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### Topic: Titration of the Unknown HCl-H3PO4 Mixture  (Read 14462 times)

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#### kizunaencounter

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##### Titration of the Unknown HCl-H3PO4 Mixture
« on: November 19, 2011, 07:39:59 PM »
For the Procedure
1. unknown HCl-H3PO4 mixture was provide in a 50 ml Erlenmeyer flask.Then 25.00ml of the sample was pipetted into a 100ml volumetric flask and diluted with deionized water to the mark.
2. The 25.00ml of the diluted mixture was pipetted into a 250ml beaker with 75ml of deionized water. Titration was carried out with a ph meter and 0.1M NaOH.
3. For each titration calculate and report the molarity of HCl and the molarity of H3PO4 in the original unknown mixture.

This is what I have so far i think I did it correctly but here it gos anyways.

First off the ratio is 1:1 ratio according to the equation.
According to the titration graph for trial 1 the first equivalence point is 24.5 ml at 4.1 ph and the 2nd equivalence point was 38.0ml at 9.1 ph. Also, the molarity of NaOH was standardized at .095375M.

For HCl
.095375M X 24.5 = molarity acid X 100ml=.023367M X 2 =.046734
For H3Po4
.095375M X 38.0ml= molarity acid X 100ml=.036243M X 2=.072485

#### Borek

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #1 on: November 20, 2011, 04:54:03 AM »
First off the ratio is 1:1 ratio according to the equation.

What equation?

Quote
According to the titration graph for trial 1 the first equivalence point is 24.5 ml at 4.1 ph and the 2nd equivalence point was 38.0ml at 9.1 ph. Also, the molarity of NaOH was standardized at .095375M.

For HCl
.095375M X 24.5 = molarity acid X 100ml=.023367M X 2 =.046734
For H3Po4
.095375M X 38.0ml= molarity acid X 100ml=.036243M X 2=.072485

What is neutralized at first equivalence point?
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#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #2 on: November 20, 2011, 10:42:31 AM »
First off the ratio is 1:1 ratio according to the equation.

What equation?

Quote
According to the titration graph for trial 1 the first equivalence point is 24.5 ml at 4.1 ph and the 2nd equivalence point was 38.0ml at 9.1 ph. Also, the molarity of NaOH was standardized at .095375M.

For HCl
.095375M X 24.5 = molarity acid X 100ml=.023367M X 2 =.046734
For H3Po4
.095375M X 38.0ml= molarity acid X 100ml=.036243M X 2=.072485

What is neutralized at first equivalence point?

the total neutralization of HCl to NaCL and H3PO4 to NaH2PO4 according to
HCl(aq) + NaOH(aq)>H2O(l)+NaCl(aq)
H3Po4(aq)+NaOH(aq)>NaH2PO4(aq)+H2O(l)

the second equivalence point is the neutralization of NaH2PO4(aq)+H2O(l)

The equation for the standardizeof NaOH with potassium hydrogen phthalate
NaOH(aq)+KHC8H4O4>Na+(aq)+K+(aq)+h2O(l)+C8H4O42-(aq)

#### Borek

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #3 on: November 20, 2011, 02:52:29 PM »
You answered in a way that it is impossible to tell which question you were answering. However, your approach is wrong.

However, you seem to be right about one thing - second equivalence point is neutralization of NaH2PO4. Can you use this information to calculate amount of phosphoric acid in the original mixture?
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#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #4 on: November 20, 2011, 07:29:53 PM »
You answered in a way that it is impossible to tell which question you were answering. However, your approach is wrong.

However, you seem to be right about one thing - second equivalence point is neutralization of NaH2PO4. Can you use this information to calculate amount of phosphoric acid in the original mixture?
This is the equations for the first equivalence point.
the total neutralization of HCl to NaCL and H3PO4 to NaH2PO4 according to
HCl(aq) + NaOH(aq)>H2O(l)+NaCl(aq)
H3Po4(aq)+NaOH(aq)>NaH2PO4(aq)+H2O(aq)

What equation? assumption according to the other equations

#### Borek

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #5 on: November 21, 2011, 04:06:00 AM »
HCl and H3PO4 don't react 1:1. If you have 1 mole of HCl and 2 moles of H3PO4 at the first equivalence point HCl neutralized 1 mole of NaOH and H3PO4 neutralized 2 moles of NaOH, so - surprise - they reacted 1:2.

Please follow hint from my previous post, I am not going to repeat myself.
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#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #6 on: November 21, 2011, 03:40:06 PM »
HCl and H3PO4 don't react 1:1. If you have 1 mole of HCl and 2 moles of H3PO4 at the first equivalence point HCl neutralized 1 mole of NaOH and H3PO4 neutralized 2 moles of NaOH, so - surprise - they reacted 1:2.

Please follow hint from my previous post, I am not going to repeat myself.

what equations says they react at 1:2. I am sorry but I don't understand. I read here(http://www.titrations.info/acid-base-titration-phosphoric-acid) that it can be either 1:1 or 2:1.

Unless those calculations I did in my earlier post was for the number of moles for NaOH but I thought they would be the number of moles for the HCl-H3Po4 cause i thought it was a 1:1 ratio but I am wrong. but if your saying its a 1:2 ratio then it would be .023367M X(1/2)=0.11684M

Well here is what I do know if it is correct; the first equivalence point is the titration of hydrogen ions from the HCl and one hydrogen ion from H3PO4. For the second equivalence point the neutralization of NaH2PO4 occurred
« Last Edit: November 21, 2011, 04:21:29 PM by kizunaencounter »

#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #7 on: November 21, 2011, 04:41:15 PM »
Lets see if i have figured this out

So for the first equivalence point

24.5Ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.023367M HCl-H3Po4 mixture solution '
because the amount of NaOH consumed up to the first equivalence is equal to the amount of H3PO4 + amount of HCL. The amount of NaOH consumed between the first and second equivalence point equals the amount of H3PO4. Therefore,at the second equivalence

38.0ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.072485M which then has to be subtracted from the amount of H3PO4 and HCL mixture to find the amount of H3PO4 which is 0.072485M - .023367M=.049118M H3PO4

#### DrCMS

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #8 on: November 21, 2011, 04:45:13 PM »
Lets see if i have figured this out

So for the first equivalence point

24.5Ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.023367M HCl-H3Po4 mixture solution '
because the amount of NaOH consumed up to the first equivalence is equal to the amount of H3PO4 + amount of HCL. The amount of NaOH consumed between the first and second equivalence point equals the amount of H3PO4.

Yes that is it.

Therefore,at the second equivalence

38.0ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.072485M which then has to be subtracted from the H3PO4 and HCL mixture to find the amount of H3PO4 which is 0.072485M - .023367M=.049118M H3PO4

No it is not!  Think about it again, how much NaOH is used to go from the 1st to 2nd equivalence point?

#### Borek

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #9 on: November 21, 2011, 04:54:31 PM »
what equations says they react at 1:2. I am sorry but I don't understand. I read here(http://www.titrations.info/acid-base-titration-phosphoric-acid) that it can be either 1:1 or 2:1.

What if there were 1 mole of HCl and 3 moles of H3PO4? Then they will react 1:3. What if you have 5 mole sof HCl and 2 moles of H3PO4? Then they will react 5:2.

Conclusion - you can;t tell anything about the ratio, as it depends on the solution composition. You have to treat each neutralization separately. So you were right about the fact first end point shows sum of both. What is left in the solution after first end point? What is titrated now?
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#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #10 on: November 21, 2011, 05:15:39 PM »
Lets see if i have figured this out

So for the first equivalence point

24.5Ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.023367M HCl-H3Po4 mixture solution '
because the amount of NaOH consumed up to the first equivalence is equal to the amount of H3PO4 + amount of HCL. The amount of NaOH consumed between the first and second equivalence point equals the amount of H3PO4.

Yes that is it.

Therefore,at the second equivalence

38.0ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.072485M which then has to be subtracted from the H3PO4 and HCL mixture to find the amount of H3PO4 which is 0.072485M - .023367M=.049118M H3PO4

No it is not!  Think about it again, how much NaOH is used to go from the 1st to 2nd equivalence point?

Ohh i see it would be .095375M X 13.5ml = molarity acid X 100ml=.012876M   therefore .023367M - .01287M=.010497

and in between the first and second equivalence point its an equimoloar mixture of H2PO4-  +  HPO4

#### kizunaencounter

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##### Re: Titration of the Unknown HCl-H3PO4 Mixture
« Reply #11 on: November 27, 2011, 12:24:39 PM »
Lets see if i have figured this out

So for the first equivalence point

24.5Ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.023367M HCl-H3Po4 mixture solution '
because the amount of NaOH consumed up to the first equivalence is equal to the amount of H3PO4 + amount of HCL. The amount of NaOH consumed between the first and second equivalence point equals the amount of H3PO4.

Yes that is it.

Therefore,at the second equivalence

38.0ml NaOH/100ml HCl-H3PO4 mixture solution(0.95375M NaOH/10^3)(1molH3Po4/1molNaOH)(10^3/1L)=.072485M which then has to be subtracted from the H3PO4 and HCL mixture to find the amount of H3PO4 which is 0.072485M - .023367M=.049118M H3PO4

No it is not!  Think about it again, how much NaOH is used to go from the 1st to 2nd equivalence point?

Ohh i see it would be .095375M X 13.5ml = molarity acid X 100ml=.012876M   therefore .023367M - .01287M=.010497

and in between the first and second equivalence point its an equimoloar mixture of H2PO4-  +  HPO4

then I have to times it by 10 since I diluted to 1/10 of the solution. Does this all make sense or am I missing something still.