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Topic: Diprotic acids/bases  (Read 6267 times)

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Offline spunkylulu

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Diprotic acids/bases
« on: March 27, 2014, 03:01:07 PM »
Can someone explain to me how I should go about solving this problem?

What is the pH of a solution of 100 mL of .1M Na2CO3 and 50 mL of .1M HCl?
When Ka1= 4.2 X 10-7 and Ka2= 4.8X10-11

If someone could explain the reactions to me, that would be most helpful.

Offline Borek

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Re: Diprotic acids/bases
« Reply #1 on: March 27, 2014, 03:04:53 PM »
What happens to the solution of carbonate, when you add strong acid (hint: take a look at stoichiometry).
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #2 on: March 27, 2014, 03:47:51 PM »
Na2CO3 +HCl --> 2NaCl +HCO3?

Offline Borek

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Re: Diprotic acids/bases
« Reply #3 on: March 27, 2014, 04:15:13 PM »
Check the stoichiometry - do you have enough HCl for the reaction to proceed as written?

I assume HCO3 to be just a typo?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #4 on: March 28, 2014, 03:00:42 PM »
I don't know.  I have trouble with reactions involving polyprotic acids and bases.  Can you point me in the right direction?

Offline Borek

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Re: Diprotic acids/bases
« Reply #5 on: March 28, 2014, 03:30:47 PM »
What is bicarbonate?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #6 on: March 28, 2014, 04:34:14 PM »
HCO3

Offline Borek

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Re: Diprotic acids/bases
« Reply #7 on: March 28, 2014, 05:17:34 PM »
No, there is no such thing as HCO3. There is a HCO3- ion, which is a first product of CO32- protonation.

Now think about what is happening when you add HCl to teh solution of carbonate. As I said earlier, think about stoichiometry - compare amount of carbonate and amount of HCl.
« Last Edit: March 28, 2014, 06:06:28 PM by Borek »
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #8 on: March 28, 2014, 06:11:58 PM »
Okay wait. 

2HCl + CO3 --> CO2 + H2O + 2Cl? 

I feel like I'm not getting anywhere!

Offline Borek

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Re: Diprotic acids/bases
« Reply #9 on: March 28, 2014, 06:53:54 PM »
2HCl + CO3 --> CO2 + H2O + 2Cl?

Don't ignore charges in your posts.

I told you to watch stoichiometry. Do you have enough HCl for the reaction you wrote?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #10 on: March 28, 2014, 07:03:53 PM »
OH.

No, because the ratio of HCl to Na2CO3 is 2:1 and we only have 50 ml of HCl which is only half the amount of Na2CO3

Offline Borek

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Re: Diprotic acids/bases
« Reply #11 on: March 29, 2014, 05:07:34 AM »
So is it possible for the reaction to proceed as you wrote it, or does it have to stop earlier?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #12 on: March 29, 2014, 09:35:00 AM »
No.  So is that why there are two dissociation constants?  But I'd only use the first one? 

Offline Borek

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Re: Diprotic acids/bases
« Reply #13 on: March 29, 2014, 11:03:10 AM »
We will get to dissociation constants later, for now we need to establish what happened in the solution. And technically solution you got is not different from the one prepared by mixing sodium bicarbonate with equimolar amount of NaCl. Do you see it?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #14 on: March 29, 2014, 05:57:31 PM »
Wait, will this reaction be:

Na2CO3 + 2HCl --> 2NaCl + H2CO3?

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