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]
<em>Where [A⁻] could be taken as moles of the sodium formate and [HA] moles of the formic acid</em>
<em />
4.25 = 3.75 + log [A⁻] / [HA]
0.5 = log [A⁻] / [HA]
3.162 = [A⁻] / [HA] <em>(1)</em>
<em></em>
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] <em>(2)</em>
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 =
<h3>15.2mL of the 0.10M sodium formate solution</h3>
[HA] = 4.805x10⁻⁴ moles * (1L / 0.10mol) = 0.0048L =
<h3>4.8mL of the 0.10M formic acid solution</h3>
