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Calculate the Concentration of Triprotic Acid

by samdiah
Tags: acid, concentration, triprotic
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samdiah
#1
Oct23-08, 12:03 AM
P: 81
1. The problem statement, all variables and given/known data

The phosphoric acid in a 100.00 mL sample of a cola drink was titrated with 0.1029 M NaOH. The first equivalence point was detected after 13.34 mL of base added, and the scond equivalence point after 28.24 mL. Calculate the concentration of H2PO4- in mol/L. (Hint: if only H3PO4 were present, where would the second equivalence point be?)

2. Relevant equations

H3PO4 + 2OH --> 2H2O + HPO4

mole ratio 1 :2 :2 : 1

3. The attempt at a solution

I subtracted the first eq Volume from the second: 28.24-13.34 ml and got 14.9 mL=0.0149L

I times this Volume by [NaOH] and got NaOH moles= 0.0149 x 0.1029= 0.001532

Since 2 moles NaOH=1 mol biphosphate--> 0.001532 x 1/2=7.666 x 10-4

I divide this by 0.1 L to get [H2PO4]=7.666 x 10-3

This is not right correct answer is 1.605 x 10-3

I have been stuck on this for many hours. If someone can tell me how to do this I will really appreciate it. I already got the answer wrong on the quiz, but I have to study for exam.

Thanks a lot.
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Borek
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Oct23-08, 03:13 AM
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Quote Quote by samdiah View Post
Since 2 moles NaOH=1 mol biphosphate--> 0.001532 x 1/2=7.666 x 10-4
1.53 mmol is amount used to neutralize ONE proton of the phosphoric acid, so ratio is 1:1.

But it doesn't give correct answer as well.

What other acids are present in coca cola? Perhaps you are neutralizing not only phosphoric acid?
samdiah
#3
Oct23-08, 09:17 AM
P: 81
This is the reaction for first endpoint:

H3PO4(s) + H2O(l) ⇌ H3O+(aq) + H2PO4(aq) Ka1= 7.510−3

This is the reaction for second endpoint:

H2PO4(aq)+ H2O(l) ⇌ H3O+(aq) + HPO42(aq) Ka2= 6.210−8

This is the third, but it cannot be seen:

HPO42(aq)+ H2O(l) ⇌ H3O+(aq) + PO43(aq) Ka3= 2.1410−13

Can someone please help me? I am stuck on this for too long of time. Thanks.

Borek
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Oct23-08, 09:29 AM
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Calculate the Concentration of Triprotic Acid

What other acids (other than phosphoric) are present in coca cola?
samdiah
#5
Oct23-08, 09:54 AM
P: 81
This is the only information given. We r not suppose consider other acid. Just the biphosphate and phosphoric acid. I am doing this question in 6 different ways and it doesn't yield the correct answer. I also got 1.533x10-3 a few times now.
Borek
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Oct23-08, 10:07 AM
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If amount of base used to neutralize first proton and the second proton differs, you have either part of the acid neutralized before you start titration, or there is other weak acid (with strength comparable to 2nd phosphoric acid proton) present in the solution. No other reason for different results.

Note that your hint question hints at other substances present in the solution.

Hint 2: are you asked to calculate the concentration of phosphoric acid?
samdiah
#7
Oct23-08, 10:25 AM
P: 81
What I was thinking of the hint was that H2PO4- is the other substance present. If there was no biphosphate then the 2nd equivalence point will be less then 2xV1, because 1 phosphoric acid=2OH.

"Hint 2: are you asked to calculate the concentration of phosphoric acid?"

No we r asked to calculate concentration of biphosphate.
GCT
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Oct23-08, 10:29 AM
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Quote Quote by samdiah View Post
This is the reaction for first endpoint:

H3PO4(s) + H2O(l) ⇌ H3O+(aq) + H2PO4–(aq) Ka1= 7.510−3

This is the reaction for second endpoint:

H2PO4–(aq)+ H2O(l) ⇌ H3O+(aq) + HPO42–(aq) Ka2= 6.210−8

This is the third, but it cannot be seen:

HPO42–(aq)+ H2O(l) ⇌ H3O+(aq) + PO43–(aq) Ka3= 2.1410−13

Can someone please help me? I am stuck on this for too long of time. Thanks.

Take into account

- After you titrate past the second equivalence point the HPO4 2- is a base - although it is amphoteric it has an equilibrium mostly with H2PO4 - . You need to utilize the equilibrium constant to calculate the latter concentration.

- Phosphoric acid is in equilibrium. However the question seems to have you conclude that only it is present. If it was in equilibrium you would need to found the relative concentrations of phosphoric acid and its conjugate base.
Borek
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Oct23-08, 10:37 AM
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Quote Quote by GCT View Post
You need to utilize the equilibrium constant to calculate the latter concentration.
No need for any equilibrium calculations. This is simple stoichiometry.

I must admit I was on the wrong track earlier.
GCT
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Oct23-08, 10:50 AM
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Quote Quote by Borek View Post
No need for any equilibrium calculations. This is simple stoichiometry.

I must admit I was on the wrong track earlier.
You need equilibrium calculations , you need to account for both HPO4 2- and PO4 3- , however most of it has to do with HPO4 2- and H2PO4 -.
Borek
#11
Oct23-08, 11:00 AM
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We will get to this later, once samdiah solve the question.
samdiah
#12
Oct23-08, 11:09 AM
P: 81
I did use stoichiometry the whole time, but keep getting all sorts of wrong answers. I subtracted the 1st eq pt from 2nd to get V required to titrate only the biphosphate. I use this V and NaOH to get moles of NaOH. Since the mole ratio is 1:1, I say that moles of bi phosphate is also same. Use this moles divide it by 0.1 L and get 0.0153321. This is not right. Wut am I doing wrong?
Borek
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Oct23-08, 11:19 AM
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Think, why amounts of base used for both parts of the titration differ. As I already wrote earlier - there is either some other acid present, or part of the phosphoric acid was neutralized before you started titration. What will be the composition of starting solution in the latter case?

Also note - you are not asked to calculate amount of phosphoric acid, but amount of H2PO4-!
GCT
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Oct23-08, 11:55 AM
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Quote Quote by samdiah View Post
I did use stoichiometry the whole time, but keep getting all sorts of wrong answers. I subtracted the 1st eq pt from 2nd to get V required to titrate only the biphosphate. I use this V and NaOH to get moles of NaOH. Since the mole ratio is 1:1, I say that moles of bi phosphate is also same. Use this moles divide it by 0.1 L and get 0.0153321. This is not right. Wut am I doing wrong?
When a strong monoprotic acid is titrated with a strong base the pH is 7 at equivalence point.

When a weak monoprotic acid is titrated with a strong base the pH is NOT 7 at equivalence point.

Explain this and you are going to be able to understand how to solve this problem.
GCT
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Oct23-08, 12:15 PM
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The original question is ambiguous - is it asking for the initial concentration or the concentration after the titration , if it is the former then Borek is leading you in the right direction.
samdiah
#16
Oct23-08, 03:30 PM
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Quote Quote by GCT View Post
The original question is ambiguous - is it asking for the initial concentration or the concentration after the titration , if it is the former then Borek is leading you in the right direction.
It is asking for initial concetration. The reason that pH is not 7 because weak acid doesn't dissacoiate fully.

Would u guys recommend me to read anything to do this. I did all the reading assigned by my Prof and they only have to do with Ka's and pH nothing about 1st and 2nd Volumes.

Thanks for all the help. I m just too stupid.
GCT
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Oct23-08, 03:37 PM
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Quote Quote by samdiah View Post
It is asking for initial concetration. The reason that pH is not 7 because weak acid doesn't dissacoiate fully.

Would u guys recommend me to read anything to do this. I did all the reading assigned by my Prof and they only have to do with Ka's and pH nothing about 1st and 2nd Volumes.

Thanks for all the help. I m just too stupid.
You are not stupid. If it is the initial concentration then Borek's method is legitimate.

The conjugate of a weak acid is a weak base ... meaning that it has a significant role as a base - when all of the weak acid is neutralized the weak base that results actually raises the pH. Not that this matters since the question is referring to the original concentration.
GCT
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Oct23-08, 03:57 PM
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Note that the volume required to titrate the first and second equivalence points are not the same ... this means that the H3PO4 had already dissociated to H2PO4 - to some extent.

However you are able to determine the total formal concentration of H3PO4 by determining the volume required for the second equivalence point. When you do this you are then going to be able to understand the discrepancy in the first equivalence point ... simply subtract to find the relative amounts of H3PO4 and H2PO4 -.

It is very simple.


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