In the titration of a weak base, such as methylamine (CH₃NH₂), with a strong acid like hydrochloric acid (HCl), it is essential to understand the underlying chemistry and calculations involved. Methylamine has a base dissociation constant (K_b) of 4.4 × 10⁻⁴, indicating that it is a weak base. When titrated with HCl, the reaction can be represented as:

CH₃NH₂ + HCl → CH₃NH₃⁺ + Cl⁻

In this scenario, we start with 50 mL of 0.150 M methylamine and 75 mL of 0.200 M hydrochloric acid. To find the pH after the titration, we first need to calculate the moles of each reactant. Converting the volumes from milliliters to liters gives:

For methylamine:
Moles = Volume (L) × Molarity (mol/L) = 0.050 L × 0.150 mol/L = 0.0075 moles

For hydrochloric acid:
Moles = Volume (L) × Molarity (mol/L) = 0.075 L × 0.200 mol/L = 0.015 moles

Since we have more moles of HCl than methylamine, the weak base will be completely consumed, leaving us with excess strong acid. The remaining moles of HCl can be calculated as:

Remaining moles of HCl = 0.015 moles (initial) - 0.0075 moles (reacted) = 0.0075 moles

Next, we determine the total volume of the solution after mixing, which is:

Total Volume = 50 mL + 75 mL = 125 mL = 0.125 L

Now, we can find the concentration of the remaining HCl:

Concentration of HCl = Remaining moles / Total volume = 0.0075 moles / 0.125 L = 0.060 M

To calculate the pH, we use the formula:

pH = -log[H⁺] = -log(0.060) ≈ 1.22

This pH value indicates that we are in the region after the equivalence point of the titration, where excess strong acid significantly influences the pH. Understanding these calculations and concepts is crucial for analyzing weak base-strong acid titrations effectively.

Struggling with Analytical Chemistry?