Question

A 25.0 mL sample of 0.100 M acetic acid is titrated with 0.125 M NaOH at 25°C. What is the pH of the solution after 10.0, 20.0, and 30.0 mL of the base have been added?

Answer 1

a)pH = 4.74

b) pH = 6.75

c) pH = 12.36

Step-by-step explanation:

In titration of strong base with a weak acid, initially we get salt of the acid and base and due to presence of weak acid in the solution this results into formation of a buffer.

On complete neutralization, the salt undergoes hydrolysis.

After complete neutralization, the solution becomes basic and the pH is due to presence of hydroxide ions.

a) on addition of 10.0 mL of NaOH

The initial moles of acetic acid present = molarity X volume = 0.1 X 0.025 =0.0025

The moles of NaOH added = molarity X volume = 0.125 X 0.01 = 0.00125

The moles of acid left after reaction with base = 0.0025 -0.00125 = 0.00125

the moles of salt formed = 0.00125

the pH of solution will be pH of the buffer formed [pKa of acetic acid =4.74]

`pH = pKa + log([salt])/([acid]) =4.74 + log [0.00125 / 0.000125] = 4.74`

b) Addition of 20.0 mL of NaOH

The initial moles of acetic acid present = molarity X volume = 0.1 X 0.025 =0.0025

The moles of NaOH added = molarity X volume = 0.125 X 0.02 = 0.0025

The moles of acid left after reaction with base = 0.0025 -0.0025 = 0

the moles of salt formed = 0.0025

The salt will undergo hydrolysis as:

`CH_(3)COO^(-)  +H_(2)O  --->  CH_(3)COOH + OH^(-)`

The Kb will be

`Kb = ([CH_(3)COOH][OH^(-)])/([CH_(3)COO^(-)])`

[salt] = moles / total volume = 0.0025 / 25+20 = 5.56 X 10⁻⁶ M

Kb = 5.56 X 10⁻¹⁰ M

therefore

5.56 X 10⁻¹⁰ M X 5.56 X 10⁻⁶ M = [OH⁻]²

[OH⁻] = 5.56 X 10⁻⁸ M

pOH = 7.25

pH = 14 -7.25 = 6.75

c) on addition of 30 mL of NaOH 20 mL will be used for complete neutralization and rest 10 mL will be of NaOH only.

moles of NaOH left = molarity X volume = 0.125 X 10 mL = 0.00125

volume of solution = 30 + 25 = 55

the [NaOH] = moles / volume = 0.023 M

pOH = 1.64

pH = 12.36