Problem 63 - General Chemistry: Principles and Modern Applications

Question

Show that under the following conditions, \(\mathrm{Ba}^{2+}(\mathrm{aq})\) can be separated from \(\mathrm{Sr}^{2+}(\mathrm{aq})\) and \(\mathrm{Ca}^{2+}(\) aq) by precipitating \(\mathrm{BaCrO}_{4}(\mathrm{s})\) with the other ions remaining in solution: $$\begin{aligned} &\left[\mathrm{Ba}^{2+}\right]=\left[\mathrm{Sr}^{2+}\right]=\left[\mathrm{Ca}^{2+}\right]=0.10 \mathrm{M}\\\ &\left[\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=\left[\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right]=1.0 \mathrm{M}\\\ &\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.0010 \mathrm{M}\\\ &\mathrm{K}_{\mathrm{sp}}\left(\mathrm{BaCrO}_{4}\right)=1.2 \times 10^{-10}\\\ &K_{\mathrm{sp}}\left(\mathrm{SrCrO}_{4}\right)=2.2 \times 10^{-5} \end{aligned}$$ Use data from this and previous chapters, as necessary.

Step-by-Step Solution

Answer

Given the provided conditions, the compound \(\mathrm{BaCrO}_{4}\) will precipitate because its Ion Product surpasses its Solubility Product. Conversely, \(\mathrm{SrCrO}_{4}\) won't precipitate as its Ion Product is less than its Solubility Product. Therefore, \(\mathrm{Ba^{2+}}\) can indeed be separated via forming a precipitate, while \(\mathrm{Sr^{2+}}\) will remain in the solution.

Step 1: Determine the concentrations of ions

The concentrations are given as \( [\mathrm{Ba}^{2+}] = [\mathrm{Sr}^{2+}] = [\mathrm{Ca}^{2+}] = 0.10 \, \mathrm{M} \), \( [\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] = 0.0010 \, \mathrm{M} \) and \( [\mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2}^{-}] = 1.0 \, \mathrm{M} \). We only consider \( [\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}] \) because the compound formed upon reaction involves Chromium and not \( \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{O}_{2} \).

Step 2: Calculate the Ion Product (IP) for \(\mathrm{BaCrO}_{4}\) and \(\mathrm{SrCrO}_{4}\)

IP is calculated as the product of the molar concentrations of the ions. Considering the reaction \(\mathrm{Ba^{2+}} + \mathrm{CrO_{7}^{2-}} \rightarrow \mathrm{BaCrO_{4}(s)}\), we have IP for \(\mathrm{BaCrO}_{4}\) as \( (0.10 \, \mathrm{M})(0.0010 \, \mathrm{M}) = 1.0 \times 10^{-4} \). Similarly for \(\mathrm{SrCrO}_{4}\), IP will be \( (0.10 \, \mathrm{M})(0.0010 \, \mathrm{M}) = 1.0 \times 10^{-4} \) also.

Step 3: Compare Ion Products with Solubility Products

The solubility product (\( K_{\mathrm{sp}} \)) for \(\mathrm{BaCrO}_{4}\) is given as \( 1.2 \times 10^{-10} \) and for \(\mathrm{SrCrO}_{4}\) is \( 2.2 \times 10^{-5} \). Comparing these values with the respective IP's, we see that for \(\mathrm{BaCrO}_{4}\) the IP exceeds \( K_{\mathrm{sp}} \) indicating formation of a precipitate. However, for \(\mathrm{SrCrO}_{4}\), the IP is less than \( K_{\mathrm{sp}} \), thus no precipitate forms here.

Key Concepts

Solubility Product Constant

The solubility product constant, commonly represented as \( K_{sp} \), is a numerical expression of a salt's solubility in water. It's a special type of equilibrium constant that applies to the dissolution of sparingly soluble salts. The value of \( K_{sp} \) is a fixed number at a given temperature and provides insight into the extent to which a compound will dissolve in solution. The smaller the \( K_{sp} \), the less soluble the compound is.

For example, when considering the compound barium chromate \( (BaCrO_4) \), if its \( K_{sp} \) is \( 1.2 \times 10^{-10} \), it suggests that at equilibrium, very little of the \( BaCrO_4 \) is dissolved in the solution, forming a minimal amount of \( Ba^{2+} \) and \( CrO_4^{2-} \) ions.

Ion Product

Ion product (IP), also known as the reaction quotient for a dissolution reaction, is similar to the solubility product constant but doesn't necessarily represent the equilibrium state. It is calculated by multiplying the actual molar concentrations of the ions in solution. If the IP equals the \( K_{sp} \), the solution is at equilibrium and saturated with respect to the salt. If the IP is greater than the \( K_{sp} \), the solution is supersaturated, and precipitation is likely to occur. In contrast, if the IP is less than \( K_{sp} \), the solution is undersaturated, and no precipitate forms.

For instance, the calculation of IP for \( BaCrO_4 \) with the provided concentrations shows an IP value of \( 1.0 \times 10^{-4} \), which suggests that a precipitation reaction may occur since it is higher than the compound's \( K_{sp} \).

Separation of Ions

Separation of ions utilizes the differing solubilities of compounds to isolate one ion from a mixture. In a solution where several ions are present, adding a reagent that forms a precipitate with one of the ions—but not with others—can achieve separation. The differences in solubility are often quantified by different \( K_{sp} \) values.

In the context of the exercise, barium \( (Ba^{2+}) \) can be separated from strontium \( (Sr^{2+}) \) and calcium \( (Ca^{2+}) \), because the product of barium with \( CrO_4^{2-} \) exceeds its \( K_{sp} \) value, whereas the corresponding products with strontium and calcium do not. This leads to the selective precipitation of \( BaCrO_4 \) while \( Sr^{2+} \) and \( Ca^{2+} \) remain in the solution.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction where two soluble salts react in solution to form one or more insoluble products—precipitates. The precipitate can be identified as the solid that emerges from the liquid phase.

Considering the exercise, when \( Ba^{2+} \) ions are mixed with \( Cr_2O_7^{2-} \) ions under the given concentrations, \( BaCrO_4 \) precipitates out of the solution because the ion product exceeds the solubility product constant. This is an indicative phenomenon of a precipitation reaction occurring in real-time, demonstrating a practical application of chemistry principles in separating different ions from a solution.