Question

Consider a buffered solution containing \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) . Which of the following statements concerning this solution is(are) true? \(\left(K_{\mathrm{a}} \text { for } \mathrm{CH}_{3} \mathrm{NH}_{3}+=2.3 \times 10^{-11}\right)\) a. A solution consisting of 0.10\(M \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and 0.10 \(\mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) would have a greater buffering capacity than one containing 1.0 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and 1.0 \(\mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) b. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\) then the \(\mathrm{pH}\) is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the pH. d. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\) then \(\mathrm{pH}<3.36\) e. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right],\) then \(\mathrm{pH}=10.64\)

Short answer

Statements b, c, and e are true. Statement a is false, and statement d is uncertain and not necessarily true.

Step-by-step solution

Calculate pKa

Calculate the pKa value using the given Ka value. The pKa is the negative logarithm of the Ka value: \[pKa = -\log_{10} (Ka)\] Plug in the given Ka value: \[pKa = -\log_{10} (2.3 \times 10^{-11}) ≈ 10.64\] The pKa value for CH3NH3+ is 10.64.

Analyze statement a

A buffered solution with higher concentrations of the conjugate acid and base has a greater buffering capacity, as it can accommodate more added acid or base without a significant change in pH. In this case, a solution containing 1.0 M CH3NH3Cl and 1.0 M CH3NH2 will have a greater buffering capacity compared to a solution with only 0.1 M of each species. Therefore, statement a is false.

Analyze statement b

Use the Henderson-Hasselbalch equation to determine the pH of the buffer: \[pH = pKa + \log_{10} \left( \frac{[CH3NH2]}{[CH3NH3+]} \right)\] If [CH3NH2] > [CH3NH3+], then the fraction inside the logarithm will be greater than 1, making the log positive. Consequently, the pH will be greater than the pKa value. Therefore, statement b is true.

Analyze statement c

Adding more CH3NH3Cl to the initial buffer solution will increase the concentration of the conjugate acid, CH3NH3+. This will drive the equilibrium towards the formation of CH3NH2 and HCl, leading to a more acidic solution and subsequently decreasing the pH. Therefore, statement c is true.

Analyze statement d

Again, use the Henderson-Hasselbalch equation: \[pH = pKa + \log_{10} \left( \frac{[CH3NH2]}{[CH3NH3+]} \right)\] If [CH3NH2] < [CH3NH3+], then the fraction inside the logarithm will be less than 1, making the log negative. Consequently, the pH will be lower than the pKa value (10.64). However, statement d gives a specific pH value (3.36), which we cannot guarantee without knowing the exact concentrations of CH3NH2 and CH3NH3+. Therefore, statement d is uncertain and not necessarily true.

Analyze statement e

Using the Henderson-Hasselbalch equation once more: \[pH = pKa + \log_{10} \left( \frac{[CH3NH2]}{[CH3NH3+]} \right)\] If [CH3NH2] = [CH3NH3+], the fraction inside the logarithm will be equal to 1, making the log zero. Consequently, the pH will be equal to the pKa value (10.64). Therefore, statement e is true. #Conclusion#: Given the buffered solution containing CH₃NH₃Cl and CH₃NH₂ with Ka 2.3 × 10⁻¹¹: - Statement a is false. - Statement b is true. - Statement c is true. - Statement d is uncertain and not necessarily true. - Statement e is true.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a simple yet powerful tool to understand the pH of buffer solutions. It connects the concentration of acid and its conjugate base to the pH and the pKa of the acid. It is expressed as:\[pH = pKa + \log_{10} \left( \frac{[A^-]}{[HA]} \right)\]where

• \([A^-]\) is the concentration of the conjugate base.
• \([HA]\) is the concentration of the acid.
• \(pKa\) is the acid dissociation constant.

In buffer solutions, the Henderson-Hasselbalch equation helps predict the pH given the concentrations of the conjugate acid-base pair. If the concentration of the conjugate base \([A^- ]\) increases, the pH will be higher than the pKa, and vice versa. This is because there is a direct relationship between the concentrations and the log function's value in the equation.

buffer capacity

Buffer capacity describes a solution's ability to resist changes in pH when small amounts of acid or base are added. A buffer can only resist pH changes effectively when there is a sufficient concentration of its acid and conjugate base present.
For buffer solutions, the rule of thumb is:

• Higher concentrations of the acid and its conjugate base yield a higher buffer capacity, meaning the buffer can handle more added acid or base before significant pH changes occur.

In the context of the problem, a buffer with 1.0 M concentrations of both the acid and base would have a greater capacity than one with only 0.1 M of each. This means it can withstand more acidic or basic additions without a dramatic pH change.

pH and pKa relationship

Understanding the pH and pKa relationship is crucial for analyzing acid-base reactions. The value of the pKa indicates the strength of the acid:

• A lower pKa value denotes a stronger acid, which can dissociate more into protons \([H^+]\) and conjugate base.

When a buffer's pH is lower than its pKa, more acid is present than its conjugate base. Conversely, when the pH is higher than the pKa, the conjugate base predominates.
This relationship forms the basis for assertions such as statement b in the exercise: if the concentration of \( \text{[CH}_3\text{NH}_2] \) exceeds \( \text{[CH}_3\text{NH}_3^+] \), then the pH surpasses the pKa value.

conjugate acid-base pair

A conjugate acid-base pair consists of two species that transform into each other by gaining or losing a proton. In the exercise, \( \text{CH}_3\text{NH}_3^+ \) and \( \text{CH}_3\text{NH}_2 \) form such a pair. When the acid loses a proton, it turns into its conjugate base, and vice versa.
This concept is essential to understanding buffer solutions. In buffers, both the acid and its conjugate base are present in significant amounts to absorb added acids \( \text{or bases} \) with minimal pH changes.

• The flexibility of easily transitioning between the acid and base forms stabilizes the pH of the solution.

When more acid \( \text{CH}_3\text{NH}_3^+ \) is added, it combines with any base in the solution to resist pH changes in a buffer system.