# Using Henderson-Hasselbalch equation, calculate the volume of 0.20 M Acetic Acid and 0.2 M Sodium Acetate needed to prepare 50 mL of 0.1 M Acetate Buffer solution (pH = 4.5). pKa of Acetic Acid is 4.74?

May 31, 2018

This can be contradictory, depending on whether the $\text{0.1 M}$ is the total species concentration or the concentration of each of the two components. I'll consider this to be the former...

${V}_{{A}^{-}} = \text{9.125 mL}$

${V}_{H A} = \text{15.875 mL}$

The Henderson-Hasselbalch equation is:

"pH" = "pK"_a + log\frac(["A"^(-)])(["HA"])

We have a $\text{pH 4.5}$ solution of acetic acid and acetate, so from there we can get the ratio of weak acid to conjugate base:

\frac(["A"^(-)])(["HA"]) = 10^("pH" - "pK"_a)

$= {10}^{4.5 - 4.74}$

$= 0.5754$

Now, if the total concentration is $\text{0.10 M}$, then:

["HA"] + overbrace(0.5754["HA"])^(["A"^(-)]) = "0.10 M"

=> ["HA"] = ("0.10 M")/(1.0000 + 0.5754) = ul"0.0635 M"

=> ["A"^(-)] = ul"0.0365 M"

and these concentrations are AFTER mixing. Since the total volume is $\text{50 mL}$, or $\text{0.050 L}$, the mols of each component (which are constant!) are:

${n}_{{A}^{-}} = \text{0.0365 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.001825 mols}$

${n}_{H A} = \text{0.0635 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.003175 mols}$

So, if both of the starting concentrations were $\text{0.20 M}$, we can find the volume they each start with:

color(blue)(V_(A^-)) = "1 L"/(0.20 cancel("mols A"^(-))) xx 0.001825 cancel("mols A"^(-))

= "0.009125 L" = color(blue)ul("9.125 mL")

color(blue)(V_(HA)) = "1 L"/(0.20 cancel("mols HA")) xx 0.003175 cancel("mols HA")

= "0.015875 L" = color(blue)ul("15.875 mL")

And this should make sense, because the total starting volume is $\text{25.000 mL}$, the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.