Problem 187 - A Complete Resource Book in Chemistry for JEE Main

Question

A solution containing \(\mathrm{NH}_{4} \mathrm{Cl}\) and \(\mathrm{NH}_{4} \mathrm{OH}\) has a hydroxide ion concentration of \(10^{-6} \mathrm{~mol}\) litre \(^{-1}\), which of the following hydroxides could be precipitated when this solution is added in equal volume to a solution containing \(0.1 \mathrm{M}\) of metal ions? (a) \(\mathrm{AgOH}\left(\mathrm{K}_{\mathrm{sp}}=5 \times 10^{-3}\right)\) (b) \(\mathrm{Cd}(\mathrm{OH})_{2}\left(\mathrm{~K}_{s p}=8 \times 10^{-6}\right)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}\left(\mathrm{~K}_{\mathrm{sp}}=3 \times 10^{-11}\right)\) (d) \(\mathrm{Fe}(\mathrm{OH})_{3}\left(\mathrm{~K}_{\mathrm{sp}}=8 \times 10^{-16}\right)\)

Step-by-Step Solution

Answer

None of the hydroxides precipitate as the ion products are less than their solubility products.

Step 1: Calculate the diluted hydroxide ion concentration

The initial concentration of hydroxide ions in the solution is \([OH^-] = 10^{-6} \, \text{mol/L}\). When this solution is mixed with an equal volume of another solution, the concentration of hydroxide ions will be halved due to dilution.Thus, the new hydroxide ion concentration is \([OH^-] = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, \text{mol/L}\).

Step 2: Determine the ion product for each metal hydroxide

The ion product, \(Q_{ip}\), for each of the metal hydroxides is calculated by multiplying the concentration of the metal ions with the concentration of the hydroxide ions.For the equal volume condition:- \(Q_{ip} \text{ for } \text{AgOH} = 0.1 \times 5 \times 10^{-7} = 5 \times 10^{-8}\)- \(Q_{ip} \text{ for } \text{Cd(OH)}_2 = 0.1 \times (5 \times 10^{-7})^2 = 2.5 \times 10^{-14}\)- \(Q_{ip} \text{ for } \text{Mg(OH)}_2 = 0.1 \times (5 \times 10^{-7})^2 = 2.5 \times 10^{-14}\)- \(Q_{ip} \text{ for } \text{Fe(OH)}_3 = 0.1 \times (5 \times 10^{-7})^3 = 1.25 \times 10^{-20}\)

Step 3: Compare Qip with Ksp for precipitation

A precipitate will form when the ion product \(Q_{ip}\) is greater than the solubility product \(K_{sp}\) for a given compound.- For \(\text{AgOH}\), \(Q_{ip} = 5 \times 10^{-8}\), \(K_{sp} = 5 \times 10^{-3}\) \(\Rightarrow Q_{ip} < K_{sp}\) (No precipitate)- For \(\text{Cd(OH)}_2\), \(Q_{ip} = 2.5 \times 10^{-14}\), \(K_{sp} = 8 \times 10^{-6}\) \(\Rightarrow Q_{ip} < K_{sp}\) (No precipitate)- For \(\text{Mg(OH)}_2\), \(Q_{ip} = 2.5 \times 10^{-14}\), \(K_{sp} = 3 \times 10^{-11}\) \(\Rightarrow Q_{ip} < K_{sp}\) (No precipitate)- For \(\text{Fe(OH)}_3\), \(Q_{ip} = 1.25 \times 10^{-20}\), \(K_{sp} = 8 \times 10^{-16}\) \(\Rightarrow Q_{ip} < K_{sp}\) (No precipitate)

Step 4: Conclusion

Based on the comparisons in Step 3, none of the hydroxides will precipitate since, in all cases, \(Q_{ip}\) is less than \(K_{sp}\).

Key Concepts

Hydroxide Ion Concentration

Understanding hydroxide ion concentration is crucial in predicting the formation of precipitates. In the given exercise, the original hydroxide ion concentration was \([OH^-] = 10^{-6} \, \text{mol/L}\). When this solution is mixed with another solution in an equal volume, the concentration halves. This dilution results in a new hydroxide ion concentration of \([OH^-] = 5 \times 10^{-7} \, \text{mol/L}\).Hydroxide ion concentration is important because it influences the solubility of various metal hydroxides. When these metal ions are introduced into the solution, they can react with hydroxide ions to potentially form solid precipitates.Knowing how to calculate and adjust for changes in hydroxide ion concentration is essential. This is especially true in laboratory settings where precise concentrations are needed to control reaction outcomes.

Ion Product

The ion product, denoted as \(Q_{ip}\), is a value that helps predict whether a precipitate will form in a solution. It is calculated by multiplying the concentrations of the involved ions. For a hydroxide ion solution, \(Q_{ip}\) involves the concentration of metal ions and hydroxide ions.For example, when the dilated hydroxide ion solution is added to a metal ion solution:

• For \(\text{AgOH}\), \(Q_{ip} = 0.1 \times 5 \times 10^{-7} = 5 \times 10^{-8}\).
• For \(\text{Cd(OH)}_2\), \(Q_{ip} = 0.1 \times (5 \times 10^{-7})^2 = 2.5 \times 10^{-14}\).
• For \(\text{Mg(OH)}_2\), \(Q_{ip} = 0.1 \times (5 \times 10^{-7})^2 = 2.5 \times 10^{-14}\).
• For \(\text{Fe(OH)}_3\), \(Q_{ip} = 0.1 \times (5 \times 10^{-7})^3 = 1.25 \times 10^{-20}\).

These calculations show how concentrations impact the ion product, which is then compared to the solubility product constant to predict precipitation.

Precipitation

Precipitation occurs when the ion product \(Q_{ip}\) exceeds the solubility product constant \(K_{sp}\). When \(Q_{ip} > K_{sp}\), the solution becomes supersaturated, and a solid precipitate forms. However, if \(Q_{ip} < K_{sp}\), as in this exercise, no precipitate will form.Comparison of ion product with solubility product for each hydroxide indicates:

• For \(\text{AgOH}\), \(Q_{ip} = 5 \times 10^{-8}, \space K_{sp} = 5 \times 10^{-3}\), hence no precipitation occurs.
• For \(\text{Cd(OH)}_2\), \(Q_{ip} = 2.5 \times 10^{-14}, \space K_{sp} = 8 \times 10^{-6}\), hence no precipitation occurs.
• For \(\text{Mg(OH)}_2\), \(Q_{ip} = 2.5 \times 10^{-14}, \space K_{sp} = 3 \times 10^{-11}\), hence no precipitation occurs.
• For \(\text{Fe(OH)}_3\), \(Q_{ip} = 1.25 \times 10^{-20}, \space K_{sp} = 8 \times 10^{-16}\), hence no precipitation occurs.

In all cases in the exercise, \(Q_{ip}\) was found to be less than \(K_{sp}\), indicating that the solutions remained unsaturated, and no precipitate forms. Understanding the relationship between these two values is key to predicting precipitation.