Chapter 10: Problem 80
What is the \(\mathrm{pH}\) of a buffer system that contains \(0.15 \mathrm{M}\) \(\mathrm{NH}_{4}^{+}\) and \(0.10 \mathrm{M} \mathrm{NH}_{3}\) ? The \(\mathrm{p} K_{\mathrm{a}}\) of \(\mathrm{NH}_{4}^{+}\) is 9.25.
Short Answer
Expert verified
The \(\mathrm{pH}\) of the buffer system is approximately 9.07.
Step by step solution
01
Identify the Buffer Components
In this problem, the buffer system contains the weak acid \(\mathrm{NH}_4^+\) and its conjugate base \(\mathrm{NH}_3\). We are given the concentrations of \(\mathrm{NH}_4^+\) which is \(0.15 \mathrm{M}\) and \(\mathrm{NH}_3\) which is \(0.10 \mathrm{M}\).
02
Recall the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is used to calculate the \(\mathrm{pH}\) of a buffer solution: \[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \left( \frac{[\mathrm{base}]}{[\mathrm{acid}]} \right) \] For this buffer, the base is \(\mathrm{NH}_3\) and the acid is \(\mathrm{NH}_4^+\).
03
Plug Values into the Henderson-Hasselbalch Equation
The provided values are \([\mathrm{NH}_3] = 0.10 \mathrm{M}\) and \([\mathrm{NH}_4^+] = 0.15 \mathrm{M}\), with \(\mathrm{p}K_{a} = 9.25\). Now substitute these into the equation: \[ \mathrm{pH} = 9.25 + \log \left( \frac{0.10}{0.15} \right) \]
04
Calculate the Logarithm
First, compute the fraction \(\frac{0.10}{0.15}\), which equals approximately \(0.6667\). Then, find \(\log(0.6667)\), which is approximately \(-0.1761\).
05
Determine the pH
Substitute the calculated logarithm back into the Henderson-Hasselbalch equation: \[ \mathrm{pH} = 9.25 - 0.1761 = 9.0739 \] Thus, the \(\mathrm{pH}\) of the buffer system is approximately 9.07.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is an essential tool in chemistry for calculating the pH of buffer solutions. A buffer solution contains a weak acid and its conjugate base, or vice versa, to resist changes in pH when small amounts of acids or bases are added. The equation is expressed as follows:
\[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]
This formula calculates the pH based on the concentration ratio of the base form to the acid form of the buffer components, along with the pK_a of the weak acid. The pK_a value is a measure of the acid’s strength, indicating how easily it donates protons. By knowing the concentrations of the buffer components and the pK_a, you can easily determine the pH of the solution using this equation.
\[ \mathrm{pH} = \mathrm{p}K_{\mathrm{a}} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]
This formula calculates the pH based on the concentration ratio of the base form to the acid form of the buffer components, along with the pK_a of the weak acid. The pK_a value is a measure of the acid’s strength, indicating how easily it donates protons. By knowing the concentrations of the buffer components and the pK_a, you can easily determine the pH of the solution using this equation.
weak acid
In a chemical context, a weak acid is one that is not fully ionized in a solution. Unlike strong acids which completely dissociate into ions, weak acids only partially dissociate.
Key characteristics of weak acids include:
Key characteristics of weak acids include:
- They have higher pK_a values indicating weaker acidity.
- They establish an equilibrium between the undissociated acid and its ions.
- Their pH levels are usually above 4.5 but below 7.
conjugate base
The conjugate base is the species formed when a weak acid loses a proton. It is a part of the acid-base pair that can accept protons, helping to stabilize the pH of the solution.
Some important points about conjugate bases include:
Some important points about conjugate bases include:
- They are typically found in equilibrium with their corresponding weak acid.
- Their presence is essential for the function of a buffer solution.
- They can neutralize small amounts of added acid by accepting protons.
pK_a
pK_a is a fundamental concept in chemistry that describes the acid dissociation constant in logarithmic form. It quantifies the strength of a weak acid:
- The lower the pK_a, the stronger the acid and the more it tends to release protons.- The higher the pK_a, the weaker the acid, with fewer protons being released.
The pK_a value is used in the Henderson-Hasselbalch equation to help determine the pH of a solution. In the exercise provided, the pK_a of ammonium ion \(\mathrm{NH}_4^+\) is 9.25. This higher pK_a value indicates that \(\mathrm{NH}_4^+\) does not release its protons easily, characterizing it as a weak acid. Understanding pK_a helps predict how changes in concentration affect the pH in buffer systems.
- The lower the pK_a, the stronger the acid and the more it tends to release protons.- The higher the pK_a, the weaker the acid, with fewer protons being released.
The pK_a value is used in the Henderson-Hasselbalch equation to help determine the pH of a solution. In the exercise provided, the pK_a of ammonium ion \(\mathrm{NH}_4^+\) is 9.25. This higher pK_a value indicates that \(\mathrm{NH}_4^+\) does not release its protons easily, characterizing it as a weak acid. Understanding pK_a helps predict how changes in concentration affect the pH in buffer systems.