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, and 0.25 Ve, 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