January 23, 2021, 10:27:00 PM
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Topic: Another question, about buffer solutions, where I'm very confused at what to do  (Read 501 times)

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Offline RaelAerosolKid

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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




Offline AWK

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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.
AWK

Offline RaelAerosolKid

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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

Offline RaelAerosolKid

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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?

Offline AWK

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Solve for buffer H2X/NaHX with the sum of the concentration 0.2000 M, then neutralize all salts with HCl.
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Offline RaelAerosolKid

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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)


Offline RaelAerosolKid

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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

Offline AWK

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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
« Last Edit: January 10, 2021, 06:08:54 PM by AWK »
AWK

Offline RaelAerosolKid

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Would this explanation assume that we have a reaction that is like "HCl + Na2X --> NaHX + NaCl" and not right away "2HCl + Na2X --> H2X + 2NaCl"?

Offline AWK

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This is exactly how many reactions go.
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Offline Borek

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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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline RaelAerosolKid

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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  ;)

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