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Chapter 15: Problem 115

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

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

Step by step solution

01

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.
02

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.
03

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.
04

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.
05

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.
06

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.

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Most popular questions from this chapter

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

The active ingredient in aspirin is acetylsalicylic acid. A 2.51 -g sample of acetylsalicylic acid required 27.36 \(\mathrm{mL}\) of 0.5106 $\mathrm{M} \mathrm{daOH}\( for complete reaction. Addition of 13.68 \)\mathrm{mL}$ of 0.5106\(M \mathrm{HCl}\) to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH \(=3.48 .\) Determine the molar mass of acetylsalicylic acid and its \(K_{2}\) value. State any assumptions you must make to reach your answer.

Sketch a pH curve for the titration of a weak acid (HA) with a strong base (NaOH). List the major species, and explain how you would go about calculating the pH of the solution at various points, including the halfway point and the equivalence point

When a diprotic acid, \(\mathrm{H}_{2} \mathrm{A},\) is titrated with NaOH, the protons on the diprotic acid are generally removed one at a time, resulting in a pH curve that has the following generic shape: a. Notice that the plot has essentially two titration curves. If the first equivalence point occurs at 100.0 mL NaOH added, what volume of NaOH added corresponds to the second equivalence point? b. For the following volumes of NaOH added, list the major species present after the OH- reacts completely. $$ \begin{array}{l}{\text { i. } 0 \mathrm{mL} \text { NaOH added }} \\ {\text { i. between } 0 \text { and } 100.0 \mathrm{mL} \text { NaOH added }}\end{array} $$ $$ \begin{array}{l}{\text { iii. } 100.0 \text { mL NaOH added }} \\ {\text { iv. between } 100.0 \text { and } 200.0 \mathrm{mL} \text { NaOH added }} \\\ {\text { v. } 200.0 \mathrm{mL} \text { NaOH added }}\end{array} $$ $$ 200.0 \mathrm{mL} $$ c. If the pH at 50.0 \(\mathrm{mL}\) NaOH added is 4.0 and the pH at 150.0 \(\mathrm{mL}\) . NaOH added is \(8.0,\) determine the values \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) for the diprotic acid.

Which of the following can be classified as buffer solutions? $$ \begin{array}{l}{\text { a. } 0.25 M \mathrm{HBr}+0.25 \mathrm{M} \mathrm{HOBr}} \\ {\text { b. } 0.15 \mathrm{M} \mathrm{HClO}_{4}+0.20 \mathrm{M} \mathrm{RbOH}} \\ {\text { c. } 0.50 \mathrm{M} \mathrm{HOCl}+0.35 \mathrm{MKOCl}}\end{array} $$ $$ \begin{array}{l}{\text { d. } 0.70 M \mathrm{KOH}+0.70 \mathrm{M} \text { HONH_ }} \\ {\text { e. } 0.85 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}+0.60 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3}}\end{array} $$
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