Chapter 19: Problem 14
What are the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and the \(\mathrm{pH}\) of a buffer that consists of \(0.55 M \mathrm{HNO}_{2}\) and \(0.75 \mathrm{M} \mathrm{KNO}_{2}\left(\mathrm{~K}_{\mathrm{a}}\right.\) of \(\left.\mathrm{HNO}_{2}=7.1 \times 10^{-4}\right) ?\)
Short Answer
Expert verified
pH = 3.28[H3O+] = 5.2 x 10^-4 M
Step by step solution
01
Write the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation relates pH, pKa, and the concentrations of an acid and its conjugate base. It is written as:\[\text{pH} = \text{p}K_a + \text{log}\frac{[\text{A}^-]}{[\text{HA}]}\]where [A^-] is the concentration of the conjugate base (in this case, \text{KNO}_2) and [HA] is the concentration of the acid (\text{HNO}_2).
02
Calculate pKa from Ka
The pKa can be calculated using the formula:\[\text{p}K_a = -\text{log}(\text{K}_a)\]Given \(K_a\text{ of }\text{HNO}_2 = 7.1 \times 10^{-4}\), calculate pKa:\[\text{p}K_a = -\text{log}(7.1 \times 10^{-4}) \ \ti\approx 3.15\]
03
Plug Values into Henderson-Hasselbalch Equation
Use the values of given concentrations and the calculated pKa to find the pH:\[\text{pH} = 3.15 + \text{log} \frac{0.75 \text{ M}}{0.55 \text{ M}} = 3.15 + \text{log}(1.36) \ \ti\approx 3.15 + 0.13 \ \ti\approx 3.28\]
04
Calculate \left[\text{H}_3\text{O}^+\right]
Use the result from the pH calculation to find \left[\text{H}_3\text{O}^+\right using the formula:\[\text{pH} = -\text{log}(\text{[H}_3\text{O}^+\text{]}) \ \tiso\ \ti\text{[H}_3\text{O}^+\text{]} = 10^{-\text{pH}}\]Substitute the value of the pH:\[\text{[H}_3\text{O}^+\text{]} = 10^{-3.28} \ \ti\approx 5.2 \times 10^{-4} \text{ M}\]
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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 essential for understanding buffer solutions. This equation links together the pH, pKa, and the concentrations of an acid and its conjugate base. The formula looks like this: \[\text{pH} = \text{p}K_a + \text{log}\frac{[\text{A}^-]}{[\text{HA}]}\] Here, \( [\text{A}^-] \) is the concentration of the conjugate base (e.g., \(\text{KNO}_2\)) and \( [\text{HA}] \) is the concentration of the acid (e.g., \( \text{HNO}_2 \)). This equation allows us to predict the pH of a solution when we know the pKa and the ratio of the concentrations of the base and the acid.
- The term \( \text{p}K_a \) is derived from the acid dissociation constant \( K_a \).
pKa Calculation
Calculating pKa from Ka is a fundamental step in using the Henderson-Hasselbalch equation. The pKa is found using the following equation: \[\text{p}K_a = -\text{log}(\text{K}_a)\] Given an acid dissociation constant \(\text{K}_a\), you can find its pKa. For our exercise, the \( \text{K}_a \) for \( \text{HNO}_2 \) is \( 7.1 \times 10^{-4} \). So, \[\text{p}K_a = -\text{log}(7.1 \times 10^{-4}) \approx 3.15\]
- This means \( \text{HNO}_2 \) has a pKa of approximately 3.15.
Acid-Base Equilibrium
Acid-base equilibrium is at the heart of understanding how buffers work. A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid. These components work together to resist changes in pH when small amounts of acid or base are added. Here’s what you need to know:
- The weak acid (e.g., \( \text{HNO}_2 \)) partially dissociates in water, releasing \( \text{H}^+ \) ions.
- The conjugate base (e.g., \( \text{KNO}_2 \)) can accept \( \text{H}^+ \) ions, neutralizing added acids.
pH Calculation
The pH of a solution indicates its acidity or alkalinity. Calculating the pH of a buffer involves a few straightforward steps:
- First, use the Henderson-Hasselbalch equation: \[\text{pH} = 3.15 + \text{log}\frac{0.75 \text{ M}}{0.55 \text{ M}} \approx 3.15 + 0.13 \approx 3.28\]
- This gives the pH of the buffer solution as approximately 3.28.
- Next, calculate the hydronium ion concentration \( \text{[H}_3\text{O}^+] \) from the pH: \[\text{[H}_3\text{O}^+] = 10^{-\text{pH}} \approx 10^{-3.28} \approx 5.2 \times 10^{-4} \text{ M}\]