A solution is prepared at 25 °C that is initially 0.11 M in dimethylamine ((CH,) NH), a weak base with K,=5.4 × 10*, and 0.44M in dimethylammonium chloride ((CH3),NH½CI). Calculate the pH of the solution. Round your answer to 2 decimal places. pH =
A solution is prepared at 25 °C that is initially 0.11 M in dimethylamine ((CH,) NH), a weak base with K,=5.4 × 10*, and 0.44M in dimethylammonium chloride ((CH3),NH½CI). Calculate the pH of the solution. Round your answer to 2 decimal places. pH =
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![### Weak Base and its Conjugate Acid Buffer Solution
**Problem Statement:**
A solution is prepared at 25 °C that is initially 0.11 M in dimethylamine \(((CH_3)_2NH)\), a weak base with \(K_b = 5.4 \times 10^{-4}\), and 0.44 M in dimethylammonium chloride \(((CH_3)_2NH_2Cl)\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
1. **Given Data:**
- Weak base: Dimethylamine \(((CH_3)_2NH)\)
- Initial concentration of dimethylamine \(((CH_3)_2NH)\): 0.11 M
- Conjugate acid: Dimethylammonium chloride \(((CH_3)_2NH_2Cl)\)
- Initial concentration of dimethylammonium chloride \(((CH_3)_2NH_2Cl)\): 0.44 M
- Base dissociation constant \( (K_b) \) for dimethylamine: \(5.4 \times 10^{-4}\)
2. **Calculation of pH:**
- First, calculate the pKa from the given Kb:
\[ K_w = 10^{-14} \]
\[ K_a \cdot K_b = K_w \]
\[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{5.4 \times 10^{-4}} = 1.85 \times 10^{-11} \]
\[ pK_a = -\log K_a = -\log(1.85 \times 10^{-11}) = 10.73 \]
- Use the Henderson-Hasselbalch equation to find the pH:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
\[ \text{pH} = 10.73 + \log \left( \frac{0.11}{0.44} \right) \]
\[ \text{pH} = 10.73 + \log (0.25) \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38d49af3-6887-43af-8a6c-6e39b942bd2d%2F046343c0-1ebd-41f7-a972-455ddb9d5ba6%2Fmtj0lb7_processed.png&w=3840&q=75)
Transcribed Image Text:### Weak Base and its Conjugate Acid Buffer Solution
**Problem Statement:**
A solution is prepared at 25 °C that is initially 0.11 M in dimethylamine \(((CH_3)_2NH)\), a weak base with \(K_b = 5.4 \times 10^{-4}\), and 0.44 M in dimethylammonium chloride \(((CH_3)_2NH_2Cl)\). Calculate the pH of the solution. Round your answer to 2 decimal places.
**Solution:**
1. **Given Data:**
- Weak base: Dimethylamine \(((CH_3)_2NH)\)
- Initial concentration of dimethylamine \(((CH_3)_2NH)\): 0.11 M
- Conjugate acid: Dimethylammonium chloride \(((CH_3)_2NH_2Cl)\)
- Initial concentration of dimethylammonium chloride \(((CH_3)_2NH_2Cl)\): 0.44 M
- Base dissociation constant \( (K_b) \) for dimethylamine: \(5.4 \times 10^{-4}\)
2. **Calculation of pH:**
- First, calculate the pKa from the given Kb:
\[ K_w = 10^{-14} \]
\[ K_a \cdot K_b = K_w \]
\[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{5.4 \times 10^{-4}} = 1.85 \times 10^{-11} \]
\[ pK_a = -\log K_a = -\log(1.85 \times 10^{-11}) = 10.73 \]
- Use the Henderson-Hasselbalch equation to find the pH:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \]
\[ \text{pH} = 10.73 + \log \left( \frac{0.11}{0.44} \right) \]
\[ \text{pH} = 10.73 + \log (0.25) \
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