Question

Calculate the pH at the halfway point and at the equivalence point for 102.8 mL of 0.19 M C2H5NH2 (Kb = 5.6 ✕ 10−4) titrated with 0.38 M HCl

Answer 1

base millimoles = 102.8 x 0.19 = 19.532

at equivalence point base millimoles = acid millimoles

19.532 = 0.38 x V

V = 51.4 mL

volume of acid = 51.4 mL

pKb = -log Kb

pKb = 3.25

A ) half way equivalence point

at this point millimoles of base = millimoles of salt formed

so pOH = pKb

pOH = 3.25

pH + pOH = 14

pH = 10.75

B) at the equivalence point

at this point base and acid consumed only salt reamains

salt millimoles = 19.532

salt concentration = millimoles / total volume = 19.532 / (102.8 + 51.4) = 0.1267 M

salt is formed from strong acid + weak base

so

pH = 7 - 1/2 [pKb +log C]

pH = 7 - 1/2 [3.25 + log 0.1267 ]

pH = 5.82