Dimethylamine, (CH3)2NH, is a weak base ( Kb = 5.4 x 10?4). What is the pH of a 0.50 M solution and what is the percent dissociation?
A) 12.2, 3.2%
B) 0.016, 3.2%
C) 8.5, 6.3 x 10?4 %
D) 1.8, 3.2%
E) 5.5, 6.3 x 10?4 %
The correct answer and explanation is :
To determine the pH and percent dissociation of a 0.50 M dimethylamine ((\text{CH}_3)_2\text{NH}), follow these steps:
Step 1: Write the Base Dissociation Equation
[
(\text{CH}_3)_2\text{NH} + H_2O \rightleftharpoons (\text{CH}_3)_2\text{NH}_2^+ + OH^-
]
Given:
- Base dissociation constant, (K_b = 5.4 \times 10^{-4})
- Initial concentration, ( C = 0.50 ) M
Step 2: Set Up ICE Table
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| ((\text{CH}_3)_2\text{NH}) | 0.50 | (-x) | (0.50 – x) |
| ((\text{CH}_3)_2\text{NH}_2^+) | 0 | (+x) | (x) |
| (\text{OH}^-) | 0 | (+x) | (x) |
Using the Kb expression:
[
K_b = \frac{[\text{OH}^-][(\text{CH}_3)_2\text{NH}_2^+]}{[(\text{CH}_3)_2\text{NH}]}
]
[
5.4 \times 10^{-4} = \frac{x^2}{0.50 – x}
]
Since ( K_b ) is small, assume ( x \ll 0.50 ), so ( 0.50 – x \approx 0.50 ):
[
5.4 \times 10^{-4} = \frac{x^2}{0.50}
]
[
x^2 = (5.4 \times 10^{-4}) \times 0.50
]
[
x^2 = 2.7 \times 10^{-4}
]
[
x = \sqrt{2.7 \times 10^{-4}}
]
[
x = 0.0164 \text{ M} = [\text{OH}^-]
]
Step 3: Find pOH and pH
[
\text{pOH} = -\log (0.0164)
]
[
\text{pOH} = 1.78
]
[
\text{pH} = 14 – 1.78 = 12.2
]
Step 4: Find Percent Dissociation
[
\% \text{Dissociation} = \left( \frac{x}{C} \right) \times 100
]
[
= \left( \frac{0.0164}{0.50} \right) \times 100
]
[
= 3.28\% \approx 3.2\%
]
Final Answer:
[
\boxed{A) \text{ pH } = 12.2, \text{ Percent Dissociation } = 3.2\%}
]
Now, here’s an image representing the dissociation of dimethylamine in water:
