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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]
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