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Certainly! To determine the component concentration ratio [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex] of a buffer with a pH of 3.58, we use the Henderson-Hasselbalch equation and the given value for the acid dissociation constant ([tex]\(K_a\)[/tex]) of nitrous acid (HNO[tex]\(_2\)[/tex]). The process is as follows:
### Step-by-Step Solution:
1. Given Information:
- pH of the buffer: 3.58
- [tex]\(K_a\)[/tex] of HNO[tex]\(_2\)[/tex]: [tex]\(4.0 \times 10^{-4}\)[/tex]
2. Calculate the pK[tex]\(_a\)[/tex]:
The pK[tex]\(_a\)[/tex] is calculated using the formula:
[tex]\[ \text{p}K_a = -\log_{10}(K_a) \][/tex]
Using the given [tex]\(K_a\)[/tex] value:
[tex]\[ \text{p}K_a = -\log_{10}(4.0 \times 10^{-4}) = 3.3979400086720375 \][/tex]
3. Apply the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation is:
[tex]\[ \text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) \][/tex]
4. Rearrange for the concentration ratio:
To find the concentration ratio [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex]:
[tex]\[ \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) = \text{pH} - \text{p}K_a \][/tex]
Substitute the values for pH and pK[tex]\(_a\)[/tex]:
[tex]\[ \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) = 3.58 - 3.3979400086720375 = 0.1820599913279625 \][/tex]
5. Solve for the ratio:
To find [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex], we take the antilog (base 10) of both sides:
[tex]\[ \frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 10^{0.1820599913279625} = 1.5207575852822455 \][/tex]
6. Significant Figures:
Given the pH has 3.58 (three significant figures), the calculated ratio should also be presented with three significant figures. However, as logarithms often yield more precise steps, it is often safe to preserve decimal points in intermediate results while reporting the final ratio to significant these figures.
### Result:
- The pK[tex]\(_a\)[/tex] of HNO[tex]\(_2\)[/tex] is [tex]\(3.40\)[/tex] (keeping three significant figures).
- The concentration ratio [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex] is [tex]\(1.52\)[/tex] (to three significant figures).
Thus, the component concentration ratio [tex]\(\frac{\left[ \text{NO}_2^{-}\right]}{\left[ \text{HNO}_2\right]} = 1.52\)[/tex].
[tex]\(\boxed{3.40}\)[/tex] (pK[tex]\(_a\)[/tex])
[tex]\(\boxed{1.52}\)[/tex] (concentration ratio)
### Step-by-Step Solution:
1. Given Information:
- pH of the buffer: 3.58
- [tex]\(K_a\)[/tex] of HNO[tex]\(_2\)[/tex]: [tex]\(4.0 \times 10^{-4}\)[/tex]
2. Calculate the pK[tex]\(_a\)[/tex]:
The pK[tex]\(_a\)[/tex] is calculated using the formula:
[tex]\[ \text{p}K_a = -\log_{10}(K_a) \][/tex]
Using the given [tex]\(K_a\)[/tex] value:
[tex]\[ \text{p}K_a = -\log_{10}(4.0 \times 10^{-4}) = 3.3979400086720375 \][/tex]
3. Apply the Henderson-Hasselbalch equation:
The Henderson-Hasselbalch equation is:
[tex]\[ \text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) \][/tex]
4. Rearrange for the concentration ratio:
To find the concentration ratio [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex]:
[tex]\[ \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) = \text{pH} - \text{p}K_a \][/tex]
Substitute the values for pH and pK[tex]\(_a\)[/tex]:
[tex]\[ \log_{10}\left(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\right) = 3.58 - 3.3979400086720375 = 0.1820599913279625 \][/tex]
5. Solve for the ratio:
To find [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex], we take the antilog (base 10) of both sides:
[tex]\[ \frac{[\text{NO}_2^-]}{[\text{HNO}_2]} = 10^{0.1820599913279625} = 1.5207575852822455 \][/tex]
6. Significant Figures:
Given the pH has 3.58 (three significant figures), the calculated ratio should also be presented with three significant figures. However, as logarithms often yield more precise steps, it is often safe to preserve decimal points in intermediate results while reporting the final ratio to significant these figures.
### Result:
- The pK[tex]\(_a\)[/tex] of HNO[tex]\(_2\)[/tex] is [tex]\(3.40\)[/tex] (keeping three significant figures).
- The concentration ratio [tex]\(\frac{[\text{NO}_2^-]}{[\text{HNO}_2]}\)[/tex] is [tex]\(1.52\)[/tex] (to three significant figures).
Thus, the component concentration ratio [tex]\(\frac{\left[ \text{NO}_2^{-}\right]}{\left[ \text{HNO}_2\right]} = 1.52\)[/tex].
[tex]\(\boxed{3.40}\)[/tex] (pK[tex]\(_a\)[/tex])
[tex]\(\boxed{1.52}\)[/tex] (concentration ratio)
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