Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acetate concentration of 250 mM and a pH of 5.0. What concentrations of acetic acid and sodium acetate should you use? Assuming you wish to make 2 liters of this buffer, how many moles of acetic acid and sodium acetate will you need? How many grams of each will you need (molecular weights: acetic acid 60. 05 g mol 1, sodium acetate, 82. 03 g mol 1)?

The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

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 H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + [tex]log\frac{[A^{-} ]}{[HA]}[/tex]

[tex]0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}][/tex] 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