Question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54 × 10−5), with 0.1000 M NaOH solution after the following additions of titrant.

a. 20.10 mL:

pH =

b. 26.00 mL:

pH =

c. 13.00 mL

pH =

Answer 1

Given , Ka = 1.54 x 10^-5

pKa = -log Ka

= -log (1.54 x 10^-5)

pKa = 4.81

a) 20.0 ml

The millimoles of acid = HA

= 20 x 0.1

= 2

The millimoles of OH- = 20.10 x 0.1

= 2.01

HA + OH- -----------------> H2O + A-

2 2.01 0        0

-0.01 0 2.01

pH = pKa + log [A-/ HA]

pH = 4.81 + log (2.01 / -0.01)

= 4.81 - 2.30

pH = 2.51

b) 26.00 mL

[OH-] = 26 x 0.1 - 20 x 0.1 / (20 + 26)

            = 2.55 M

pOH = -log [OH-]

= -log (2.55)

= 0.406

pH + pOH = 14

pH = 14 - pOH

= 14 - 0.406

pH = 13.59

c) 13.00 mL

[OH-]   = 1.3 - 2.1 / (20+13)

= 1.237 M

pOH = -log(OH^-)

= -log (1.237)

= - 0.0923

pH + pOH =14

pH = 14 + 0.0923

pH  = 14.09