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Topic: Diprotic acids/bases  (Read 6704 times)

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Offline Borek

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Re: Diprotic acids/bases
« Reply #15 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.
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #16 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

Offline Borek

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Re: Diprotic acids/bases
« Reply #17 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?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #18 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?

Offline Borek

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Re: Diprotic acids/bases
« Reply #19 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?
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #20 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. 

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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #22 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)

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Re: Diprotic acids/bases
« Reply #23 on: March 29, 2014, 08:23:45 PM »
I have posted link to page that contains formula you need.
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Offline spunkylulu

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Re: Diprotic acids/bases
« Reply #24 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)

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Re: Diprotic acids/bases
« Reply #25 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.

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