Connect with a community that values knowledge and expertise on IDNLearn.com. Ask anything and receive prompt, well-informed answers from our community of knowledgeable experts.
Sagot :
To determine the concentration of [tex]\(\left[ H_3O^+ \right]\)[/tex] and the pH of the buffer solution consisting of [tex]\(0.32 \, \text{M} \, \text{HNO}_2\)[/tex] and [tex]\(0.66 \, \text{M} \, \text{KNO}_2\)[/tex], we need to follow a detailed step-by-step approach.
### Step-by-Step Solution
1. Identify the given information:
- The concentration of [tex]\(\text{HNO}_2\)[/tex] (a weak acid) is [tex]\(0.32 \, \text{M}\)[/tex].
- The concentration of [tex]\(\text{KNO}_2\)[/tex] (which provides the conjugate base [tex]\(\text{NO}_2^-\)[/tex]) is [tex]\(0.66 \, \text{M}\)[/tex].
- The acid dissociation constant [tex]\(K_a\)[/tex] for [tex]\(\text{HNO}_2\)[/tex] at [tex]\(25^{\circ} \text{C}\)[/tex] is [tex]\(4.5 \times 10^{-4}\)[/tex].
2. Use the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation for a buffer solution is:
[tex]\[ \text{pH} = \text{p}K_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
where:
- [tex]\([\text{HA}]\)[/tex] is the concentration of the weak acid [tex]\(\text{HNO}_2\)[/tex].
- [tex]\([\text{A}^-]\)[/tex] is the concentration of the conjugate base [tex]\(\text{NO}_2^-\)[/tex].
- [tex]\(\text{p}K_a\)[/tex] is the negative logarithm of the acid dissociation constant: [tex]\(\text{p}K_a = -\log{K_a}\)[/tex].
3. Calculate [tex]\(\text{p}K_a\)[/tex]:
[tex]\[ \text{p}K_a = -\log(4.5 \times 10^{-4}) \approx 3.35 \][/tex]
4. Calculate the [tex]\(\text{pH}\)[/tex] of the buffer:
[tex]\[ \text{pH} = 3.35 + \log\left( \frac{0.66}{0.32} \right) \][/tex]
Calculate the ratio:
[tex]\[ \frac{0.66}{0.32} \approx 2.0625 \][/tex]
Take the logarithm:
[tex]\[ \log(2.0625) \approx 0.314 \][/tex]
Now, sum this with [tex]\(\text{p}K_a\)[/tex]:
[tex]\[ \text{pH} = 3.35 + 0.314 \approx 3.66 \][/tex]
5. Determine the [tex]\(\left[ H_3O^+ \right]\)[/tex]:
The relationship between [tex]\(\left[ H_3O^+ \right]\)[/tex] and pH is given by:
[tex]\[ \text{pH} = -\log[\text{H}_3\text{O}^+] \][/tex]
Rearrange to solve for [tex]\(\left[ H_3O^+ \right]\)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \][/tex]
Substitute the pH value:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-3.66} \approx 2.18 \times 10^{-4} \, \text{M} \][/tex]
### Final Answers
- The concentration of [tex]\(\left[ H_3O^+ \right]\)[/tex] is [tex]\(2.18 \times 10^{-4} \, \text{M}\)[/tex].
- The pH of the buffer solution is approximately [tex]\(3.66\)[/tex].
So, the answers with the correct significant figures are:
[tex]\[ \left[ H_3O^+ \right] = 2.18 \times 10^{-4} \, \text{M} \][/tex]
[tex]\[ \text{pH} \approx 3.66 \][/tex]
### Step-by-Step Solution
1. Identify the given information:
- The concentration of [tex]\(\text{HNO}_2\)[/tex] (a weak acid) is [tex]\(0.32 \, \text{M}\)[/tex].
- The concentration of [tex]\(\text{KNO}_2\)[/tex] (which provides the conjugate base [tex]\(\text{NO}_2^-\)[/tex]) is [tex]\(0.66 \, \text{M}\)[/tex].
- The acid dissociation constant [tex]\(K_a\)[/tex] for [tex]\(\text{HNO}_2\)[/tex] at [tex]\(25^{\circ} \text{C}\)[/tex] is [tex]\(4.5 \times 10^{-4}\)[/tex].
2. Use the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation for a buffer solution is:
[tex]\[ \text{pH} = \text{p}K_a + \log\left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \][/tex]
where:
- [tex]\([\text{HA}]\)[/tex] is the concentration of the weak acid [tex]\(\text{HNO}_2\)[/tex].
- [tex]\([\text{A}^-]\)[/tex] is the concentration of the conjugate base [tex]\(\text{NO}_2^-\)[/tex].
- [tex]\(\text{p}K_a\)[/tex] is the negative logarithm of the acid dissociation constant: [tex]\(\text{p}K_a = -\log{K_a}\)[/tex].
3. Calculate [tex]\(\text{p}K_a\)[/tex]:
[tex]\[ \text{p}K_a = -\log(4.5 \times 10^{-4}) \approx 3.35 \][/tex]
4. Calculate the [tex]\(\text{pH}\)[/tex] of the buffer:
[tex]\[ \text{pH} = 3.35 + \log\left( \frac{0.66}{0.32} \right) \][/tex]
Calculate the ratio:
[tex]\[ \frac{0.66}{0.32} \approx 2.0625 \][/tex]
Take the logarithm:
[tex]\[ \log(2.0625) \approx 0.314 \][/tex]
Now, sum this with [tex]\(\text{p}K_a\)[/tex]:
[tex]\[ \text{pH} = 3.35 + 0.314 \approx 3.66 \][/tex]
5. Determine the [tex]\(\left[ H_3O^+ \right]\)[/tex]:
The relationship between [tex]\(\left[ H_3O^+ \right]\)[/tex] and pH is given by:
[tex]\[ \text{pH} = -\log[\text{H}_3\text{O}^+] \][/tex]
Rearrange to solve for [tex]\(\left[ H_3O^+ \right]\)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-\text{pH}} \][/tex]
Substitute the pH value:
[tex]\[ [\text{H}_3\text{O}^+] = 10^{-3.66} \approx 2.18 \times 10^{-4} \, \text{M} \][/tex]
### Final Answers
- The concentration of [tex]\(\left[ H_3O^+ \right]\)[/tex] is [tex]\(2.18 \times 10^{-4} \, \text{M}\)[/tex].
- The pH of the buffer solution is approximately [tex]\(3.66\)[/tex].
So, the answers with the correct significant figures are:
[tex]\[ \left[ H_3O^+ \right] = 2.18 \times 10^{-4} \, \text{M} \][/tex]
[tex]\[ \text{pH} \approx 3.66 \][/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.