Potentiometric Titration

Question

Generate a titration curve for the titration of 50mL methyl amine (0.02M) with hydrochloric acid (0.02). All calculations with 5mL increment for pre and post equivalence and 0.2 mL increment at 5ml before the equivalence point. A table would suffice.

Answer 1

Reaction between methylamine (a base) and hydrochloric acid is as follows:

CH₃NH₂(aq) + HCl(aq) ==> CH₃NH₃⁺(aq) + Cl^-(aq)

Note that with each addition of HCl, a buffer is formed containing the original base (CH₃NH₂) plus the conjugate acid (CH₃NH₃⁺). So, at each 5 ml interval of HCl (until 5 ml prior to equivalence), and then for each 0.2 ml interval until equivalence, and the for each 5 ml HCl after equivalence, you must calculate the pH. This can be accomplished by using the Henderson Hasselbalch equation:

pOH = pKb + log [conj.acid] / [base]. Solve for pOH and then convert to pH. You'll have to look up the pKb (or Kb) for methylamine. Make a table of mls HCl vs pH.

The calculation for the FIRST 5 mls of HCl is shown below as an example. You will have to do this type of calculation for the other additions of HCl

Initial moles of CH₃NH₂ = 0.05 L x 0.02 mol/L = 0.001 mols CH₃NH₂

Moles HCl added in the first 5 mls = 0.005 L x 0.02 mol/L = 0.0001 mols HCl

Moles of CH₃NH₃⁺ formed = 0.0001 mols CH₃NH₃⁺

Moles CH₃NH₂ remaining = 0.001 - 0.0001 = 0.0009 mol CH₃NH₂

pOH = pKb + log (0.0001/0.001) after looking up pKb, solve for pOH and then pH

Repeat this for each addition of HCl as described in the question.