Question

1.When a 24.8 mL sample of a 0.469 M aqueous hydrofluoric acid solution is titrated with a 0.395 M aqueous sodium hydroxide solution, what is the pH at the midpoint in the titration?

2.What is the pH at the equivalence point in the titration of a 26.8 mL sample of a 0.436 M aqueous acetic acid solution with a 0.384 M aqueous barium hydroxide solution?

3.A 41.4 mL sample of a 0.378 M aqueous hydrofluoric acid solution is titrated with a 0.461 M aqueous barium hydroxide solution. What is the pH after 5.48 mL of base have been added?

4.A 22.7 mL sample of 0.301 M ammonia, NH₃, is titrated with 0.245 M hydroiodic acid.At the titration midpoint, the pH is

5.A 22.9 mL sample of 0.345 M triethylamine, (C₂H₅)₃N, is titrated with 0.337 M hydroiodic acid.At the equivalence point, the pH is

6. A 22.2 mL sample of 0.243 M ethylamine, C₂H₅NH₂, is titrated with 0.241 M nitric acid.After adding 8.84 mL of nitric acid, the pH is

Answer 1

1)

mmoles of HF = 24.8 x 0.469 = 11.63

At titation midpoint : pH = pKa

pKa of HF = 3.20

pH = 3.20

2)

2 CH3COOH + Ba(OH)2   --------> Ba(CH3COO)2 + 2 H2O

M1V1 / n1 = M2 V2 / n2

0.436 x 26.8 / 2 = 0.384 x V2 / 1

V2 = 15.2 mL

At equivalence point salt reamins.

salt concentraiton = 11.685 / 26.8 + 15.2 = 0.278 M

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (4.74 + log 0.278)

pH = 9.09

3)

mmoles of HF = 41.4 x 0.378 = 15.65

mmoles of Ba(OH)2 = 2.526

2 HF + Ba(OH)2   -------------> BaF2 + 2 H2O

15.65       2.526                           0               0

10.60         0                               2.526

pH = pKa + log [salt / acid]

    = 3.20 + log [2.526 / 10.60]

pH = 2.58

4)

pH = 9.26