Question

Which of the following mixtures would result in buffered solutions when \(1.0 \mathrm{~L}\) of each of the two solutions are mixed? a. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) c. \(0.2 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) d. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\)

Short answer

The mixture (a) \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{3} \mathrm{Cl}\) will result in a buffered solution when mixed.

Step-by-step solution

Identify weak acids/bases and their conjugates

First, we need to identify the weak acids and bases present in the solutions and their conjugate acids/bases. The weak base present in all the given mixtures is \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) (methylamine), and its conjugate acid is \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) (methylammonium ion). KOH, on the other hand, is a strong base, and its conjugate acid is water, which doesn't participate in the buffering effect.

Analyzing each mixture

Now that we've identified the weak base and its conjugate, let's analyze each given mixture to determine if it forms a buffered solution: a. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{3} \mathrm{Cl}\): In this mixture, KOH will react with the weak acid \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\), forming the weak base \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and water. The final solution will contain equal concentrations of the weak base and its conjugate acid, forming a buffered solution. b. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{2}\): In this case, the KOH will react with a part of the weak base \(\mathrm{CH}_{3}\mathrm{NH}_{2}\) and form its conjugate acid \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\). After the reaction, the concentrations of the weak base and its conjugate acid will not be equal. Thus, this mixture will not result in a buffered solution. c. \(0.2 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{3} \mathrm{Cl}\): In this mixture, the strong base KOH will overwhelm the weak acid, leading to \(0.1 \mathrm{M}\) of the weak base and its conjugate acid after the reaction. This will not result in a buffered solution. d. \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.2 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{3} \mathrm{Cl}\): In this case, KOH will react with some of the weak acid \(\mathrm{CH}_{3}\mathrm{NH}_{3}^{+}\) to form the weak base \(\mathrm{CH}_{3}\mathrm{NH}_{2}\), ultimately resulting in an unequal concentration of the weak base and its conjugate acid. Thus, this mixture will not result in a buffered solution.

Conclusion

Based on the analysis of each mixture, only the mixture (a) \(0.1 \mathrm{M} \mathrm{KOH}\) and \(0.1 \mathrm{M} \mathrm{CH}_{3}\mathrm{NH}_{3} \mathrm{Cl}\) will result in a buffered solution when mixed.