Simple Ka question

[stester77s]

When determining the Ka with this formula:
HA ---> H+ + A-

Ka = [H+]*[A-]/[HA]

does [H+] in this equation refer to the amount of H+ in the solution including the amount that comes from water (10^-7 mol/L from water + the amount from the acid) or does it not include the amount that would also be there from water? Does it only refer to the concentration of [H+] that comes directly from the acid, HA?

Thanks!

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[chenzt]

You are welcome to include the H+ concentration of water, but soon enough you will realize how little it changes things. Anything that is weak enough that water can significantly change the H+ concentration in your calculation probably aren't considered acids.

 

[stester77s]

You are welcome to include the H+ concentration of water, but soon enough you will realize how little it changes things. Anything that is weak enough that water can significantly change the H+ concentration in your calculation probably aren't considered acids.

Please consider the case when you are dealing with a very dilute concentration of a weak acid. Say you add 1*10^-8 M of acetic acid to a liter of water. Whether you include the amount of H3O from water will affect your calculation. The amount of H3o from water is significant not because the strength of the acid, but because of how little acid you added.

As you can see, this will affect your calculation - so could you or someone please tell me by the book whether you are supposed to include the concentration of H3O from water when doing this calculation? I am pretty sure you're not supposed to, but I am having trouble finding an answer to clarify this.

 

[chenzt]

When in such dilute concentration, the usual Ka = [H+][A-]/[HA] doesn't apply anymore, since that equation makes a few assumptions, such as HA >>> H+ and H+ >>> OH-, which is not true for this case.

So you can't completely ignore the H+ from water, because it is significant, but can't just add to it either. Because once H+ from acetic acid is added to the system, you are stressing the system.

H+ + OH- <---> H2O

H+ will re-establish equilibrium by moving to the right, or reacting itself with OH- and turning into H2O.

http://www.chem1.com/acad/webtext/pdf/c1xacid2.pdf

The actual formula will require equation 24 on page 14 of the above link, which is a cubic equation, and way too complex for MCAT.

This is why the pH of 1x10^-10M of HCl isn't 10 (it wouldn't make sense that adding an acid will make a solution basic). You would think that's the answer if you simply do -log[H+] but it's a little more complicated than that.