Question

Please show your work. Thanks in advance.

A 100.00 mL volume of 0.0400 M propionic acid (CH3CH2COOH; Ka = 1.34 ´ 10-5 ) was titrated with 0.0837 M NaOH.

Calculate the pH after the addition of: 0 Ve, 0.25 Ve, Ve, and 1.1 Ve mL of titrant where Ve is the volume of NaOH required to reach the equivalence point.

Answer 1

1)

millimoles of propinoic acid = 100.00 x 0.0400 = 4.00

Ka = 1.34 x 10^-5

pKa = 4.87

a) 0 Ve :

pH = 1/2 (pKa - log C)

    = 1/2 (4.87 - log 0.04)

pH = 3.14

b)

at equivalence point

mmoles of acid = mmoles of base

4.00 = 0.0837 x V

Ve = 47.79 mL

0.25 Ve = 0.25 x 47.79 = 11.95

mmoles of NaOH = 11.95 x 0.0837 = 1

CH3CH2COOH    + NaOH    ---------------> CH3CH2COO-   +   H2O

      4                            1                                             0                    0

     3                            0                                                1

pH = pKa + log [salt / acid]

    = 4.87 + log [1 / 4]

pH = 4.27

c)

At equivalence point. salt only remains.

salt concentration = 4 / (100 + 47.79) = 0.027 M

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

      = 7 + 1/2 (4.87 + log 0.027)

pH = 8.65

d)

volume = 1.1 Ve = 1.1 x 47.79 = 52.57 mL

mmoles of NaOH = 52.57 x 0.0837 = 4.4

[OH-] = 4.4 - 4.0 / 100 + 52.57 = 2.62 x 10^-3 M

pOH = -log (2.62 x 10^-3) = 2.58

pH = 11.42