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### Topic: Problem of the week - 31/12/2012  (Read 10502 times)

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#### Borek

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##### Problem of the week - 31/12/2012
« on: December 31, 2012, 07:37:34 AM »
Back to work...

Explain how you would prepare the acetate buffer solution with pH 4.0 and still liquid at -1°C.

Assume pKa=4.76. You can also assume molarity and molality to be equal to each other - they are not, but the solution will be diluted enough that the error will be below 3%. But if you want to go the hard way - assume solution density of 1.0004 g/mL.
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#### curiouscat

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##### Re: Problem of the week - 31/12/2012
« Reply #1 on: January 02, 2013, 07:41:53 AM »
Add 0.4 mol Acetic Acid and 0.0696 mol Sodium Acetate to H2O to make 1 L total solution and then chill?

Assuming I did it right, anything above these concentrations seems to work so long as the Ratio of 5.75 is maintained (Ratio of mols of Acetic Acid to Sodium Acetate).

#### curiouscat

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##### Re: Problem of the week - 31/12/2012
« Reply #2 on: January 03, 2013, 08:30:29 AM »
Would it be an acceptable answer to just prepare a regular buffer and then just add in plenty of, say, ethylene glycol in there? Glycol shouldn't interfere with the buffering action, right?

Instead of Glycol, say fructose or sucrose  or some such soluble, inert, spectator.

#### Borek

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##### Re: Problem of the week - 31/12/2012
« Reply #3 on: January 06, 2013, 07:33:53 AM »
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?

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.
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#### curiouscat

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##### Re: Problem of the week - 31/12/2012
« Reply #4 on: January 06, 2013, 08:09:13 AM »
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 C0HA with salt of conc. C0A-

By definition of Ka
$$K_a=\frac{[H^+](C_0^{A^-}+[H^+])}{C_0^{HA}-[H^+]} \\$$

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)

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

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

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

Again H+ is negligible with respect to other terms.

This and Eq. (a) are two equations in two unknowns C0HA and C0A-. 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.

Quote
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."