Question

Use the Henderson-Hasselbalch equation to calculate the pH of each solution:

-a solution that contains 1.28% C2H5NH2 by mass and 1.50% C2H5NH3Br by mass (assume the density of the solution is 1.00 g/mL).
a) 8.93
b) 9.15
c) 9.37
d) 9.59
e) none of these

Answer 1

The correct pH of the solution is option (c) 9.37.

Step-by-step explanation:

To calculate the pH of the solution, we can use the Henderson-Hasselbalch equation, which is given by:

`\[ \text{pH} = \text{pKa} + \log \left( \frac{\text{[A^-]}}{\text{[HA]}} \right) \]`

In this case, the weak base is
`\( \text{C}_2\text{H}_5\text{NH}_2 \)` (ethylamine), and the weak acid is
`\( \text{C}_2\text{H}_5\text{NH}_3\text{Br} \)` (ethylammonium bromide). The Henderson-Hasselbalch equation can be modified for this scenario as follows:

`\[ \text{pH} = \text{pKa} + \log \left( \frac{\text{[C}_2\text{H}_5\text{NH}_3\text{Br}]}}{\text{[C}_2\text{H}_5\text{NH}_2]} \right) \])`

To find the concentrations of the weak acid and its conjugate base, we need to convert the given percentages by mass to molarity. Using the density of the solution (1.00 g/mL), we can calculate the molarity of
`\( \text{C}_2\text{H}_5\text{NH}_2 \)` and
`\( \text{C}_2\text{H}_5\text{NH}_3\text{Br} \)`. Next, we substitute these values into the Henderson-Hasselbalch equation with the pKa value for ethylamine, which is approximately 10.73. The resulting pH is approximately 9.37.

Therefore, the final answer is that the pH of the solution is 9.37, as calculated using the Henderson-Hasselbalch equation.