Activity coefficients - Revisited

VoG II 21:23 05 Sep 2005
Locked

Some time ago I posted a question about this and I have now found the solution (pun intended).

Example:

Forum Editor 00:18 06 Sep 2005

Although I'm still non the wiser.

wiz-king 05:12 06 Sep 2005

May the (ionic) force be with you.

sattman 20:46 06 Sep 2005

I think that Vog is building a Kylstron in his front room.

wiz-king 21:12 06 Sep 2005

The klyston predates him ay a decade or so. I havn't seen one in use for about 50 years now!

Jackcoms 21:27 06 Sep 2005

Fascinating.

Now, for those of us still on Planet Earth, would you care to explain (in English), please?

Phphred 21:53 06 Sep 2005

Or Chinese perhaps as no doubt we shall all be none the wiser!

VoG II 21:58 06 Sep 2005

With the indulgence of the Forum Editor I will do my best.

Not everything in chemistry has to be done by experiment. A lot of things can be calculated based upon published information. In this example I need to calculate whether limescale (calcium carbonate, CaCO3) will precipitate in a particular solution.

There is published information on the solubility product (K) of CaCO3 and

K = [Ca] x [CO3]

where [X] denotes concentration.

So in very simple terms if I know [Ca] and [CO3] and multiplying them together exceeds K then CaCO3 will precipitate. The calculation in this case is more complex than that, but I hope that you get the general idea so far.

The problem is that published values of K (for anything) refer to a hypothetical state of 'infinite dilution'. In solutions containing ions the effective concentration (or activity) of each ion is depressed by the cloud of other ions surrounding it. The activity is concentration that has been corrected for the non-ideal behaviour in real solutions. So in the 'K' calculations above, one needs to use activity rather than concentration.

activity = concentration x activity coefficient.

The corrections are fairly minor in very dilute solutions such as drinking water (Ionic Strength <0.01 mol/l). Also, there are equations to calculate activity coefficients for dilute solutions with which I am familiar. My problem was knowing how to calculate these for much more concentrated solutions, with which I am unfamiliar.

The complexity is in the Excel spreadsheet that is used to perform the solubility calculations - it is not quite as the simple explanation above as it requires the solution of a 6th order polynomial equation. The reason for this is that if CaCO3 precipitates then other things change as well and the calculation has to take account of that.

Anyway, I can now use the above program and use a lookup table in Excel to choose an appropriate value for activity coefficient.

=VLOOKUP(I8, IUPAC!A8:B107, 2, TRUE)

to interpolate a value corresponding to my value for Ionic Strength (in I8).

Phphred 22:01 06 Sep 2005

You are better than the average bear Booboo!

gudgulf 22:29 06 Sep 2005

The chemistry explaination I can follow....but the Excel bit....WTF? (profuse apologies,"Language Timothy!).

Sorry VoG™ but it really does serve people right for asking,lol.

Kev.Ifty 22:45 06 Sep 2005

But does the calculation take into account the ambient temperature?

;-0

But my Dad's got a bigger test tube than yours....

This thread is now locked and can not be replied to.

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