Question

A buffer consists of 0.91-M propanoic acid ( Ka= 1.4 × 10^-5) and 0.76-M sodium propanoate.

a) Calculate the pH of this buffer.

b)Calculate the pH after the addition of 1.2 mL of 0.10–M HCl to 0.010 L of the buffer.

c) Calculate the pH after the addition of 1.8 mL of 1.0–M HCl to 0.010 L of the buffer.

Answer 1

a) Calculate the pH of this buffer.

Initial concentrations:
Buffer of Propanoic acid(AH) and its conjugate base, A- in salt sodium propanoate, NaA
AH = A- + H+
Concentration of sodium propanoate, [A-] = 0.76 M ;
concentration of propanoic acid, [AH] = 0.91 M ;
pKa = -Log(ka)= -Log(1.4E-5) = 4.85
From Henderson-Hasselbach equation for the buffer of acid, AH and base, A-.
pH = pKa + Log([Base]/[Acid])
pH = pKa + Log([A-]/[AH]) = 4.85 + Log(0.76/0.91) = 4.77

b)Calculate the pH after the addition of 1.2 mL of 0.10–M HCl to 0.010 L of the buffer.
ANSWER:

Volume of buffer = 0.01 L
The following reaction will go to completion to form the weak acid, AH
A- + H+ -> AH

Stoichiometry calculation:
Before reaction: 0.01L of a buffer solution has [A-] = 0.76 M and [AH] = 0.91 M
[AH] = 0.01*0.91 = 0.0091 mol
[A-] = 0.01*0.76 = 0.0076 mol
1.2 mL or 0.0012L of 0.10–M HCl solution was added
[H+] = [HCl] = 0.0012*0.1 = 0.00012 mol

After reaction:
Stoichiometry reaction A- + H+ -> AH will be controlled by
limiting reactant present as [H+] = 0.00012 mol

[AH] = 0.0091 + 0.00012 = 0.00922 mol
[A-] = 0.0076 - 0.00012 = 0.00748 mol
[H+] = [HCl] = 0.00012 -0.00012 = 0 mol

Total volume of 0.01L of buffer and 0.0012Lof HCl solution mixture = 0.01+0.0012 = 0.0112 L
Molar concentrations in 0.0112 L of solution:
[Acid] = [AH] = 0.00922/ 0.0112 = 0.823 M
[Base] = [A-] =   0.00748/ 0.0112 = 0.67 M

from Henderson-Hasselbach equation:
pH = 4.85 + Log([A-]/[AH]) = 4.85 + Log(0.67/ 0.823) = 4.76

c) Calculate the pH after the addition of 1.8 mL of 1.0–M HCl to 0.010 L of the buffer.
ANSWER:

1.8 mL or 0.0012L of 1.0–M HCl solution was added
[H+] = [HCl] = 0.0018*1 = 0.0018 mol

After reaction:
Stoichiometry reaction A- + H+ -> AH will be controlled by
limiting reactant present as [H+] = 0.0018 mol

[AH] = 0.0091 + 0.0018 = 0.0109 mol
[A-] = 0.0076 - 0.0018 = 0.0058 mol
[H+] = [HCl] = 0.0018 -0.0018 = 0 mol

Total volume of 0.01L of buffer and 0.0018Lof HCl solution mixture = 0.01+0.0018 = 0.0118 L
Molar concentrations in 0.0112 L of solution:
[Acid] = [AH] = 0.0109/ 0.0118 = 0.924 M
[Base] = [A-] =   0.0058/ 0.0118 = 0.49 M

from Henderson-Hasselbach equation:
pH = 4.85 + Log([A-]/[AH]) = 4.85 + Log(0.48/ 0.924) = 4.6