Problem 73 - Chemistry: The Central Science in SI Units, Global Edition

Question

A solution contains \(1.0 \times 10^{-4} \mathrm{Ca}^{2+}(a q)\) and \(1.0 \times 10^{-4}\) \(\mathrm{La}^{3+}(a q) .\) If \(\mathrm{NaF}\) is added, will \(\mathrm{CaF}_{2}\left(K_{s p}=3.9 \times 10^{-11}\right)\) or \(\mathrm{LaF}_{3}\left(K_{s p}=2 \times 10^{-19}\right)\) precipitate first? Specify the concentration of \(\mathrm{F}^{-}(a q)\) needed to begin precipitation.

Step-by-Step Solution

Answer

When adding \(\mathrm{NaF}\), \(\mathrm{LaF}_3\) will precipitate first. The concentration of \(\mathrm{F}^-\) needed to start the precipitation is \(1.26 \times 10^{-5} \mathrm{M}\).

Step 1: Write Solubility Product Expressions

Write the solubility product expressions for the two salts. For \(\mathrm{CaF}_2\), it is: \[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2\] For \(\mathrm{LaF}_3\), it is: \[K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^-]^3\]

Step 2: Calculate the concentration of \(\mathrm{F}^-\) at which each salt will precipitate

Using the solubility product expressions, find the concentration of \(\mathrm{F}^-\) needed for precipitation to start for each salt. For \(\mathrm{CaF}_2\): \[3.9 \times 10^{-11} = (1.0 \times 10^{-4})[\mathrm{F}^-]^2\] Solve for \([\mathrm{F}^-]\): \[[\mathrm{F}^-]_1 = \sqrt{\frac{3.9 \times 10^{-11}}{1.0 \times 10^{-4}}} = 6.24 \times 10^{-4} \mathrm{M}\] For \(\mathrm{LaF}_3\): \[2 \times 10^{-19} = (1.0 \times 10^{-4})[\mathrm{F}^-]^3\] Solve for \([\mathrm{F}^-]\): \[[\mathrm{F}^-]_2 = \sqrt[3]{\frac{2 \times 10^{-19}}{1.0 \times 10^{-4}}} = 1.26 \times 10^{-5} \mathrm{M}\]

Step 3: Compare the concentration values and determine which salt will precipitate first

Comparing the \(\mathrm{F}^-\) concentration values, we find that the concentration needed to start precipitation for \(\mathrm{CaF}_2\) is \(6.24 \times 10^{-4} \mathrm{M}\) and for \(\mathrm{LaF}_3\) is \(1.26 \times 10^{-5} \mathrm{M}\). Since the concentration required for \(\mathrm{LaF}_3\) to precipitate is lower, it will precipitate first. The concentration of \(\mathrm{F}^-\) needed to start the precipitation is \(1.26 \times 10^{-5} \mathrm{M}\).

Key Concepts

Solubility Product

The solubility product, represented as \(K_{sp}\), is a constant that describes the equilibrium between a solid and its respective ions in a saturated solution. In simpler terms, it signifies the extent to which a salt can dissolve in water. Every slightly soluble salt has its unique \(K_{sp}\) value. Understanding how to use solubility product expressions is crucial in predicting whether a salt will precipitate from solution.

For example, when dealing with calcium fluoride, \(CaF_2\), the solubility product expression is:

• \[K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{F}^-]^2\]

Similarly, for lanthanum fluoride, \(LaF_3\), the expression is:

• \[K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^-]^3\]

These expressions reflect the ratio of product ions, and their respective powers in the equation depend on the stoichiometry of the compound in concern. By calculating these expressions, one can predict the point at which the salt will begin to precipitate.

Calcium Fluoride Precipitation

When predicting precipitation, calcium fluoride, \(CaF_2\), is a compound of interest. It starts to precipitate when the concentration of ions in the solution exceeds its \(K_{sp}\). In our case, we use the equation:

• \[3.9 \times 10^{-11} = (1.0 \times 10^{-4})[\mathrm{F}^-]^2\]

To find the critical concentration of fluoride ions, \([\mathrm{F}^-]\), needed for precipitation, we solve for \([\mathrm{F}^-]\):

• \[[\mathrm{F}^-] = \sqrt{\frac{3.9 \times 10^{-11}}{1.0 \times 10^{-4}}} = 6.24 \times 10^{-4} \mathrm{M}\]

This computation tells us that calcium fluoride will not start precipitating until fluoride ions reach the concentration of \(6.24 \times 10^{-4} \mathrm{M}\). It signals when the solution becomes saturated with respect to \(CaF_2\). This important threshold helps understand the behavior of \(CaF_2\) in mixed ion solutions.

Lanthanum Fluoride Precipitation

Lanthanum fluoride, \(LaF_3\), starts to precipitate at a different concentration compared to calcium fluoride due to its distinct \(K_{sp}\). The expression explaining \(LaF_3\) precipitation is:

• \[2 \times 10^{-19} = (1.0 \times 10^{-4})[\mathrm{F}^-]^3\]

To determine when precipitation occurs, solve for the fluoride ion concentration:

• \[[\mathrm{F}^-] = \sqrt[3]{\frac{2 \times 10^{-19}}{1.0 \times 10^{-4}}} = 1.26 \times 10^{-5} \mathrm{M}\]

This result informs us that lanthanum fluoride requires a much lower concentration of fluoride ions, \(1.26 \times 10^{-5} \mathrm{M}\), to precipitate. Because this concentration is lower than that needed for calcium fluoride, \(LaF_3\) will precipitate first when fluoride ions are slowly added to the solution.

Fluoride Ion Concentration

The concentration of fluoride ions, \([\mathrm{F}^-]\), plays a critical role in determining the order of salt precipitation. The different salts in a solution will start to precipitate at particular fluoride concentrations. In this specific scenario, as we increase the concentration of \(\mathrm{F}^-\) ions, lanthanum fluoride precipitates first because it's \(K_{sp}\) is much lower compared to calcium fluoride.

The calculations show:

• \(1.26 \times 10^{-5} \mathrm{M}\) for \(LaF_3\)
• \(6.24 \times 10^{-4} \mathrm{M}\) for \(CaF_2\)

The critical insight is that the salt with the lower \(K_{sp}\) or requiring the lower concentration of \([\mathrm{F}^-]\) will precipitate first. This concept aids in practical applications such as separating ions through selective precipitation, predicting the outcomes in metabolic pathways, and many other chemical processes.