Question

What compound, if any, will precipitate when \(80 \mathrm{~mL}\) of \(1.0 \times 10^{-5} \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is added to \(20 \mathrm{~mL}\) of \(1.0 \times 10^{-5} \mathrm{M}\) \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} ?\)

Short answer

Fe(OH)₃ will precipitate.

Step-by-step solution

Calculate concentrations after mixing

When two solutions are mixed, their concentrations change. First, calculate the new concentration of \( \mathrm{Ba(OH)}_2 \):\[ C_{\text{Ba(OH)}_2} = \frac{(80 \text{ mL}) \times (1.0 \times 10^{-5} \mathrm{M})}{100 \text{ mL}} = 8.0 \times 10^{-6} \mathrm{M}\] Next, calculate the concentration of \( \mathrm{Fe}_2(\mathrm{SO}_4)_3 \):\[ C_{\mathrm{Fe}_2(\mathrm{SO}_4)_3} = \frac{(20 \text{ mL}) \times (1.0 \times 10^{-5} \mathrm{M})}{100 \text{ mL}} = 2.0 \times 10^{-6} \mathrm{M}\] The total volume of the new solution is \( 100 \text{ mL} \).

Identify possible precipitates

The possible precipitate forms if any combination of ions exceeds the solubility product (Ksp). For this system, possible compounds are \( \mathrm{BaSO}_4 \) and \( \mathrm{Fe(OH)}_3 \). We will check both using their respective reaction equations: \[ \mathrm{Ba}^{2+} + \mathrm{SO}_4^{2-} \rightarrow \mathrm{BaSO}_4 \]\[ \mathrm{Fe}^{3+} + 3\mathrm{OH}^- \rightarrow \mathrm{Fe(OH)}_3 \]

Calculate ion product for BaSO4

First, calculate the ion products (Q) for \( \mathrm{BaSO}_4 \). The ion product \( Q_{\mathrm{BaSO}_4} \) is given by:\[ Q_{\mathrm{BaSO}_4} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}] \]Both ions have concentrations of \( 8.0 \times 10^{-6} \mathrm{M} \) and \( 2.0 \times 10^{-6} \mathrm{M} \) respectively. Thus:\[ Q_{\mathrm{BaSO}_4} = (8.0 \times 10^{-6})(2.0 \times 10^{-6}) = 1.6 \times 10^{-11} \] Compare \( Q_{\mathrm{BaSO}_4} \) to the known \( K_{sp} \) of \( \mathrm{BaSO}_4 \) which is \( 1.1 \times 10^{-10} \). Since \( Q < K_{sp} \), \( \mathrm{BaSO}_4 \) does not precipitate.

Calculate ion product for Fe(OH)3

Next, calculate the ion product for \( \mathrm{Fe(OH)}_3 \). The ion product \( Q_{\mathrm{Fe(OH)}_3} \) is given by:\[ Q_{\mathrm{Fe(OH)}_3} = [\mathrm{Fe}^{3+}][\mathrm{OH}^-]^3 \] The concentrations are \( 2.0 \times 10^{-6} \mathrm{M} \) for \( \mathrm{Fe}^{3+} \) and \( 2 \times 8.0 \times 10^{-6} \mathrm{M} = 1.6 \times 10^{-5} \mathrm{M} \) for \( \mathrm{OH}^- \) since each \( \mathrm{Ba(OH)}_2 \) provides 2 \( \mathrm{OH}^- \). Thus:\[ Q_{\mathrm{Fe(OH)}_3} = (2.0 \times 10^{-6})(1.6 \times 10^{-5})^3 = 8.192 \times 10^{-21} \]Compare \( Q_{\mathrm{Fe(OH)}_3} \) to the known \( K_{sp} \) of \( \mathrm{Fe(OH)}_3 \) which is \( 4.0 \times 10^{-38} \). Since \( Q > K_{sp} \), \( \mathrm{Fe(OH)}_3 \) precipitates.

Key concepts

Ion Product

The ion product is an important concept in chemistry, particularly in the study of precipitation reactions. To understand this, think of the ion product, often designated as \( Q \), as a snapshot of the current concentrations of ions in a solution. It provides a way to predict whether a precipitate will form when different salts are mixed.

• If \( Q < K_{sp} \), the solution is unsaturated, meaning no precipitate will form. The ions remain dissolved.
• If \( Q = K_{sp} \), the system is at equilibrium, and the solution is saturated, which means it is on the verge of forming a precipitate.
• If \( Q > K_{sp} \), the solution is supersaturated, leading to the formation of a precipitate as the excess ions will combine to form a solid.

When combining solutions, calculating \( Q \) helps chemists determine the fate of ions and whether they will come together to form a solid or remain as individual ions in a solution. Therefore, looking at \( Q \) helps in understanding what happens at the molecular level during chemical reactions, particularly precipitation reactions.

Solubility Product (Ksp)

The solubility product, denoted as \( K_{sp} \), is a vital concept in predicting and understanding precipitation reactions. It is a constant that helps describe the solubility of a compound at a given temperature. In simple terms, \( K_{sp} \) tells you how much of a compound can dissolve in a certain amount of solvent before no more can dissolve.
The solubility product is determined experimentally and used to calculate a compound's solubility by relating the concentrations of the dissolved ions. For example, for a compound such as \( \mathrm{BaSO}_4 \), \( K_{sp} \) is defined by the equilibrium expression:
\[ K_{sp} = [\mathrm{Ba}^{2+}][\mathrm{SO}_4^{2-}] \]
Understanding \( K_{sp} \) is critical because it helps you ascertain whether a precipitation reaction will occur when two solutions are mixed. It serves as the benchmark with which the ion product (\( Q \)) is compared. If the ion product exceeds the solubility product, precipitation occurs, resulting in a decrease in the concentration of free ions as they combine to form a solid.

Concentration Calculations

Concentration calculations are a fundamental step in solving precipitation reaction problems. Before figuring out whether a precipitate forms, the concentrations of ions in the solution after mixing must be accurately calculated. This involves determining the new concentrations when two solutions are combined.
By using the formula:
\[ C = \frac{C_1 V_1 + C_2 V_2}{V_{\text{total}}} \]
You calculate the concentration \( C \) after mixing, where \( C_1 \) and \( C_2 \) are initial concentrations, \( V_1 \) and \( V_2 \) are volumes of the respective solutions, and \( V_{\text{total}} \) is the total volume.
Accurately calculating the concentrations is key because it directly feeds into finding the ion product, \( Q \). It ensures you have the correct values to predict if the ion concentrations will exceed the \( K_{sp} \) and thus form a precipitate. This fundamental step in precipitation reactions keeps the rest of your calculations on track and helps ensure a clear understanding of the chemical dynamics at play.