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Trial question that I'd like opinions on (1 Viewer)

CM_Tutor

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I came across this question as an MCQ and don't agree with any of the answers, so I'd like to hear what you think would be the answer.

A student wishes to determine the concentration of a solution of lead(II) cations by using a precipitation titration. A 25.00 mL sample of the solution is titrated against 0.0978 mol L-1 hydrochloric acid. The average titre for the formation of a faint precipitate of lead(II) chloride is 36.75 mL. What is the concentration of lead(II) ions in the original sample, in mol L-1?
 

jazz519

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I came across this question as an MCQ and don't agree with any of the answers, so I'd like to hear what you think would be the answer.

A student wishes to determine the concentration of a solution of lead(II) cations by using a precipitation titration. A 25.00 mL sample of the solution is titrated against 0.0978 mol L-1 hydrochloric acid. The average titre for the formation of a faint precipitate of lead(II) chloride is 36.75 mL. What is the concentration of lead(II) ions in the original sample, in mol L-1?

What is it that you don't agree with?

because it seems like a straight forward question in terms of what is commonly asked in the HSC trial papers.
Pb2+(aq) + 2HCl(aq) --> PbCl2(s) + 2H+(aq)
find n(HCl) = cv = (0.0979)(36.75/1000)
molar ratio to the Pb2+ and then find conc using c = n(Pb2+) / 0.025
 

cbatostudy

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Is this question 20 of CSSA 2020? I remember doing it and none of the answers match up. Although C(0.00502) is the answer if it was of the final solution and not the aliquot. I think I ended up getting 0.0124 mol/l
 

Gloucestershire

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Total volume of sample = 61.75 mL.
Added Cl- is 0.00359415 mol, i.e [Cl-] = 0.05820.
Ksp = 1.7e-5 = [Pb2+][Cl-]^2
So [Pb2+] = 0.005018 M

The initial concentration of lead is then calculated as 0.01239 M, so same answer as cbatostudy. Formation of other complexes is neglected on assumption that it was shown be negligible at question development time.
 

CM_Tutor

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At first, I approached the question as Jazz has, getting [Pb2+] = 0.0720 mol L-1 (3 sig fig).

However, this was not an option provided, and I was suspicious as this has a final mixture with [Pb2+] = 0.02937 mol L-1 and [Cl-] = 0.05820... mol L-1 and Qsp = 0.02937 x 0.05820...2 = 9.949... x 10-5, which is nearly five times the value of Ksp = 1.7 x 10-5.

Taking Gloucestershire's approach, we get [Pb2+] = 0.00502 mol L-1 (3 sig fig) in the solution when precipitation occurs and thus [Pb2+] = 0.01239 mol L-1 in the initial sample, but that wasn't an answer either.

The answer 0.00502 mol L-1 is provided, but that is not the initial sample.

So, I wanted to check that I wasn't missing something.

I am surprised that the 0.0720 answer wasn't included as a distractor, and nor was the correct answer provided... I'm glad that it is just a case of poor question writing and not me :)
 

jazz519

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At first, I approached the question as Jazz has, getting [Pb2+] = 0.0720 mol L-1 (3 sig fig).

However, this was not an option provided, and I was suspicious as this has a final mixture with [Pb2+] = 0.02937 mol L-1 and [Cl-] = 0.05820... mol L-1 and Qsp = 0.02937 x 0.05820...2 = 9.949... x 10-5, which is nearly five times the value of Ksp = 1.7 x 10-5.

Taking Gloucestershire's approach, we get [Pb2+] = 0.00502 mol L-1 (3 sig fig) in the solution when precipitation occurs and thus [Pb2+] = 0.01239 mol L-1 in the initial sample, but that wasn't an answer either.

The answer 0.00502 mol L-1 is provided, but that is not the initial sample.

So, I wanted to check that I wasn't missing something.

I am surprised that the 0.0720 answer wasn't included as a distractor, and nor was the correct answer provided... I'm glad that it is just a case of poor question writing and not me :)
In that same exam by the way there was another wrong question. So I think they just didn't check their paper properly before releasing it for students to do
 

CM_Tutor

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In that same exam by the way there was another wrong question. So I think they just didn't check their paper properly before releasing it for students to do
Which one? I haven't got their solutions for it, unfortunately
 

jazz519

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Which one? I haven't got their solutions for it, unfortunately
Question 7 the one about functional group isomers. The answers said that B was the correct answer which were I and III. These were both alcohols so they can't be functional group isomers.
 

CM_Tutor

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Question 7 the one about functional group isomers. The answers said that B was the correct answer which were I and III. These were both alcohols so they can't be functional group isomers.
Are we looking at the same paper? I have Q4 as a functional group isomer of butanoic acid where the answer is the ester, D, as the alkanediol B is not an isomer at all. Q7 is about the copper(II) / ammonia complex.

Q2 had no correct answer due to a typo in D.

And, I had doubts about Q16
 

idkkdi

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Are we looking at the same paper? I have Q4 as a functional group isomer of butanoic acid where the answer is the ester, D, as the alkanediol B is not an isomer at all. Q7 is about the copper(II) / ammonia complex.

Q2 had no correct answer due to a typo in D.

And, I had doubts about Q16
have a look at cssa 2021. that one was full of errors haha.
 

jazz519

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Are we looking at the same paper? I have Q4 as a functional group isomer of butanoic acid where the answer is the ester, D, as the alkanediol B is not an isomer at all. Q7 is about the copper(II) / ammonia complex.

Q2 had no correct answer due to a typo in D.

And, I had doubts about Q16
Must be a different year for the CSSA
 

CM_Tutor

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Note: I have an errata note for the 2020 paper that says that q7 was meant to say position isomer.

And Q18 was wrong in the answers to 2020 too.
 
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idkkdi

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that's the one I'm looking at... did you notice there is no correct answer to Q2?
ye, I didn't bother doing the paper after I saw q2 didn't have a single correct equation haha.
 

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