Problem 30 - Chemistry: Atoms First

Question

Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for \(K_{\mathrm{sp}}\) values.) (a) \(\mathrm{KClO}_{4}:\left[\mathrm{K}^{+}\right]=0.01 \mathrm{M},\left[\mathrm{ClO}_{4}^{-}\right]=0.01 \mathrm{M}\) (b) \(\mathrm{K}_{2} \mathrm{Pt} \mathrm{Cl}_{6}:\left[\mathrm{K}^{+}\right]=0.01 M,\left[\mathrm{PtCl}_{6}^{2-}\right]=0.01 \mathrm{M}\) (c) \(\mathrm{PbI}_{2}:\left[\mathrm{Pb}^{2+}\right]=0.003 \mathrm{M},\left[\mathrm{I}^{-}\right]=1.3 \times 10^{-3} \mathrm{M}\) (d) \(\mathrm{Ag}_{2} \mathrm{S}:\left[\mathrm{Ag}^{+}\right]=1 \times 10^{-10} \mathrm{M},\left[\mathrm{S}^{2-}\right]=1 \times 10^{-13} \mathrm{M}\)

Step-by-Step Solution

Answer

To find out which compounds precipitate, compare the calculated ionic products for each compound with their respective K_{sp} values. Compounds with ionic products exceeding K_{sp} values will precipitate.

Step 1 - Understanding the Problem

Determine the possibility of a precipitate forming by comparing the ionic product (Q) of given ion concentrations in a solution to the known solubility product constant (K_{sp}) for each of the compounds. If the ionic product exceeds the K_{sp}, the compound will precipitate.

Step 2 - Calculating Ionic Products for Each Compound

Calculate the ionic product (Q) for each compound using the formula: Q = [cation]^m * [anion]^n, where m and n are the stoichiometric coefficients from the compound's formula.

Step 3 - Comparing Ionic Product with K_{sp}

Compare the calculated ionic product (Q) to the K_{sp} for each compound. If Q > K_{sp}, it indicates that the compound will precipitate from the solution as the solution is supersaturated.

Step 4 - Applying the Steps to Each Compound

Using Appendix J for K_{sp} values, apply the above steps to each compound: (a) KClO_4, (b) K_2PtCl_6, (c) PbI_2, and (d) Ag_2S.

Step 5 - Conclusion

Based on the K_{sp} values and the calculated ionic products, determine which compound(s), if any, will precipitate out of the solutions provided.

Key Concepts

Ionic Product

When considering the solubility of compounds in a solution, it's crucial to understand the concept of the ionic product, often symbolized as Q. The ionic product refers to the product of the concentrations of the ions in the solution, raised to the power of their coefficients in the balanced equation for the dissolution of the ionic compound.

For a generic ionic compound that dissociates into cations and anions, the expression of the ionic product is given by: \[ Q = [\text{cation}]^m \times [\text{anion}]^n \] where \([\text{cation}]\) and \([\text{anion}]\) are the molar concentrations of the ions, and \(m\) and \(n\) represent the stoichiometric coefficients of the cations and anions, respectively.

The significance of the ionic product lies in its comparison with the solubility product constant (\(K_{sp}\)). This comparison indicates whether a solution is saturated with ions and whether precipitation will occur. In simpler terms, if a solution's ionic product exceeds the solubility product constant, the excess ions will begin to form a solid precipitate, coming out of the solution.

Precipitation Reactions

Precipitation reactions occur when ions in a solution react to form an insoluble compound which settles out of the solution as a solid. These reactions are quite common in the study of chemistry and they can be easily visualized as the formation of a cloudiness or a solid in a previously clear solution.

To predict if a precipitation reaction will occur, one must know the solubility rules and the concentrations of the reacting ions in solution. The fundamental principle in predicting precipitation is that if the ionic product (Q) is greater than the solubility product constant (\(K_{sp}\)), insoluble precipitates will form. This is because the maximum solubility limit has been exceeded and the solution cannot hold the ions in a dissolved state.

For instance, a reaction in which silver ions (\(Ag^+\)) and chloride ions (\(Cl^-\)) are mixed in a solution might result in the formation of silver chloride (\(AgCl\)) precipitate, provided their ionic product exceeds the \(K_{sp}\) value of silver chloride.

Ksp Values

The solubility product constant, denoted as \(K_{sp}\), is an equilibrium constant that applies to the dissolution of sparingly soluble ionic compounds. Each sparingly soluble compound has a unique \(K_{sp}\) value that reflects its intrinsic solubility in water.

The \(K_{sp}\) value is determined experimentally and is temperature-dependent. The value provides insight into the extent to which a compound will dissolve in water. In general, a higher \(K_{sp}\) indicates greater solubility. For example, a compound with a \(K_{sp}\) of \(1\times10^{-5}\) is more soluble than one with a \(K_{sp}\) of \(1\times10^{-10}\).

When the ionic product of a solution is equal to the \(K_{sp}\), the solution is at equilibrium, and no further dissolution or precipitation will occur. However, if the ionic product exceeds the \(K_{sp}\), it suggests that more solid has dissolved than the solution can theoretically hold at equilibrium, leading to precipitation.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass where the number of atoms of each element involved in a reaction must be the same before and after the reaction takes place.

In the context of a precipitation reaction and solubility products, stoichiometry is used to determine the ratio in which the ions combine to form the solid precipitate. The stoichiometric coefficients from the balanced chemical equation for the dissolution of the compound are essential in this process as they are used to calculate the ionic product (Q).

For example, if we consider the dissolution of lead iodide (\(PbI_2\)), which breaks down into one lead ion (\(Pb^{2+}\)) and two iodide ions (\(I^-\)), the stoichiometric coefficients 1 and 2 would be used to calculate Q as follows: \[ Q=[Pb^{2+}]^{1}\times[I^-]^{2} \] Understanding the stoichiometric relationships allows for precise predictions about whether a precipitate will form under specific concentration conditions.