Homework Help: Chemistry (A level)

Posted by Benjy on Saturday, December 6, 2008 at 9:49am.

I was given a Chemistry question on the titration of a strong base being added to a weak acid. The question was:
"Calculate the pH in a titration when 10.0cm^3 of a 0.10moldm^-3 solution of NaOH is added to a 10.0cm^3 of 0.25moldm^-3 solution of ethanioc acid? (Ka = 1.76 x 10^-5)

I'm completely lost on how to do this. I can work it out for strong acid + strong base (vise versa) but i'm not very good working with Ka. Please can you explain everything in full as i'm not amazing at working out what to put where and in which equation.
Thanks!

Chemistry (A level) - DrBob222, Saturday, December 6, 2008 at 11:33am
The trick to these problem is to recognize what you have in the solution. Let's call ethanoic acid Het.

Then Het + NaOH ==> Naet + HOH
So all of the NaOH is used, some of the Het remains unreacted and some Naet is formed. This is a buffered solution of Het/Naet (a weak acid and its salt).

How much NaOH did we start with?
0.1 mol/dm^3 x 0.01 dm^3 = 0.001 mols NaOH.
How much Het? That is 0.25 mol/dm^3 x 0.01 = 0.0025 mols Het.

Looking at the equation, you can see that all of the NaOH is used but we have some Het remaining. How much is that.
0.0025 mol - 0.001 mol = 0.0015 mols Het remaining.
How much product (Naet) was formed? Of course that is 0.001 mol Naet.

All of that is in a total volume of 10+10 = 20 cc = 0.020 dm^3 (or 0.020 liters).
So final concn of Naet = (0.001/0.020) = ??
Final concn of Het = 0.0015/0.020
(Check these numbers--I get bleary eyed when reading them.)

Now use the Henderson-Hasselbalch equation to solve.
pH = pKa + log (base/acid) = ??

The base is (et) which we calculated above and the acid (Het) we have above also. Just plug and chug.
Chemistry (A level) - Benjy, Saturday, December 6, 2008 at 6:16pm
Aaaaah, thank you DrBob222!
A lot clearer than my chemistry teacher explained it!
Shall work my way through it and see what I can make of it.
Cheers for such a fast response :)

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