_2\)} also dissociates to give \texttt{Ca\(^{2+}\)} ions, the increased concentration of \texttt{Ca\(^{2+}\)} from \texttt{Ca(NO\(_3\))\(_2\)} will suppress further dissolution of \texttt{CaF\(_2\)}. This is the common ion effect in action, which is critical when predicting and understanding solubility behavior in various chemical contexts.

Le Chatelier's Principle

A cornerstone of chemical equilibrium, Le Chatelier's principle, states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. This principle helps us understand how the equilibrium will shift when concentrations of products or reactants are altered, when the temperature changes, or when pressure is modified for gases.

In the dissolution example, when a common ion is added to the solution where \texttt{CaF\(_2\)} is dissolving, Le Chatelier's principle predicts that the equilibrium will shift to reduce the disturbance caused by the increase in ion concentration—in this case, shifting towards the left to form more undissolved \texttt{CaF\(_2\)}. By applying this principle, we identified that the addition of \texttt{NaF} introduces \texttt{F\(^-\)} ions, causing the equilibrium to shift and reduce the solubility of \texttt{CaF\(_2\)}. Le Chatelier's principle, in tandem with the common ion effect, provides a powerful explanation of solubility behaviors.

Equilibrium Concentration

Equilibrium concentration refers to the concentrations of reactants and products in a reversible reaction at equilibrium. When a substance is dissolved in water until no more can dissolve, and the rates at which it dissolves and precipitates are equal, we have reached an equilibrium state. At this point, the concentrations of the dissolved ions remain constant over time.

In our original problem, we analyzed solutions with various initial concentrations of ions to determine which would allow the greatest solubility of \texttt{CaF\(_2\)}. It is crucial to understand that the less the \texttt{F\(^-\)} or \texttt{Ca\(^{2+}\)} ions present initially, the more \texttt{CaF\(_2\)} can dissolve until the new equilibrium concentrations are established. Therefore, by knowing the initial concentrations and how they shift to achieve equilibrium, we can predict the solubility of salts in different solutions.

Ksp Expression

The Ksp expression is the mathematical representation of the solubility product constant, symbolized as \texttt{Ksp}. It is a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. The \texttt{Ksp} value indicates the extent to which a salt can dissolve in water and is specific to each salt at a given temperature.

For \texttt{CaF\(_2\)}, the \texttt{Ksp} expression is written as \texttt{Ksp = [Ca\(^{2+}\)][F\(^-\)]\(^2\)}. In this formula, \texttt{[Ca\(^{2+}\)]} and \texttt{[F\(^-\)]} represent the equilibrium concentrations of calcium and fluoride ions, respectively. The exponents reflect the stoichiometry of the dissolved ions in the balanced chemical equation. The \texttt{Ksp} helps us understand how soluble the compound is under various conditions. In practice, comparing the ion concentrations influenced by the presence of common ions to the \texttt{Ksp} value allows us to predict whether a precipitate will form or dissolve, further aiding in the comprehension of solubility and dissolution processes.