MF16014: What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M methylamine wiht 25.00 mL of 0.10 M methylammonium chloride? Assume that volumes are additive and that Kb = 3.70 x 10-4 for CH3NH2.

how do i go about this question?

If you know how to use the Henderson-Hasselbalch equation, this is the time.
(base) = 0.025*0.1/0.050
(acid) = 0.025*0.01/0.050

To solve this question, you can follow these steps:

1. Determine the concentration of the base (methylamine) and the acid (methylammonium chloride) solutions.
- Base concentration (Methylamine): 0.10 M
- Acid concentration (Methylammonium chloride): 0.10 M

2. Calculate the moles of base and acid used in the mixture.
- Moles of base: Volume of base solution (25.00 mL) * Base concentration (0.10 M)
- Moles of acid: Volume of acid solution (25.00 mL) * Acid concentration (0.10 M)

3. Determine the total volume of the mixture.
- Total volume: Volume of base solution (25.00 mL) + Volume of acid solution (25.00 mL)

4. Calculate the combined concentration of the base and acid in the mixture.
- Combined concentration: Moles of base / Total volume

5. Use the Henderson-Hasselbalch equation to find the pH of the solution.
- pH = pKa + log10([base]/[acid])

6. For this question, we need to make a small modification to the Henderson-Hasselbalch equation. Since we have the base and acid concentrations, we can use the equation:
- pH = pKa + log10([base concentration]/[acid concentration])
- pKa is determined by the Ka or Kb value given for the reaction. In this case, it is Kb = 3.70 x 10^-4 for CH3NH2.

7. Plug in the values into the equation and calculate the pH of the solution.

By following these steps, you should be able to find the pH of the solution prepared by mixing the two solutions.