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Certainly! Let's calculate the pH at the equivalence point for the titration of 0.230 M methylamine [tex]\((\text{CH}_3\text{NH}_2)\)[/tex] with 0.230 M HCl.
### Step-by-Step Solution:
1. Given:
- Concentration of methylamine: [tex]\(0.230 \, \text{M}\)[/tex]
- Concentration of HCl: [tex]\(0.230 \, \text{M}\)[/tex]
- [tex]\(K_b\)[/tex] of methylamine: [tex]\(5.0 \times 10^{-4}\)[/tex]
2. Determine [tex]\(K_a\)[/tex] for the Conjugate Acid:
Methylamine is a weak base, and at the equivalence point, we need to consider its conjugate acid. The [tex]\(K_a\)[/tex] value of the conjugate acid [tex]\((\text{CH}_3\text{NH}_3^+)\)[/tex] can be found using the relationship between [tex]\(K_a\)[/tex] and [tex]\(K_b\)[/tex]:
[tex]\[ K_w = 1.0 \times 10^{-14} \][/tex]
[tex]\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-4}} = 2.0 \times 10^{-11} \][/tex]
3. Calculate [tex]\(pK_a\)[/tex]:
[tex]\[ pK_a = -\log_{10}(K_a) \][/tex]
[tex]\[ pK_a = -\log_{10}(2.0 \times 10^{-11}) \approx 10.70 \][/tex]
4. Consider the Equivalence Point:
At the equivalence point, the methylamine has been neutralized by HCl to form its conjugate acid [tex]\((\text{CH}_3\text{NH}_3^+)\)[/tex]. The concentration of [tex]\(\text{CH}_3\text{NH}_3^+\)[/tex] will be 0.230 M (same as the initial concentration of methylamine).
5. Set Up the ICE Table for the Conjugate Acid Dissociation:
[tex]\[ \text{CH}_3\text{NH}_3^+ \leftrightharpoons \text{CH}_3\text{NH}_2 + \text{H}^+ \][/tex]
- Initial: [tex]\([\text{CH}_3\text{NH}_3^+] = 0.230 \, \text{M}\)[/tex], [[tex]\(\text{H}^+\)[/tex]] = 0, [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] = 0
- Change: [tex]\([\text{CH}_3\text{NH}_3^+]\)[/tex] decreases by [tex]\(x\)[/tex], [[tex]\(\text{H}^+\)[/tex]] increases by [tex]\(x\)[/tex], [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] increases by [tex]\(x\)[/tex]
- Equilibrium: [tex]\([\text{CH}_3\text{NH}_3^+] = 0.230 - x\)[/tex], [[tex]\(\text{H}^+\)[/tex]] = [tex]\(x\)[/tex], [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] = [tex]\(x\)[/tex]
The [tex]\(K_a\)[/tex] expression is:
[tex]\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3^+]} = \frac{x^2}{(0.230 - x)} \][/tex]
6. Assume [tex]\(x\)[/tex] is Small Compared to 0.230 M:
[tex]\[ K_a \approx \frac{x^2}{0.230} \][/tex]
[tex]\[ 2.0 \times 10^{-11} = \frac{x^2}{0.230} \][/tex]
[tex]\[ x^2 = (2.0 \times 10^{-11}) \times 0.230 \][/tex]
[tex]\[ x \approx \sqrt{(2.0 \times 10^{-11}) \times 0.230} \approx 2.1447610589527217 \times 10^{-6} \, \text{M} \][/tex]
Here, [tex]\(x\)[/tex] represents the [tex]\([\text{H}^+]\)[/tex] concentration.
7. Calculate pH:
[tex]\[ \text{pH} = -\log_{10}(x) \][/tex]
[tex]\[ \text{pH} = -\log_{10}(2.1447610589527217 \times 10^{-6}) \approx 5.67 \][/tex]
### Final Answer:
The pH at the equivalence point for the titration of 0.230 M methylamine with 0.230 M HCl is approximately 5.67.
### Step-by-Step Solution:
1. Given:
- Concentration of methylamine: [tex]\(0.230 \, \text{M}\)[/tex]
- Concentration of HCl: [tex]\(0.230 \, \text{M}\)[/tex]
- [tex]\(K_b\)[/tex] of methylamine: [tex]\(5.0 \times 10^{-4}\)[/tex]
2. Determine [tex]\(K_a\)[/tex] for the Conjugate Acid:
Methylamine is a weak base, and at the equivalence point, we need to consider its conjugate acid. The [tex]\(K_a\)[/tex] value of the conjugate acid [tex]\((\text{CH}_3\text{NH}_3^+)\)[/tex] can be found using the relationship between [tex]\(K_a\)[/tex] and [tex]\(K_b\)[/tex]:
[tex]\[ K_w = 1.0 \times 10^{-14} \][/tex]
[tex]\[ K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{5.0 \times 10^{-4}} = 2.0 \times 10^{-11} \][/tex]
3. Calculate [tex]\(pK_a\)[/tex]:
[tex]\[ pK_a = -\log_{10}(K_a) \][/tex]
[tex]\[ pK_a = -\log_{10}(2.0 \times 10^{-11}) \approx 10.70 \][/tex]
4. Consider the Equivalence Point:
At the equivalence point, the methylamine has been neutralized by HCl to form its conjugate acid [tex]\((\text{CH}_3\text{NH}_3^+)\)[/tex]. The concentration of [tex]\(\text{CH}_3\text{NH}_3^+\)[/tex] will be 0.230 M (same as the initial concentration of methylamine).
5. Set Up the ICE Table for the Conjugate Acid Dissociation:
[tex]\[ \text{CH}_3\text{NH}_3^+ \leftrightharpoons \text{CH}_3\text{NH}_2 + \text{H}^+ \][/tex]
- Initial: [tex]\([\text{CH}_3\text{NH}_3^+] = 0.230 \, \text{M}\)[/tex], [[tex]\(\text{H}^+\)[/tex]] = 0, [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] = 0
- Change: [tex]\([\text{CH}_3\text{NH}_3^+]\)[/tex] decreases by [tex]\(x\)[/tex], [[tex]\(\text{H}^+\)[/tex]] increases by [tex]\(x\)[/tex], [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] increases by [tex]\(x\)[/tex]
- Equilibrium: [tex]\([\text{CH}_3\text{NH}_3^+] = 0.230 - x\)[/tex], [[tex]\(\text{H}^+\)[/tex]] = [tex]\(x\)[/tex], [[tex]\(\text{CH}_3\text{NH}_2\)[/tex]] = [tex]\(x\)[/tex]
The [tex]\(K_a\)[/tex] expression is:
[tex]\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{NH}_2]}{[\text{CH}_3\text{NH}_3^+]} = \frac{x^2}{(0.230 - x)} \][/tex]
6. Assume [tex]\(x\)[/tex] is Small Compared to 0.230 M:
[tex]\[ K_a \approx \frac{x^2}{0.230} \][/tex]
[tex]\[ 2.0 \times 10^{-11} = \frac{x^2}{0.230} \][/tex]
[tex]\[ x^2 = (2.0 \times 10^{-11}) \times 0.230 \][/tex]
[tex]\[ x \approx \sqrt{(2.0 \times 10^{-11}) \times 0.230} \approx 2.1447610589527217 \times 10^{-6} \, \text{M} \][/tex]
Here, [tex]\(x\)[/tex] represents the [tex]\([\text{H}^+]\)[/tex] concentration.
7. Calculate pH:
[tex]\[ \text{pH} = -\log_{10}(x) \][/tex]
[tex]\[ \text{pH} = -\log_{10}(2.1447610589527217 \times 10^{-6}) \approx 5.67 \][/tex]
### Final Answer:
The pH at the equivalence point for the titration of 0.230 M methylamine with 0.230 M HCl is approximately 5.67.
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