Question

Please show your work.

Find the pH during the titration of 20.00 ml of0.1000 M butanoic acid, CH3CH2CH2COOH (Ka=1.54*10^-5), with 0.1000 M NaOH solution after the following additions of titrant.

a) 13.00 ml?pH=?

b) 20.50 ml?pH=?

c) 27.00 ml?pH=?

Answer 1

Let HA be the shorthand for butanoic acid


At 13.0, 20.50, and 27 mL NaOH added we will have a buffer solution composed of HA and A^-1. we will use Henderson-Hasselbalch equation-
pH = pKa + log(A-/HA)

at 13 mL NaOH added

moles of [HA] = 20 X 0.1 = 2 millimoles

moles of [NaOH] = 1.3 millimoles

HA + NaOH --> Na+A-   + H2O

so moles of salt = 1.3 millimoles

so moles of acid = 0.7 millimoles
pH =pKa + log [salt / acid] = 4.8 + log [1.3 / 0.7] = 4.8 + 0.26 = 5.06

at 20.50 mL NaOH added, so millimoles = 20.5 X 0.1 = 2.05 millimoles

The whole acid is neutralized by base and 0.05 millimoles of NaOH will be left.

pOH = -log [OH-] = 4.3

pH = 9.6
3)

at 27 mL NaOH added, so millimoles = 27X 0.1 = 2.7 millimoles

The whole acid is neutralized by base and 7 millimoles of NaOH will be left.

pOH = -log [OH-] = 2.15

pH = 11.84