[Jules18]

Hey guys, this is my question, I got it wrong on a lab test:

10ml portion of vinegar sample diluted with water to 100mL.  This solution was titrated with a 1.000M NaOH standard sol'n.  If 11.66mL of NaOH was needed to neutralize the solution, what was the acetic acid content of the original vinegar sample?
(K_a acetic acid = 1.8E-5)

My answer key says the answer is 1.116M, but when I do it I keep getting like 754M.  

I figured that there was 0.01166mol of H+, and so I plugged that into this equation:

1.8E-5 = 0.01166²/(M-0.01166)

When I solve for X I get 7.541 mol of acetic acid.  What am I doing wrong?

[cpncoop]

Hi,

I think you're overthinking this a bit.  You correctly identified that you had .01166 moles of H+ in solution based on the amt of NaOH added (11.66 mL of 1 M NaOH = 11.66 mmol = .01166 mol).  Although you diluted the material by a factor of 10, you didn't change the number of moles in it.  You had .01166 mols H+ in your initial sample of 10 mL.  To convert that to the concentration of the initial solution:

.01166 moles/.01 L = 1.166 M concentration of original solution.  This is the molarity of H+ in solution.  If you want to calculate the molarity of undissociated MeCOOH in solution, you could then use the Ka equation..

Hope this helps,

 

[Jules18]

Hi,

I think you're overthinking this a bit.  You correctly identified that you had .01166 moles of H+ in solution based on the amt of NaOH added (11.66 mL of 1 M NaOH = 11.66 mmol = .01166 mol).  Although you diluted the material by a factor of 10, you didn't change the number of moles in it.  You had .01166 mols H+ in your initial sample of 10 mL.  To convert that to the concentration of the initial solution:

.01166 moles/.01 L = 1.166 M concentration of original solution.  This is the molarity of H+ in solution.  If you want to calculate the molarity of undissociated MeCOOH in solution, you could then use the Ka equation..

Hope this helps,

thanks.  The question was badly worded.  I think they were looking for the H+ concentration in the orig. sol'n, even though the question actually says acetic acid content.  >

[savy2020]

When reacting with NaOH all of the acetic acid will be neutralised.
In the original vinegar solution, though all of the acetic acid is not dissociated into H+, as the neutralisation goes on and H+ is being consumed the dissociation of acetic acid is shifted in forward direction acc. to Le Chatelier's principle.

EDIT: The question is correctly worded.

[Borek]

I figured that there was 0.01166mol of H+, and so I plugged that into this equation:

That's the amount of H⁺ neutralized, it doesn't mean there were as much H⁺ present in the initial solution. See savy's post.

[Jules18]

When reacting with NaOH all of the acetic acid will be neutralised.
In the original vinegar solution, though all of the acetic acid is not dissociated into H+, as the neutralisation goes on and H+ is being consumed the dissociation of acetic acid is shifted in forward direction acc. to Le Chatelier's principle.

EDIT: The question is correctly worded.

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

[DrCMS]

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

You are wrong, the question asks for the acetic acid concentration of the 100ml sample and the answer key gives the answer to that question correctly.  The Ka is a red herring that has no bearing on the question or answer.

[savy2020]

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

. H+ concentration after the acetic acid dissociated completely...
That H+ conc. would be equal to the Acetic acid concentration of course. Since 1mol HAc gives 1 mol H+ on complete dissociation.

[Jules18]

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

You are wrong, the question asks for the acetic acid concentration of the 100ml sample and the answer key gives the answer to that question correctly.  The Ka is a red herring that has no bearing on the question or answer.

The question asks for the concentration in the original vinegar sol'n, which I thought would refer to the 10 mL sample.  Is that what you meant? 

[DrCMS]

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

You are wrong, the question asks for the acetic acid concentration of the 100ml sample and the answer key gives the answer to that question correctly.  The Ka is a red herring that has no bearing on the question or answer.

The question asks for the concentration in the original vinegar sol'n, which I thought would refer to the 10 mL sample.  Is that what you meant?  

Yes the acetic acid concentration is the same in the 10ml portion titrated as it is in the 100ml original sample the 10ml was taken from.  You worked it out and then tried to calculate the concentration of H⁺ which the question did not ask for.

[Borek]

Yes the acetic acid concentration is the same in the 10ml portion titrated as it is in the 100ml original sample the 10ml was taken from.

I think you misread the question:

10ml portion of vinegar sample diluted with water to 100mL.

100 mL volume is completely meaningless, it is number of moles of acetic acid and 10 mL volume that counts.

[DrCMS]

Yes the acetic acid concentration is the same in the 10ml portion titrated as it is in the 100ml original sample the 10ml was taken from.

I think you misread the question:

Yeh I forgot to re-read it.  The answer key gives the correct answer to the question asked and the Ka given has no bearing on the correct answer.

[Jules18]

But it leads the student to believe it's asking him/her for the initial concentration of acetic acid, when in fact the answer key suggests that it was looking for H+ concentration after the acetic acid dissociated.  Am I wrong?

. H+ concentration after the acetic acid dissociated completely...
That H+ conc. would be equal to the Acetic acid concentration of course. Since 1mol HAc gives 1 mol H+ on complete dissociation.

acetic acid wouldn't dissociate completely.  That's why the Ka is given.

[Borek]

acetic acid wouldn't dissociate completely.  That's why the Ka is given.

You were told several times that Ka is not important for the final answer. Now you are just trolling. Topic locked.