Problem 58 - Chemistry

Question

When 100.0 \(\mathrm{mL}\) of 2.00 \(\mathrm{M} \mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{3}\) is added to 100.0 \(\mathrm{mL}\) of \(3.00 \mathrm{M} \mathrm{KIO}_{3},\) a precipitate of \(\mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}(s)\) forms. Calculate the equilibrium concentrations of \(\mathrm{Ce}^{3+}\) and \(\mathrm{IO}_{3}^{-}\) in this solution. \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Ce}\left(\mathrm{IO}_{3}\right)_{3}=3.2 \times 10^{-10} .\right]\)

Step-by-Step Solution

Answer

The equilibrium concentrations are \([\mathrm{Ce}^{3+}]_{eq} = 0.9897\ \mathrm{M}\) and \([\mathrm{IO}_3^-]_{eq} = 1.468\ \mathrm{M}\).

Step 1: Write the balanced chemical equation for the precipitation reaction.

The balanced chemical equation for the precipitation reaction is: \[ \mathrm{Ce}^{3+}(aq) + 3 \mathrm{IO}_3^-(aq) \rightleftharpoons \mathrm{Ce(IO_3)_3}(s) \]

Step 2: Calculate the initial concentrations for \(\mathrm{Ce}^{3+}\) and \(\mathrm{IO}_3^-\) ions.

To find the initial concentrations, we need to use the fact that the volumes of the two solutions are the same (100.0 mL) and that moles are additive. We can calculate the moles of each ion in the mixture: Moles of \(\mathrm{Ce}^{3+} = 100.0 \mathrm{mL} \times \frac{2.00 \mathrm{mol}}{1,000 \mathrm{mL}} = 0.200 \mathrm{mol}\) Moles of \(\mathrm{IO}_3^- = 100.0 \mathrm{mL} \times \frac{3.00 \mathrm{mol}}{1,000 \mathrm{mL}} = 0.300 \mathrm{mol}\) The volume of the mixture is the sum of the volumes of the two solutions: \(v = 100.0\mathrm{mL} + 100.0\mathrm{mL} = 200.0\mathrm{mL}\). The initial concentrations are calculated as: \[[\mathrm{Ce}^{3+}]_{initial} = \frac{0.200\ \mathrm{mol}}{200.0\ \mathrm{mL}} \times \frac{1,000\ \mathrm{mL}}{1\ \mathrm{mol}} = 1.00\ \mathrm{M}\] \[[\mathrm{IO}_3^-]_{initial} = \frac{0.300\ \mathrm{mol}}{200.0\ \mathrm{mL}} \times \frac{1,000\ \mathrm{mL}}{1\ \mathrm{mol}} = 1.50\ \mathrm{M}\]

Step 3: Set up the ICE table.

Using the initial concentrations, we can set up an ICE table for the reaction: | | \(\mathrm{Ce}^{3+}(aq)\) | \(3\mathrm{IO}_3^-(aq)\) | \(\mathrm{Ce(IO_3)_3}(s)\) | | ------- | ---------------------- | --------------------- | ----------------------- | | Initial | 1.00 M | 1.50 M | solid | | Change | -x | -3x | +x | | Equilibrium | 1.00 M - x | 1.50 M - 3x | solid |

Step 4: Write the expression for \(K_{sp}\) and solve for the equilibrium concentrations.

The solubility product constant \(K_{sp}\) expression is: \(K_{sp} = [\mathrm{Ce}^{3+}][\mathrm{IO}_3^-]^3\) Now, use the given value of \(K_{sp} = 3.2\times 10^{-10}\), and the values from the ICE table to solve for the equilibrium concentrations: \(3.2 \times 10^{-10} = (1.00\ \mathrm{M} - x) \times (1.50\ \mathrm{M} - 3x)^3\) By making the assumption that x is small enough compared to the initial values, we can simplify the equation to: \(3.2 \times 10^{-10} = \left(1.00\right) \times \left(1.50 - 3x\right)^3\) Now, solve for \(x\): \(x = 1.03\times10^{-2}\ \mathrm{M}\) Based on the ICE table, we have: \([\mathrm{Ce}^{3+}]_{eq} = 1.00\ \mathrm{M} - x = 1.00\ \mathrm{M} - 1.03\times10^{-2}\ \mathrm{M} = 0.9897\ \mathrm{M}\) \([\mathrm{IO}_3^-]_{eq} = 1.50\ \mathrm{M} - 3x = 1.50\ \mathrm{M} - 3(1.03\times10^{-2}\ \mathrm{M}) = 1.468\ \mathrm{M}\) So, the equilibrium concentrations are \([\mathrm{Ce}^{3+}]_{eq} = 0.9897\ \mathrm{M}\) and \([\mathrm{IO}_3^-]_{eq} = 1.468\ \mathrm{M}\).

Key Concepts

Solubility Product Constant \(K_{sp}\)

The solubility product constant, commonly abbreviated as \(K_{sp}\), is a crucial concept when dealing with ionic compounds in solutions. It’s an equilibrium constant specific to the dissolution of sparingly soluble salts in water. Think of it as a measurement of the extent to which a compound dissolves in solution to form its constituent ions.
A higher \(K_{sp}\) value means a compound is more soluble, while a lower \(K_{sp}\) indicates less solubility. For example, if the \(K_{sp}\) of a compound is very small, it forms very few ions in solution and thus is relatively insoluble.
For the equilibrium expression of solubility, involving a salt such as \(\text{Ce(IO}_3\text{)}_3\), \(K_{sp}\) can be written as follows:

• \([\text{Ce}^{3+}][\text{IO}_3^-]^3\)

This indicates that the solubility product is a multiplication of the molar concentrations of the ions, each raised to the power of their coefficients from the balanced equation. Using the known value of \(K_{sp}\), one can determine the concentrations of the ions at equilibrium, which provides insight into the solubility of the compound in a given reaction.

Precipitation Reaction

A precipitation reaction occurs when two aqueous solutions combine to form an insoluble solid, or precipitate. It's like a mix-and-match of ions in a solution to find the perfect pairing that results in solid formation. In this particular reaction, mixing \(\text{Ce(NO}_3\text{)}_3\) with \(\text{KIO}_3\) leads to the formation of \(\text{Ce(IO}_3\text{)}_3\) as a precipitate.
The beauty of a precipitation reaction lies in the fact that it helps in understanding the solubility levels of different ionic compounds. Whenever the product of the concentrations of the ions exceeds the \(K_{sp}\) value at a given temperature, a precipitate forms.
This reaction gives us valuable information about which combinations of ions will lead to the formation of an insoluble compound. In practical applications, observing a precipitate can help determine the saturation point of a solution, adjust concentrations, or even separate components from mixtures.

ICE Table Method

The ICE table method is a systematic approach to understand equilibrium processes. ICE stands for Initial, Change, and Equilibrium, which are the stages one assesses when analyzing a chemical equilibrium. An ICE table is especially useful with reactions involving soluble and insoluble species, like precipitation reactions.

To set up an ICE table, one follows these steps:

• Identify and write down the initial concentrations of reactants and products before the reaction reaches equilibrium.
• Determine the change in concentrations as the system moves toward equilibrium. This is often expressed as \(x\), representing the amount that changes.
• Calculate the new equilibrium concentrations by applying the changes to the initial concentrations.

In the reaction under discussion, \(\text{Ce}^{3+}\) and \(\text{IO}_3^-\) ions are initially mixed in known concentrations. As these ions combine to form a solid \(\text{Ce(IO}_3\text{)}_3\) precipitate, their concentrations decrease. Each decrease is represented by \(-x\) (and for \(\text{IO}_3^-\) as \(-3x\) based on its stoichiometry).
Understanding and applying the ICE table method provides a clear picture of what’s happening in the solution, how the reaction progresses, and predictably calculates the equilibrium concentrations of reactants and products.