Question

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?

Answer 1

This can be contradictory, depending on whether the `"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^(-)) = "9.125 mL"`

`V_(HA) = "15.875 mL"`

---

The Henderson-Hasselbalch equation is:

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

We have a `"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 `"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 `"50 mL"`, or `"0.050 L"`, the mols of each component (which are constant!) are:

`n_(A^-) = "0.0365 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.001825 mols"`

`n_(HA) = "0.0635 mol"/cancel"L" xx 0.050 cancel"L" = ul"0.003175 mols"`

So, if both of the starting concentrations were `"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 `"25.000 mL"`, the total ending volume is twice as large; the total species concentration is half the concentration that both species started with.