Add 0.4 mol Acetic Acid and 0.0696 mol Sodium Acetate to H2O to make 1 L total solution and then chill?

That's the answer I was looking for, can you explain how you got it?

(Rxn 1) HA

H

^{+} + A

^{-}(Rxn 2) MA

M

^{+} + A

^{-}Assume Rxn 1 progresses to an extent that yields a concentration x of [H+] and [A-]

*by that reaction alone.*Rxn 2 can be assumed to go to completion.

Since pH is given to be 4.0. Hence [H+]=10

^{-4}. All concentrations in gmol/L henceforth. Assume we need to mix acid of concentration C0

^{HA} with salt of conc. C0

^{A-}By definition of Ka

[tex]

K_a=\frac{[H^+](C_0^{A^-}+[H^+])}{C_0^{HA}-[H^+]} \\

[/tex]

We know [H+]=10

^{-4} which is small compared to other terms. Also pKa=4.76. Hence Ka=1.7378E-5. Hence we get the following equation (let's call it Eq. a)

[tex]

C_0^{A^-}=0.1738 \times C_0^{HA}

[/tex]

Using Freezing Point Depression:

ΔTF = KF · b · i

For water KF=1.853

Ignoring the contortions of VantHoff factors etc. , the equation really means this:

ΔTF /= KF = Sum of all dissolved entity concentrations

Here RHS=[HA]+[H+]+[A-]+[M+]

We get

[tex]

\frac{\delta T_F}{K_F} = C_0^{HA} + 2 C_0^{A^-} + [H+]

[/tex]

Again H+ is negligible with respect to other terms.

This and Eq. (a) are two equations in two unknowns C0

^{HA} and C0

^{A-}. Solve simultaneously to get the answers reported.

Was your solution similar? I was mired in the complexities of α (Vant Hoff Factor) for a bit before I decided to go this way.

As for the other comments: yes, anything more concentrated will work and adding something else to lower the freezing point should work as well. But using just acetate buffer seems to be the simplest approach (in terms of solution preparation, not necessarily in terms of calculating the buffer). Besides, usually it is best to use the simplest possible approach to limit the number of (possibly) interfering factors.

Agreed.

One, probably nitpicking, suggestion might be to word the question as:

* "Explain how you would prepare the acetate buffer solution with pH 4.0 and that would freeze exactly at -1°C."*