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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: RaelAerosolKid on January 09, 2021, 11:39:37 PM

Title: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 09, 2021, 11:39:37 PM
So my teacher gave us an exercise list and one of the questions (translated) was this:

"A solution (A) contains an alkaline mixture of Na2X 0.2000M [H2X (pKa1 = 5.5) and HX-
(pKa2 = 11.0)] and 0.2500 M NaY [HY (pKa1 = 11.0)].
A 10.00 mL aliquot of solution (A) was titrated against 0.2000 M HCl solution.

What volume of 1.5000 M HCl solution should be added to 2.00 L of solution (A) for
to obtain a buffer solution pH = 7.0?"

I was taught how to calculate the volume when I had a simple weak acid and salt, not this mess.

I thought about giving priority to the strongest acid, or to the first Ka, or the most concentrated salt but I never came to a solid conclusion. I have searched "buffer solution with mixture of 2 weak acid salts" and found nothing.

Is this question even solvable? xD

Thanks in advance



Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: AWK on January 10, 2021, 04:33:56 AM
Quote
I was taught how to calculate the volume when I had a simple weak acid and salt, not this mess.
Calculate it independently for both salts and add the volumes of acid.
Quote
What volume of 1.5000 M HCl solution should be added to 2.00 L of solution (A) for to obtain a buffer solution pH = 7.0?"
Frankly saying, chemists will not consider this solution after neutralization as a buffer solution. But roughly you can think of it as a buffer solution based on the acid dissociation constant Ka1.
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 10, 2021, 01:58:40 PM
Don't know if this is against the rules, but could you be more direct in how to solve this question?

The first part makes total sense, thanks.

But when I think about the Henderson–Hasselbalch equation to find the pH,
(it would be "pH = Pka1 - log(([H2X] + moles of HCl)/([HX-] - moles of HCl))", right?)
 I don't know the concentration of [H2X] or [HX-] to begin with so therefore I also wouldn't know what amount of HCl to add

Am I missing something? Damn this subject is hard
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 10, 2021, 02:49:51 PM
Oh I think I have a clue, I just have to calculate the pH of the salt solution alone and then I think there is no need to do the Henderson–Hasselbalch equation.

So I would have, say, X2- from the salt + H2O -> HX- + OH-

That means I would have to use the Ka2 to then transform it into a Kb and then find the concentration

Is this correct?
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: AWK on January 10, 2021, 03:37:47 PM
Solve for buffer H2X/NaHX with the sum of the concentration 0.2000 M, then neutralize all salts with HCl.
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 10, 2021, 03:38:39 PM
Nope, didn't work. I found that a 2L solution of any of the 2 salts had a pH of about 12.19. That would require an absurdly low amount of HCl to be added to change the pH to 7

FU&*K.

(I wrote this before your reply)

Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 10, 2021, 04:32:19 PM
Solve for buffer H2X/NaHX with the sum of the concentration 0.2000 M, then neutralize all salts with HCl.
Ok so I did "7 = 5.5 - (log ([H2X]/[NaHX]))" and "[H2X] + [NaHX] = 0.2"
The result was that [H2X] = 0.00613 and [NaHX] = 0.19387

I thought about the reaction 2HCl + Na2X --> H2X + 2NaCl

Does that mean I would have to do stoichiometry and get a amount of 0.00613 of H2X if I want to get a pH of 7? (And then do the same in the other salt)

Sorry for the amount of questions, my teacher doesn't really explain this really well
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: AWK on January 10, 2021, 04:38:52 PM
Neutralize NaY completely
Neutralize Na2X to NaHX
Then neutralize 0.19387 M to H2X should 0.006
Add all HCl used (remember about 2 liters of solution.
I estimated volume HCl between 0.85 - 0.87 L without a calculator. should be ~0.6
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 10, 2021, 09:59:59 PM
Would this explanation assume that we have a reaction that is like "HCl + Na2X --> NaHX + NaCl" and not right away "2HCl + Na2X --> H2X + 2NaCl"?
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: AWK on January 11, 2021, 01:42:28 AM
This is exactly how many reactions go.
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: Borek on January 11, 2021, 03:46:29 AM
Would this explanation assume that we have a reaction that is like "HCl + Na2X --> NaHX + NaCl" and not right away "2HCl + Na2X --> H2X + 2NaCl"?

You had a very similar conceptual problem with Ca(OH)2 question. Protonation of X is a two step process which can be written as two consecutive steps or as an overall reaction. Both approaches have their applications, neither is "right" or "better" in general.
Title: Re: Another question, about buffer solutions, where I'm very confused at what to do
Post by: RaelAerosolKid on January 12, 2021, 12:14:11 PM
Would this explanation assume that we have a reaction that is like "HCl + Na2X --> NaHX + NaCl" and not right away "2HCl + Na2X --> H2X + 2NaCl"?

You had a very similar conceptual problem with Ca(OH)2 question. Protonation of X is a two step process which can be written as two consecutive steps or as an overall reaction. Both approaches have their applications, neither is "right" or "better" in general.

Ok I think I managed to do it, thank you so much for your help

More helpful than my own teacher  ;)