Question

A 27.8 mL sample of 0.373 M triethylamine, (C₂H₅)₃N, is titrated with 0.383 M hydrobromic acid.

At the equivalence point, the pH is ?

Answer 1

No.of moles of (C₂H₅)₃N = (27.8) x (0.373) = 10.37 m.moles

At equivalence point, No.of moles of acid = 10.37 m.moles

Hence, Volume of acid added = (10.37 m.moles) / (0.383) = 27.07 mL

Hence, Concentration of salt = (10.37) / (27.07 + 27.8) = 0.18897 M

Therefore, pH of the solution is given by:

pH = 7 - 1/2 (pKb + log [salt])

pH = 7 - 0.5*(3.284 + log (0.18897))

pH = 5.72 ------- (answer)