Question

Your task is to create a buffered solution. You are provided with 0.10 M solutions of formic acid and sodium formate. Formic acid has a pKa of 3.75. 2. Create approximately 20 mL of buffer solution with a pH of 4.25.

Answer 1

Answer:

15.2mL of the 0.10M sodium formate solution and 4.8mL of the 0.10M formic acid solution.

Explanation:

To find the pH of a buffer based on the concentration of the acid and conjugate base we must use Henderson-Hasselbalch equation:

pH = pKa + log [A⁻] / [HA]

Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid

4.25 = 3.75 + log [A⁻] / [HA]

0.5 = log [A⁻] / [HA]

3.162 = [A⁻] / [HA] (1)

As both solutions are 0.10M and you want to create 20mL of the buffer, the moles are:

0.10M  * 20x10⁻³L =

2x10⁻³moles = [A⁻] + [HA] (2)

Replacing (2) in (1):

3.162 = 2x10⁻³moles - [HA] / [HA]

3.162 [HA] = 2x10⁻³moles - [HA]

4.162[HA] = 2x10⁻³moles

[HA] = 4.805x10⁻⁴ moles

[A⁻] = 2x10⁻³moles - 4.805x10⁻⁴ moles = 1.5195x10⁻³moles

That means, to create the buffer you must add:

[A⁻] = 1.5195x10⁻³moles * (1L / 0.10mol) = 0.0152L =

15.2mL of the 0.10M sodium formate solution

[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =

4.8mL of the 0.10M formic acid solution