Chemical Forums

Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: spunkylulu on March 27, 2014, 03:01:07 PM

Title: Diprotic acids/bases
Post by: spunkylulu 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.
Title: Re: Diprotic acids/bases
Post by: Borek 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).
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 27, 2014, 03:47:51 PM
Na2CO3 +HCl --> 2NaCl +HCO3?
Title: Re: Diprotic acids/bases
Post by: Borek 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?
Title: Re: Diprotic acids/bases
Post by: spunkylulu 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?
Title: Re: Diprotic acids/bases
Post by: Borek on March 28, 2014, 03:30:47 PM
What is bicarbonate?
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 28, 2014, 04:34:14 PM
HCO3
Title: Re: Diprotic acids/bases
Post by: Borek 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.
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 28, 2014, 06:11:58 PM
Okay wait. 

2HCl + CO3 --> CO2 + H2O + 2Cl? 

I feel like I'm not getting anywhere!
Title: Re: Diprotic acids/bases
Post by: Borek 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?
Title: Re: Diprotic acids/bases
Post by: spunkylulu 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
Title: Re: Diprotic acids/bases
Post by: Borek 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?
Title: Re: Diprotic acids/bases
Post by: spunkylulu 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? 
Title: Re: Diprotic acids/bases
Post by: Borek 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?
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 05:57:31 PM
Wait, will this reaction be:

Na2CO3 + 2HCl --> 2NaCl + H2CO3?
Title: Re: Diprotic acids/bases
Post by: Borek on March 29, 2014, 06:42:41 PM
Wait, will this reaction be:

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

No.

Several hours ago you checked that you don't have enough acid for this reaction:

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

It even looked for a moment that you knew this is not a correct reaction:

So is it possible for the reaction to proceed as you wrote it

No.

But now we are back to the square one.
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 06:43:21 PM
No no I got it

Na2CO3 + HCl -->NaHCO3 + NaCl

and then the second part of the reaction would be

NaHCO3 +HCl --> NaCl + H2O +CO2
Title: Re: Diprotic acids/bases
Post by: Borek on March 29, 2014, 06:56:21 PM
Na2CO3 + HCl -->NaHCO3 + NaCl

OK

Quote
and then the second part of the reaction would be

NaHCO3 +HCl --> NaCl + H2O +CO2

Would be, but we won't go there, as there is not enough acid, agreed?

Now, it seems like the problem can be reworded: we need to calculate pH of the solution of NaHCO3, do you see it?
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 07:07:54 PM
Yes.

So we start with 0.01 mol of Na2CO3 and 0.005 mol HCl, right?  Then what?
Title: Re: Diprotic acids/bases
Post by: Borek on March 29, 2014, 07:17:26 PM
We got to the point when we have a solution that is equivalent to the one prepared by mixing NaHCO3 and NaCl, do you see it, or not?

Do you know how to calculate pH of the solution of an amphiprotic salt?
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 07:28:14 PM
Yes.  I think what gives me the most trouble is understanding the reaction and figuring out what reaction to use when calculating pH. 
Title: Re: Diprotic acids/bases
Post by: Borek on March 29, 2014, 07:42:50 PM
http://www.chembuddy.com/?left=pH-calculation&right=pH-amphiprotic-salt
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 08:13:35 PM
That was helpful, thank you.

The only thing I continue to struggle with is knowing which Ka value to use in my calculation.  Since the reaction doesn't proceed to completion, I would think I would use the first Ka value, right?

The second reaction I have to do is NaHCO3 and HCl (which, again, won't reach completion because there isn't enough HCl) and I would think to use Ka2 (since NaHCO3 comes from the second dissociation)
Title: Re: Diprotic acids/bases
Post by: Borek on March 29, 2014, 08:23:45 PM
I have posted link to page that contains formula you need.
Title: Re: Diprotic acids/bases
Post by: spunkylulu on March 29, 2014, 08:48:12 PM
So, based on what I have gathered, I'll use Ka2 for this reaction.  Is that correct?  (I think it's Ka2 because Ka2 relates to the dissociation of bicarbonate, and that's the product of this reaction)
Title: Re: Diprotic acids/bases
Post by: Borek on March 30, 2014, 03:30:31 AM
Have you read the page I linked to? Derivation can be difficult to follow, but it yields a simple formula.