Question

When equal volumes of the following solutions are mixed, precipitation of \(\mathrm{AgCl}\left(K_{\mathrm{pp}}=1.8 \times 10^{10}\right)\) will occur only with (1) \(10^{-4} \mathrm{M}\left(\mathrm{Ag}^{-}\right)\) and \(10^{-4} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (2) \(10^{-5} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-5} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (3) \(10^{-6} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-6} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\) (4) \(10^{-10} \mathrm{M}\left(\Lambda \mathrm{g}^{-}\right)\) and \(10^{-10} \mathrm{M}\left(\mathrm{Cl}^{-}\right)\)

Short answer

(1) \(10^{-4} \ \text{M}[\mathrm{Ag}^-]\) and \(10^{-4} \text{M}[\mathrm{Cl}^-]\)

Step-by-step solution

- Understand the Problem

The goal is to determine under which conditions a precipitate of \(\mathrm{AgCl}\) will form. Precipitation occurs when the product of the ion concentrations in the solution exceeds the solubility product constant \(K_{\mathrm{sp}}\).

- Know the Solubility Product Constant (\(K_{\mathrm{sp}}\))

Here, the solubility product constant for \(\mathrm{AgCl}\) is given as \(K_{\mathrm{sp}} = 1.8 \times 10^{-10}\). Comparisons will be made between the ion products and this value.

- Determine the Ion Product

For each pair of concentrations given, calculate the ion product \([\mathrm{Ag}^+][\mathrm{Cl}^-]\). If this value is greater than \(K_{\mathrm{sp}}\), precipitation will occur.

- Calculation for Option (1)

For \([\mathrm{Ag}^+] = 10^{-4} \ \text{M}\), and \([\mathrm{Cl}^-] = 10^{-4} \ \text{M}\), the ion product is \((10^{-4})(10^{-4}) = 10^{-8}\). This value is greater than \(K_{\mathrm{sp}}\). So, precipitation will occur.

- Calculation for Option (2)

For \([\mathrm{Ag}^+] = 10^{-5} \ \text{M}\), and \([\mathrm{Cl}^-] = 10^{-5} \ \text{M}\), the ion product is \((10^{-5})(10^{-5}) = 10^{-10}\). This value equals \(K_{\mathrm{sp}}\), so precipitation will not occur.

- Calculation for Option (3)

For \([\mathrm{Ag}^+] = 10^{-6} \ \text{M}\), and \([\mathrm{Cl}^-] = 10^{-6} \ \text{M}\), the ion product is \((10^{-6})(10^{-6}) = 10^{-12}\). This value is less than \(K_{\mathrm{sp}}\), so precipitation will not occur.

- Calculation for Option (4)

For \([\mathrm{Ag}^+] = 10^{-10} \ \text{M}\), and \([\mathrm{Cl}^-] = 10^{-10} \ \text{M}\), the ion product is \((10^{-10})(10^{-10}) = 10^{-20}\). This value is much less than \(K_{\mathrm{sp}}\), so precipitation will not occur.

- Conclusion

Only the ion product from option (1) exceeds the \(K_{\mathrm{sp}}\), so precipitation will only occur with \(10^{-4} \ \text{M}[\mathrm{Ag}^+]\) and \(10^{-4} \ \text{M}[\mathrm{Cl}^-]\).

Key concepts

Precipitation Reactions

Precipitation reactions occur when ions in a solution combine to form an insoluble compound, called a precipitate. This process is crucial in various fields, from chemistry labs to waste-water treatment. For instance, when a solution of silver ions \( \mathrm{Ag}^+ \) mixes with chloride ions \( \mathrm{Cl}^- \), they can form insoluble silver chloride (\( \mathrm{AgCl} \)). Understanding when and why precipitation happens helps in predicting reaction outcomes and in designing chemical processes.
To determine whether a precipitate will form, we compare the ion product of the reacting ions in the solution with the solubility product constant (\( K_{\mathrm{sp}} \)) of the potential precipitate. If the ion product exceeds the \( K_{\mathrm{sp}} \), precipitation occurs.

Ion Product

The ion product is fundamental in deciding whether a precipitate will form in a solution. It's obtained by multiplying the molar concentrations of the ions involved in the precipitation reaction. For our example of \( \mathrm{AgCl} \), the ion product is calculated as: \[ [\mathrm{Ag}^+][\mathrm{Cl}^-] \]
During calculations, you must compare this value with the \( K_{\mathrm{sp}} \) of the substance. If the ion product exceeds the \( K_{\mathrm{sp}} \), a precipitate forms. Otherwise, the ions remain dissolved in the solution. This method allows chemists to predict and control the formation of precipitates in various reactions.

Ksp Calculation

The solubility product constant, \( K_{\mathrm{sp}} \), is a specific type of equilibrium constant. It represents the maximum product of the ion concentrations that a solution can hold without forming a precipitate. For silver chloride (\( \mathrm{AgCl} \)), the given \( K_{\mathrm{sp}} \) is: \[ K_{\mathrm{sp}} = 1.8 \times 10^{-10} \]
To determine if \( \mathrm{AgCl} \) will precipitate out of a solution, you compare the calculated ion product to this value. For example, if \( [\mathrm{Ag}^+] = 10^{-4} \ \text{M} \) and \( [\mathrm{Cl}^-] = 10^{-4} \ \text{M} \), the ion product will be: \[ (10^{-4})(10^{-4}) = 10^{-8} \]
Since \( 10^{-8} \) is greater than \( 1.8 \times 10^{-10} \), precipitation will occur.

Solution Concentrations

Understanding and calculating solution concentrations is key in predicting precipitation reactions. The concentration of each ion in the solution must be considered. In the exercise, various solutions of \( \mathrm{Ag}^+ \) and \( \mathrm{Cl}^- \) are given with different molarities:

• 10^{-4} \ \text{M} \left(\mathrm{Ag}^+\right) and 10^{-4} \ \text{M} \left(\mathrm{Cl}^-\right)
• 10^{-5} \ \text{M} \left(\mathrm{Ag}^+\right) and 10^{-5} \ \text{M} \left(\mathrm{Cl}^-\right)
• 10^{-6} \ \text{M} \left(\mathrm{Ag}^+\right) and 10^{-6} \ \text{M} \left(\mathrm{Cl}^-\right)
• 10^{-10} \ \text{M} \left(\mathrm{Ag}^+\right) and 10^{-10} \ \text{M} \left(\mathrm{Cl}^-\right)

By using these concentrations to calculate their ion products and comparing them with \( K_{\mathrm{sp}} \), we determine if and when \( \mathrm{AgCl} \) will precipitate. This technique is essential for predicting and controlling chemical reactions in both academic and industrial settings.