[RespectfulStudent]

Hi,
I am trying to determine the percent yield of an unknown acid, I am preparing for a titration lab exam.

I made sample numbers to try recreating the procedures that I'd have to follow in the lab.
My generated info:
*the acid is solid
*The acid is diprotic.
*The molar mass of the acid is 200.00 g/mol (made-up number, actual value is given during my examination)
*250mL of acid is titrated with 24.3 mL of 0.110M base NaOH(during my exam I will be using NaOH for the base)
--> I want to dissolve my acid in water and then add it to a beaker where I will fill it up to the 50mL line. If I have, say, 1g of acid (from subtracting acid and beaker mass from the empty beaker mass to get 1g, would this be a dilution by a factor of 1/50?

I used the n=cV equation to get the moles of NaOH
Division by 1000 was to convert mL to L
nNaOH = (24.3mL/1000)(0.110 mol NaOh)
=0.002673 mol NaOh (I believe that there should be 4 sf in the answer., 0.002673 has 4 significant figures. However, I don't know which value to base this from: 1000?)

n of acid = m/M
=1.80g/200.00g/mol
=0.009 g

I am unsure of my next step in finding percent purity of the acid, and am confused as to how I can achieve a balanced chemical equation. Online I read about several different methods that I have not yet learned (I am in my third year of high school) such as equivalence points.

My equation for the diprotic acid currently looks like this(unbalanced):
H₂X + NaOH --> NaX + H₂O
websites have led me to believe that I am incorrectly forming my equation, and that it should look like this:
H₂X + NaOH --> Na₂X + H₂O
I don't understand how sodium could have a charge of 2-+ in the equation above

Slightly unrelated: Could X contain a hydrogen ion?

Any attempt to clear up my confusion would be greatly appreciated.
Thanks,
respectfulstudent

[Borek]

*The molar mass of the acid is 200.00 g/mol (made-up number, actual value is given during my examination)

OK

*250mL of acid is titrated with 24.3 mL of 0.110M base NaOH(during my exam I will be using NaOH for the base)

Are these made up? Will you be given NaOH concentration?

--> I want to dissolve my acid in water and then add it to a beaker where I will fill it up to the 50mL line.

You need to know the volume precisely, beaker is not good for this. You should use a volumetric flask.

If I have, say, 1g of acid (from subtracting acid and beaker mass from the empty beaker mass to get 1g, would this be a dilution by a factor of 1/50?

No. Dilution factor is when you mix solution with solvent, not when you dissolve a solid.

I used the n=cV equation to get the moles of NaOH
Division by 1000 was to convert mL to L
nNaOH = (24.3mL/1000)(0.110 mol NaOh)
=0.002673 mol NaOh (I believe that there should be 4 sf in the answer., 0.002673 has 4 significant figures. However, I don't know which value to base this from: 1000?)

1000 is a conversion factor between L and mL and as such is an exact number (with infinite number of SF).

n of acid = m/M
=1.80g/200.00g/mol
=0.009 g

No idea where you got the 1.8 g from (and it is not 0.009 g, but 0.009 moles),

I am unsure of my next step in finding percent purity of the acid, and am confused as to how I can achieve a balanced chemical equation.

Basically it is all a simple stoichiometry, you said the acid is diprotic, that's all that matters for the neutralization reaction equation.

Online I read about several different methods that I have not yet learned (I am in my third year of high school) such as equivalence points.

You can't do titration without determining the end point (let's assume for now that end point and equivalence point are the same thing, they are not, but they are very close to each other), so if you are expected to do the titration, you should be told what it means. Look here for a basic definitions:

http://www.titrations.info/titration-basic-terms

My equation for the diprotic acid currently looks like this(unbalanced):
H₂X + NaOH --> NaX + H₂O

That's a good direction, even if I prefer H₂A (with A standing for acid). But there is a problem with your equation - if the formula is H₂X, what is the charge of X after the neutralization (removal of two H⁺)? How many Na⁺ are required for the molecule of the salt to be neutral?

Or, to make things even simpler: when you replace H⁺ with Na⁺, how many Na⁺ you need to replace each H⁺? How many H⁺ in the molecule of H₂X?

websites have led me to believe that I am incorrectly forming my equation, and that it should look like this:
H₂X + NaOH --> Na₂X + H₂O
I don't understand how sodium could have a charge of 2-+ in the equation above

It doesn't, you are apparently mistaken about something. See my previous comment.

Slightly unrelated: Could X contain a hydrogen ion?

If the acid is diprotic - no. But in general acid may behave as diprotic in the titration even if it contains more acidic protons. Don't bother, it is tricky, assume the acid is diprotic and stop there.

[Arkcon]

Also, this bit right here:

I am unsure of my next step in finding percent purity of the acid, and am confused as to how I can achieve a balanced chemical equation. Online I read about several different methods that I have not yet learned (I am in my third year of high school) such as equivalence points.

And the part that follows is a common new student mistake.  That somehow the "purity" is a somehow a "coefficient" to be applied to the balanced equation.  That's not what the chemical equation is used for.

Try to work backward, so that you learn the concepts, even if that's not your experiment:

You have a diprotic acid (that means what?) of 200 MW.  You have 100 g of it.  How much NaOH would be needed to neutralize 100% pure acid?  If it were 5% sand or sawdust it would take 5% less NaOH.  Just be ready, to do the algebra on your titration result to determine the exact purity.

[RespectfulStudent]

You have a diprotic acid (that means what?)

The acid has two hydrogen atoms. Mono-one, Di-two, Tri-three. During my exam I'll be given a monoprotic, diprotic, or triprotic acid (according to the label on the solid acid container).

Is the mole to mole ratio dependent on the number of hydrogen atoms? Would the diprotic acid have a 1:2 ratio that I would have to use as a conversion factor?

----
the molarity of NaOH is provided to us on the day of the lab exam.


Thanks  :)for the link on the titration definitions, I checked it out and understand a bit more than before.
How would I find the endpoint when making sample figures(endpoint- when the phenolphthalein indicator that we are to use turns the solution pink for about 30 seconds)?

I'll restate what I was given because I made a few errors doing so earlier and I want to enhance clarity:

1) molar mass of acid = 200 g/mol. | on the lab exam: the molar mass is provided

2) 1.80g of the impure acid is dissolved in water.  | on the lab exam: we are given a certain amount of solid acid and we have to conduct three trials with a specific amount to be weighed each time.

3) 250mL of acid titrated with 24.3 mL of 0.110 N base NaOh | on the lab exam: the base is to be NaOH. The molarity of the base will be provided; it will be close to 0.08M NaOH.
-> is 24.3 mL the endpoint?

4) I only have access to:  balances, Erlenmeyer flasks, funnels, beakers, and burettes. The indicator will be phenolphthalein. The acid is soluble in water. The acid may be monoprotic, diprotic, or triprotic. There is a note on the sheet that the titration will involve a strong base. Does this affect the balanced chemical equation structure?

_____

if the formula is H2X, what is the charge of X after the neutralization (removal of two H+)? How many Na+ are required for the molecule of the salt to be neutral?

Can the charge be 0 for X?
I thought that the neutralization reaction can only be successful if there are the same amount of hydrogen and hydroxide (ions?) in a 1:1 ratio. Why would the charge of Na+ be related to the charges of H+ and OH- (ions?).

when you replace H+ with Na+, how many Na+ you need to replace each H+? How many H+ in the molecule of H2X?

There are 2 H+ (ions?) in the molecule of H2X. I'm not sure, however, as to how this ties in to the subscript/charge that I should have on X as depicted in the website's equation: the website shows Na2X.



If someone could point me towards a link that explains the concept, it'd be very helpful.

[Borek]

if the formula is H2X, what is the charge of X after the neutralization (removal of two H+)? How many Na+ are required for the molecule of the salt to be neutral?

Can the charge be 0 for X?

No, Try to write dissociation reaction equation. And remember, that the charge is conserved just like atoms are - if total charge on the left is 0, total charge on the right must be zero as well.

I thought that the neutralization reaction can only be successful if there are the same amount of hydrogen and hydroxide (ions?) in a 1:1 ratio.

Correct.

Why would the charge of Na+ be related to the charges of H+ and OH- (ions?).

Salt molecules are electrically neutral. When dissociation removes H⁺ from the acid molecule what is left becomes negatively charged, Na⁺ must make it neutral again.

when you replace H+ with Na+, how many Na+ you need to replace each H+? How many H+ in the molecule of H2X?

There are 2 H+ (ions?) in the molecule of H2X. I'm not sure, however, as to how this ties in to the subscript/charge that I should have on X as depicted in the website's equation: the website shows Na2X.

I believe my answers above address the question.

[RespectfulStudent]

Hello,
Thanks for the response.
You mention that:

Salt molecules are electrically neutral.

is NaOH a salt molecule? I assume that neutral means that there is an equal amount of +:-ve charges in the Na atom (or is it an ion?) and the OH ion.
If so, here is my attempt at the dissociation equation:

Na+(aq) + OH-(aq) + 2H+(aq) +X^1,3,5?-(aq) <--> Na+(aq) + X-(aq) + 2H+(l)+ OH-(l)
NaOH(aq) +H2X(aq)<--> NaX(aq) + 2H2O(l)
I don't think the equation is right, so I am going to try applying information that I learned in class:
"diprotic and triprotic acids dissociate into their respective number of H+ ions."

Does this mean that the H2X acid, when dissociates, looks like this?
2H+(aq) + X-(aq) <--> 2H+(aq) + 2X(aq)
therefore
Na+(aq) + OH-(aq) + 2H+(aq) +2 X-(aq) <--> Na+(aq) + 2X-(aq) + 2H+(l)+ OH-(l)

where the bolded 2 coefficient left of X corresponds with the number of H ions in the original acid; H2X
: NaOH(aq) + H2X(aq) <--> Na2X(aq) + H2O(l)
Balanced:
2NaOH(aq) + H2X(aq) <--> Na2X(aq) + 2H2O(l)

My basic understanding of dissociation is that a molecule breaks into ions or atoms. If the note is applicable to the formulation of the equation, does this mean that the acids with (ie.) two H+ (ions?), such as H2X, dissociate into H2+ and X2- ions?

I also don't understand what it means to find the percent purity of the acid. Is this different than the concentration? Is it necessary to use -[log] ph calculations ?

Is my reasoning correct?
Thanks for the help,
RespectfulStudent

[Borek]

It looks like you are confused about formulas and meaning of numbers in them, that makes you wrong when it comes to understanding dissociation and balancing neutralization equations.

Let's start with the sulfuric acid, H₂SO₄ (which is an example of a diprotic acid, H₂A). When it dissociates, its protons leave the main molecule:

H₂SO₄ 2H⁺ + SO₄^2-

Notice several things:

1. There were two acidic protons in each molecule (H₂SO₄), they produced two protons in the solution (2H⁺) - two makes two, number of atoms doesn't change (we say it is conserved).

2. Each acid molecule, apart from two acidic protons, contained the rest of the acid (H₂SO₄). After the dissociation one such group must be present in the solution - remember, mass is conserved, atoms don't appear from nowhere nor disappear. Two hydrogens were removed, that means on the right side you must have the remaining group of SO₄ atoms (it will be charged, see below). There will be one such group, as there was one such group in the original acid molecule.

3. Initial molecule of the acid was electrically neutral. Charge is conserved (just like mass is) - but there is an additional possibility here, number of atoms is always a small positive integer, charges can be negative and positive, what matters is their sum. So, when the neutral molecule dissociates producing 2H⁺, rest of the molecule (or all other dissociation products, it doesn't have to be a single ion) must carry the same, but opposite charge of 2- - this will guarantee total charge after the dissociation is identical to the initial charge before the dissociation. So the other product of the dissociation is SO₄^2-.

See if you can apply these information to the dissociation of H₂A.

Actually all molecules are electrically neutral.

[RespectfulStudent]

I really understand your explanation, thanks for being so in-depth.

--> put simply, the #hydgrogens are the same as the charge of the acid
therefore, phosphate, PO43-, could be H3PO4 if it were in the form of a triprotic acid
My end goal: Determine % purity of the unknown solid acid, H2X
Now that I have the balanced chemical equation:
2NaOH(aq) + H2X(aq) --> Na2X(aq) + 2H2O(l)
given: Standard solution of NaOH. concentration of NaOH = 0.110g/mol (M)
I have the moles(n value) of NaOH
nNaOH = (24.3mL/1000)(0.110 mol NaOh)
=0.002673 mol NaOH

I have the moles(n value)
n of acid = m/M
=1.80g/200.00g/mol
=0.009 mol X
---------------
I am using a 1.80g sample of the solid acid, I don't want to use 1g of the sample.
---------------
What exactly is percent purity of the acid?
is it the same thing as % purity?
Thanks,
respectfulstudent

[Borek]

--> put simply, the #hydgrogens are the same as the charge of the acid
therefore, phosphate, PO43-, could be H3PO4 if it were in the form of a triprotic acid

I am not convinced this approach would not lead to some problems later, but for now - yes, looks like you got it.

Now that I have the balanced chemical equation:
2NaOH(aq) + H2X(aq) --> Na2X(aq) + 2H2O(l)

OK

given: Standard solution of NaOH. concentration of NaOH = 0.110g/mol (M)
I have the moles(n value) of NaOH
nNaOH = (24.3mL/1000)(0.110 mol NaOh)
=0.002673 mol NaOH

OK

I have the moles(n value)
n of acid = m/M
=1.80g/200.00g/mol
=0.009 mol X

I assume you mean 1.8 g sample of acid was weighted, yes? If so, no. That's how much you would have if the acid was pure, but it is not.

What you can (and should) do is to follow the stoichiometry (you have the correct reaction equation above) and calculate amount of acid from the titration result. Then you will know mass of the mixture (1/8 g) and mass of the acid, and calculating purity will be just plug and chug.

What exactly is percent purity of the acid?
is it the same thing as % purity?

Yes.

[RespectfulStudent]

at the endpoint :
2(C_H2A )(V_H2A) = (C_NaOH)(V_NaOH)
the 2 is there because there are 2 moles of acid (diprotic) for each mole of NaOH.

I am going to try to show how I would conduct the lab with emphasis on steps and how I obtain information for my calculations:

1) I have a container of solid acid. It is labelled as diprotic. (for my lab exam: my teacher will give us the numerical value of how much acid we should use for each experiment in the lab. I will pretend, in this scenario, that 1.80 g of the acid was the numerical value to be weighed out).
2) I have access to a balance* (*this is true on my lab exam; I have access to a balance). I weigh an empty beaker. It weighs 10.00 grams.  Suppose that it rounds to two decimal places (the balance). I want the final weight of the beaker filled with solid acid to weigh 11.80 grams;g: (10.00g+1.80g=11.80g) I shake a little acid into the beaker and weigh it. I get 11.0 grams. I add more acid. 12.00 grams. I dispose of some acid by shaking it into a waste beaker(? Is this dangerous). I weigh again. 11.70 grams. I add more acid. 11.80 grams. (question: If I were to get 11.82 grams, would it be smarter to write down 11.82 grams on my sheet of paper or 11.80 grams?  [I think that I will get a safe range ie. having the weight as +-0.07 g from 11.80g is acceptable.]
For the sake of maintaining the original figures in my question, I weighed exactly 1.80g of acid. (11.80g-10.00g = 1.80g acid)

(note: I think the amount should be around 50mL,not 250mL... but I don't want to change my original question) .
3) I have an Erlenmeyer flask and fill it with 250mL of water. Then I pour the beaker with SOLID ACID; 1.80g of H2A, into the 250mL of water. I shake the Erlenmeyer flask until the acid is no longer visible. Therefore I have 1.80g H2A/250mL H2O.
n acid=m/M
n=1.80g/(200.00g/mol)
=0.009 mol H2A.

250mL H2O/1000 = amount in litres? What can I do from these data?

Is the above procedure for step 3 correct?

4) I added few drops of phenolphthalein to the H2A +H2O solution in the erlenmeyer flask before each trial (total #trials = 3) was carried out.
 

5)
I use a white piece of paper as the meniscus reader. I mark down the starting amount of NaOH in the burette on my table("Titration Data") at the beginning of each trial.

6)Then, I use the switch at the bottom of the burette to release NaOH. Some time later, my NaOH solution begins to turn the acid pink-- temporarily. I begin to add the base drop by drop, slowing the rate of addition of NaOH. I swirl the erlenmeyer flask to clear the pink before adding more base. The drop that turns the solution pink for 30 seconds is signification of complete neutralization. I mark down the remaining amount of base in the burette in the "Final Volume" section of my table. I empty the neutralized solution in the waste beaker.


Lab Results (made-(note: I think the amount of H2O in the erlenmeyer flask should be around 50mL,not 250mL... but I don't want to change my original question) .up)
-NaOH had a molarity of 0.110M
-I think I would use this type of table in my lab when carrying out my titrations.
"Final Volume" is that of NaOH in the burette after the solution is pink for 30 seconds. "Volume used, subtraction," is of NaOH 0.110M
Table 1.

Titration Data

Test # | Initial value of NaOH in burette,mL|Final Volume| Volume used, subtraction|
1         |      80mL                                    |      55.7 mL |     24.3 mL                    | 
2         |      55.7 mL                                |     31.3mL   |      24.4mL                    |
3         |      31.3mL                                 |     7.1mL     |      24.2mL                    |
V_avg = ((24.3mL) + (24.4mL) + (24.2mL))
                             ____________________
                                      3 trials
= 72.9mL/3 trials
=24.3 mL
therefore V_NaOH necessary for the "complete" neutralization of H2A  is 24.3 mL. Is 24.3 mL NaOH what is called the endpoint?
-I expect for my trials to be more dissimilar to one another during my titration than depicted by the made-up trial information.






Restate the titration lab problem/question to answer: Titration of 1.80g of the diprotic acid required 24.3mL of 0.110 M NaOH. Find % purity
--> Does this question^^^ make sense?


n=cV
NaOH: H2A = 2 NaOH : 1H2A
I don't know how to proceed from here.

I seem to have found the moles of H2A; acid, if the acid were pure:

That's how much you would have if the acid was pure, but it is not.

you will know mass of the mixture (1/8 g)

Do you mean 1.8g? (confused)

Does my procedure look good?

Is it important to find what is in excess for my calculations? (limiting reagent)

If I have concentration of the acid (concentration=molarity), can I find percent yield of the acid from there?
Thanks,
Respectfulstudent

[Borek]

I want the final weight of the beaker filled with solid acid to weigh 11.80 grams;g: (10.00g+1.80g=11.80g) I shake a little acid into the beaker and weigh it. I get 11.0 grams. I add more acid. 12.00 grams. I dispose of some acid by shaking it into a waste beaker(? Is this dangerous).

In general it depends on the acid, but I doubt you will be given anything harmful.

I weigh again. 11.70 grams. I add more acid. 11.80 grams. (question: If I were to get 11.82 grams, would it be smarter to write down 11.82 grams on my sheet of paper or 11.80 grams?  [I think that I will get a safe range ie. having the weight as +-0.07 g from 11.80g is acceptable.]

Yes, it is about getting something close to the mass suggested, but don't even try to waste your time to get exact amount. ±10% should be OK. But: you have to write the exact mass, as you will use it in calculations.

Note: you DON'T have 1.8 g of the H₂A. You have 1.8 g of a MIXTURE containing H₂A.

(note: I think the amount should be around 50mL,not 250mL... but I don't want to change my original question) .
3) I have an Erlenmeyer flask and fill it with 250mL of water. Then I pour the beaker with SOLID ACID; 1.80g of H2A, into the 250mL of water. I shake the Erlenmeyer flask until the acid is no longer visible. Therefore I have 1.80g H2A/250mL H2O.
n acid=m/M
n=1.80g/(200.00g/mol)
=0.009 mol H2A.

See above,

Is the above procedure for step 3 correct?

No, especially in the context of the rest of your post.

You should not dissolve the acid (mixture) in the Erlenmeyer flask, but in a volumetric flask (filled to about 80%, and filled up to the mark after the acid has been dissolved). 250 mL measured with the Erlenmeyer flask is pretty inaccurate, 250 mL measured in the volumetric flask is highly accurate. You want to know the volume as precisely as possible, as you will be taking part of it for titration.

Again, when taking part of the solution you need to use a single volume pipette, these are highly accurate.

Compare http://www.titrations.info/pipette-burette

4) I added few drops of phenolphthalein to the H2A +H2O solution in the erlenmeyer flask before each trial (total #trials = 3) was carried out.

Yes, you should transfer samples to three Erlenmeyer flasks for titration, that's OK.
 

5)
I use a white piece of paper as the meniscus reader.

Piece of black paper is better, see the page linked to above.

I mark down the starting amount of NaOH in the burette on my table("Titration Data") at the beginning of each trial.

6)Then, I use the switch at the bottom of the burette to release NaOH. Some time later, my NaOH solution begins to turn the acid pink-- temporarily.

More or less - yes, although it is very easy to overshot this way, unfortunately, first titration should be done very slowly.

I begin to add the base drop by drop, slowing the rate of addition of NaOH. I swirl the erlenmeyer flask to clear the pink before adding more base. The drop that turns the solution pink for 30 seconds is signification of complete neutralization. I mark down the remaining amount of base in the burette in the "Final Volume" section of my table. I empty the neutralized solution in the waste beaker.

OK

Lab Results (made-(note: I think the amount of H2O in the erlenmeyer flask should be around 50mL,not 250mL... but I don't want to change my original question) .up)

As explained above, you should not use Erlenmeyer flask for dissolving the sample.

-NaOH had a molarity of 0.110M
-I think I would use this type of table in my lab when carrying out my titrations.
"Final Volume" is that of NaOH in the burette after the solution is pink for 30 seconds. "Volume used, subtraction," is of NaOH 0.110M
Table 1.

Titration Data

Test # | Initial value of NaOH in burette,mL|Final Volume| Volume used, subtraction|
1         |      80mL                                    |      55.7 mL |     24.3 mL                    | 
2         |      55.7 mL                                |     31.3mL   |      24.4mL                    |
3         |      31.3mL                                 |     7.1mL     |      24.2mL                    |
V_avg = ((24.3mL) + (24.4mL) + (24.2mL))
                             ____________________
                                      3 trials
= 72.9mL/3 trials
=24.3 mL
therefore V_NaOH necessary for the "complete" neutralization of H2A  is 24.3 mL. Is 24.3 mL NaOH what is called the endpoint?

"Endpoint" is not a volume, endpoint is when you finish the titration. 24.3 is a volume required to reach the endpoint.

-I expect for my trials to be more dissimilar to one another during my titration than depicted by the made-up trial information.

Quite likely.

Restate the titration lab problem/question to answer: Titration of 1.80g of the diprotic acid required 24.3mL of 0.110 M NaOH. Find % purity
--> Does this question^^^ make sense?

Yes.

n=cV
NaOH: H2A = 2 NaOH : 1H2A
I don't know how to proceed from here.

I seem to have found the moles of H2A; acid, if the acid were pure:

That's how much you would have if the acid was pure, but it is not.

you will know mass of the mixture (1/8 g)

Do you mean 1.8g? (confused)

Yes, 1.8 g, sorry for the typo.

Does my procedure look good?

More or less.

Is it important to find what is in excess for my calculations? (limiting reagent)

No.

If I have concentration of the acid (concentration=molarity), can I find percent yield of the acid from there?

Yes. You know mass of the sample (1.8 g). Once you know acid molarity you can calculate mass of the acid. These two numbers are necessary (and perfectly enough) to calculate purity.

[RespectfulStudent]

taking part of the solution you need to use a single volume pipette, these are highly accurate.

In our lab, we aren't allowed to use a pipette. It's supposed to be an additional challenge to the experiment.
We have access to : balances, Erlenmeyer flasks, funnels, beakers, and burettes.

-->Do you have any suggestions for a workaround to the flask using the equipment that I am allowed to use?

-->Is the procedure for step 3 correct if an alternative measuring container is used?

--> Furthermore, If I measure 250mL of water, then add 1.8 g of H2A (which I will keep in mind is not necessarily pure, do I have to mention this in the work or is it implied), would the concentration be 1.8gH2A/250mL H2O?

-->

"Endpoint" is not a volume, endpoint is when you finish the titration. 24.3 is a volume required to reach the endpoint.

is the endpoint when neutralization is complete, where pH 7 is the endpoint? How is endpoint expressed?

Piece of black paper is better

I'll do just that, thank you for the advice.

I am grateful for your continued support,
RespectfulStudent

[Borek]

taking part of the solution you need to use a single volume pipette, these are highly accurate.

In our lab, we aren't allowed to use a pipette. It's supposed to be an additional challenge to the experiment.

Then you should weight three samples of the acid, one for each titration. Otherwise you won't be able to do the experiment accurately.

-->Is the procedure for step 3 correct if an alternative measuring container is used?

General idea of dissolving the sample to prepare known volume of the solution is OK.

--> Furthermore, If I measure 250mL of water, then add 1.8 g of H2A (which I will keep in mind is not necessarily pure, do I have to mention this in the work or is it implied), would the concentration be 1.8gH2A/250mL H2O?

If the acid is not pure, the concentration is not 1.8 g per whatever volume it was dissolved in.

I feel like you are still missing the general idea of the experiment.

You have a 1.8 g sample of a mixture. It contains x g of the substance H₂A. Your task is to titrate the sample to measure number of moles of the H₂A. Then you are expected to convert the measured number of moles to x, and use x and the total sample mass (1.8 g) to find out what is the percentage of H₂A in the sample.

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"Endpoint" is not a volume, endpoint is when you finish the titration. 24.3 is a volume required to reach the endpoint.

is the endpoint when neutralization is complete, where pH 7 is the endpoint? How is endpoint expressed?

See the link I posted earlier.

For strong acids equivalence point (which is when the amount of base is stoichiometrically equivalent to the amount of acid present) is at pH 7.00, for weak acids (and you will be titrating a weak acid), pH at the equivalence point is different. Don't bother, details are most likely way over what you are expected to know).

[RespectfulStudent]

You have a 1.8 g sample of a mixture. It contains x g of the substance H2A. Your task is to titrate the sample to measure number of moles of the H2A. Then you are expected to convert the measured number of moles to x, and use x and the total sample mass (1.8 g) to find out what is the percentage of H2A in the sample.

Do I need to perform an additional step in the procedure to obtain a value for x (to clarify: X is not equal to A, is this correct?), or can I find x with the information that I already have?

Thanks,
RespectfulStudent

[Borek]

Do I need to perform an additional step in the procedure to obtain a value for x (to clarify: X is not equal to A, is this correct?), or can I find x with the information that I already have?

Not sure what you really mean by A.

No need for any additional steps, you are doing the titration to find out x.

Imagine you have a 1.80 g sample containing oxalic acid (diprotic, molar mass of 90 g/mol). You have titrated it, reaching endpoint with 20.00 mL of 1 M NaOH.

How many moles of NaOH were used?

How many moles of the oxalic acid were present in the sample?

What is the mass of the oxalic acid in the sample?

What is the original sample purity?

(Do the calculations using these numbers, now).