Answer:
0.182 moles of acetic acid are needed, this means 10.93 g.
0.318 moles of sodium acetate are needed, this means 26.08 g.
Explanation:
The Henderson–Hasselbalch (<em>H-H</em>) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:
pH = pka +
In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.
From the problem we know that
[A⁻] + [HA] = 250 mM = 0.250 M eq. 1
We use the <em>H-H</em> equation, using the data we know, to describe [A⁻] in terms of [HA]:
5.0 = 4.76 +
eq.2
Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:
1.74 [HA] + [HA] = 0.250 M
[HA] = 0.091 M
Then we calculate [A⁻]:
[A⁻] + 0.091 M = 0.250 M
[A⁻] = 0.159 M
Using the volume, we can calculate the moles of each substance:
- moles of acetic acid = 0.091 M * 2 L = 0.182 moles
- moles of sodium acetate = 0.159 M * 2 L = 0.318 moles
Using the molecular weight, we can calculate the grams of each substance:
- grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
- grams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 g