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I don't get this pKa pH problem!

by DJSkopelitis
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Sep27-08, 02:37 PM
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A Compound (not specified) has 2 ionizable groups. One is at pKa 4.2 and the other pKa is between 7 and 10. If you have 10.0 mL of a 1.0M solution at a pH of 8.17 and you add 10.0mL of 1.0M HCl which changes the pH to 4.2 what is the second pKa?

I tried working this problem out a few different ways and keep getting different answers. The last calculation I tried gave me a pKa of 4.44.

I'm using pH = pKa + log [A-]/[HA] I used pH = pKa-log[HA] / 2 to get an HA of 1.38. I plugged 8.17 and 4.2 in respectively as pH and pKa. Looking at it now that doesn't seem right.

But where I get stuck is how to add the 0.01 moles of HCl into the equation. I also was wondering if the answer is much more simple than I think it is. At the second ionizable group's pKa it would equal the pH right? So if that's the case is the pKa just 8.17? Or am I just completely wrong?
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Sep27-08, 05:06 PM
Borek's Avatar
P: 23,406
8.17 it is

I like simplicity of that question.
Sep27-08, 05:15 PM
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epenguin's Avatar
P: 1,967
Maybe the question is not quite clear. I think it is not meant to say that 'the substance' has that pH when dissolved, rather it has been brought to that pH. Say someone dissolved some Na2X and already added some acid. Or they had dissolved some H2X and added an amount of NaOH which brought to pH up to 8.17.

It is convenient to remind yourself, and I think you are supposed to use it here, that when a group is half titrated half dissociated) then the pH = its pK and vice versa.

Then there are no very complex calculations at all here.

There is another related but not identical problem right now which you should understand if you have to be conversant and at ease with these problems.

For cases I expect you will soon have to deal with where you do need more calculation it is always done from 3 and a half principles which are here

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