• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

PLEASE CHECK - are my titration calculations correct? (1 Viewer)

Lordsion

Member
Joined
Jul 3, 2007
Messages
105
Gender
Female
HSC
2008
Forever in debt if you answer this!!!!


Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.

I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.

The vinegar was DILUTED 1 in 5.

I HAVE TO FIND (A) AND (B)

(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)

(b) Molarity of original vinegar solution (undiluted 50ml)

MY SOLUTIONS ====== ARE THEY RIGHT METHOD?

(a) C1 X V 1 = C2 X V2

0.10 (concentration of NaOH) X 25 (ml of NaOH) = C2 (? concentration of diluted vinegar) X 21 (titre of vinegar used)

0.10 X 25 / 21 = C2

0.1190 ..... mol -1


(b) since i diluted the vinegar 4 times its volume .....

0.1190 .... X 4 (or is it 5? ) = 0.476 (this is the concentration of undiluted vinegar)



is this right ?

Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???
 

dan.121212

Member
Joined
Nov 11, 2006
Messages
49
Gender
Undisclosed
HSC
2008
i believe first part is correct, but second part - times by five, not four (i believe)

also, there is no such thing as "0.10 Mol -1" or "0.100 mol -1", - there IS such a thing as: 0.10 mol, 0.100 mol, 0.100 mol. L^-1, 0.10 mol.L^-1

the first two are number of moles, the latter two are measures of concentration in mol per litre

as for answering your Q - i answered it already on your 'Titration - are my steps right?' forum
 

undalay

Active Member
Joined
Dec 14, 2006
Messages
1,002
Location
Ashfield
Gender
Male
HSC
2008
Lordsion said:
Also, is 0.10 Mol -1 the same as 0.100 mol -1 ???
I assume you are refering to moles per litre.
In which case the first number has less significant figures then the second.
This may or may not alter your final result, however generally in HSC it doesn't matter that much.
 

Lordsion

Member
Joined
Jul 3, 2007
Messages
105
Gender
Female
HSC
2008
dan.121212 said:
i believe first part is correct, but second part - times by five, not four (i believe)

also, there is no such thing as "0.10 Mol -1" or "0.100 mol -1", - there IS such a thing as: 0.10 mol, 0.100 mol, 0.100 mol. L^-1, 0.10 mol.L^-1

the first two are number of moles, the latter two are measures of concentration in mol per litre

as for answering your Q - i answered it already on your 'Titration - are my steps right?' forum
THANK YOU SO MUCH - i have an assesment tom/ow and i am going through the questions making sure i have them right.

One more question though:

"How would the concentration of the NaOH solution in this experiment have been determined?"

I am gessing this:

The lab assistant would determine the amount of grams of NaOH needed to make up the concentration of 0.1 mol L^-1 by using this formula:

m = M x Mr x V. (Mass of substance = concentration of NaOH X molecular weight of NaOH X volume wanted ... say 250 ml) ?

I really need help with this one!
 

Lordsion

Member
Joined
Jul 3, 2007
Messages
105
Gender
Female
HSC
2008
And also - that calculation in my initial post is worth 10 marks !!

Will i get ten makrs for those calculations - like, they dont seem like enough to get 10 marks.
 

Lordsion

Member
Joined
Jul 3, 2007
Messages
105
Gender
Female
HSC
2008
dan.121212 said:
DONT PANIC - ull be fine

But what do you think - about how it (concentration of NaOH) is determined?

I forget if any other rules are involved.
 
Joined
Jan 16, 2008
Messages
214
Location
Down On The Upside
Gender
Male
HSC
2008
Lordsion said:
But what do you think - about how it (concentration of NaOH) is determined?

I forget if any other rules are involved.
Since your test was the next day it's already done but ill let you know anyway.

The concentration of the solution would either be tested by titrating it with a primary standard solution.

Or if it was made to be the primary standard solution (and it wouldve been checked with a titration anyway) the lab assistan would have done the maths (can't remember) then weighed the sample to the appropriate weight, added it to the appropriate volume of water and shakin' and stired.

I think that's your answer.

Cheers.
 

imnotkarl

New Member
Joined
Jun 15, 2007
Messages
9
Gender
Male
HSC
2008
Lordsion said:
Forever in debt if you answer this!!!!


Titration of NaOH with Vinegar (acetic acid). Vinegar is in the burette, NaOH in the flask.

I used NaOH with concentration 0.10 mol -1 and i used 25 ml of it.
My Titre of vinegar used to reach equivalence point of NaOH was 21 ml.

The vinegar was DILUTED 1 in 5.

I HAVE TO FIND (A) AND (B)

(a) molarity of diluted vinegar solution (which is 1 in 5 - i did 50 ml vinegar with 200 ml water = 250 ml)

(b) Molarity of original vinegar solution (undiluted 50ml)

MY SOLUTIONS ====== ARE THEY RIGHT METHOD?

(a) C1 X V 1 = C2 X V2
Ok Hold up, this is not tiratraton, this is dilution.
C1V1=C2V2 is ONLY for dilution.

It will give you the answer to your question but is not the correct method to acheive a band 6 response.

Follow these steps for the correct TITRATION procedure:

Step 1: Write a Balanced Chemical Equation for the Reaction (of acid+base...B+A=S+W+CD)

Step 2: Calculate the moles of the known solution (n=m/M or n=C/V)

Step 3: Use the mole ratio to work out the moles of the unknown solution in the flask.

Step 4: Determine the concentration of the unknown solution using C=n/V.

Done presto!

Using the Dilution equation can only be used for the primamry standard but cannot yeild accurate results for the secondry standard.

In questions that give solutions with grams of acids diluted, it does not work.

cheers,
pat
 
Joined
Dec 17, 2006
Messages
433
Gender
Undisclosed
HSC
N/A
imnotkarl said:
Ok Hold up, this is not tiratraton, this is dilution.
Ok; hold up; this is titration and dilution.

Thread-starter has a question where volumes of an acid and a base reacting, with one's concentration known and the other to be found.
C1V1=C2V2 is ONLY for dilution.
Depends actually; what do subscripts '1' and '2' stand for? If you let '1' and '2' equal 'initial' and 'final', then it's dilution; if it's 'a' and 'b' (for 'acid' and 'base'), then it could be titration.
Lordsion said:
Will i get ten makrs for those calculations - like, they dont seem like enough to get 10 marks.
No, no you won't.
imnotkarl said:
It will give you the answer to your question but is not the correct method to acheive a band 6 response.
'Band-6 response' tends to apply to extended-response questions.[/quote]
dan.121212 said:
DONT PANIC - ull be fine
Generic advice is needless. Console yourself with it when you're panicking.

OP, your calculations for the reacting vinegar are correct. Your calc.'s for the initial concentration of the vinegar, prior to dilution, it would be multiplied by five.
(civi = cfvf, where c's are concentrations, and v's are (overall) volumes.)

Hey, good job with answering the OP's questions, people.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top