Question

A 25.0 mL sample of 0.100 M HC2H3O2 (Ka = 1.8 x 10^-5) is titrated with a 0.100 M NaOH solution. What is the pH after the addition of 12.5 mL of NaOH?

Answer 1

Solution-

Given-

[CH3COOH] = 0.10 M

Volume of CH3COOH = 25.0 mL = 0.025 L

[NaOH] = 0.10 M

Volume of NaOH = 12.5 mL = 0.0125 L

Ka = 1.8 x 10^-5

Dissociation of Acetic acid

CH3COOH_(aq) → CH3COO−_(aq) + H+_(aq)

Ka = [H+] [CH3COO-] / [CH3COOH]
we know for the weak acid (Acetic acid) [H+] = [CH3COO-]

Ka = [H+] ²/ [CH3COOH]
let’s insert the value in equation

1.8 x 10^-5 = [H+]² / 0.1 M

[H+]² = (1.8 x 10^-5) x 0.1
[H+]² = 1.8 x10^-6
[H+] = √(1.8 x 10^-6)
[H+] = 1.341 x 10^-3
pH = -log [H+]
pH = -log (1.341 x 10^-3)
pH = 2.46

When NaOH added to the CH3COOH solution

The equation becomes

CH3COOH_(aq) + NaOH → CH3COONa_(salt) + H2O_(aq)

The pH of the buffer solution is calculated using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
pKa = -log Ka = - log (1.8 x 10^-5) = 4.74

We know the

# of mole = Molarity x Volume in L
Mol CH3COOH in 25mL of 0.1M solution = 0.1 M x 0.025 L = 0.0025 mol
Mol NaOH in 12.5 mL of 0.1M solution = 0.1 M x 0.012 L = 0.0012 mol
The NaOH reacts with the CH3COOH in 1:1 ratio
Therefore salt produce = 0.0012 mol CH3COONa
Remaining unreacted = 0.0025 mol - 0.0012 mol = 0.0013 mol CH3COOH
In total volume 37.5 mL = 0.0375 L
Molarity of CH3COONa = 0.0012 mol /0.0375 L = 0.032M
Molarity of CH3COOH = 0.0013 mol / 0.0375 L = 0.034M
Now using Henderson - Hasselbalch equation
pH = pKa + log ([salt] /[acid])
pH = 4.74 + log ( (0.032/0.034)
pH = 4.74 + log 0.9411
pH = 4.74 + (-0.02636)
pH = 4.71

Answer- pH after the addition of 12.5 mL of NaOH = 4.71