Titration problem need solutions! (1 Viewer)

[The Bograt]

Just a couple of questions I wasn't sure i got right: If some of you could answer them and tell me what you got I would be very happy!

1. What is the molarity of a solution of H2SO4 if 42.7cm3 is required to exactly titrate 27.5cm3 of 0.61M NaOH
here is what i got:

H2SO4 + 2NaOH --> Na2SO4 + 2H2O(l)
Mole ratio of H2SO4 to NaOH = 1:2
therefore moles NaOH = 0.61 x 27.5
= 16.775 mol
therefore moles of H2SO4 needed = 16.775/2
= 8.3875 moles
conc = mol/vol
M Na2SO4 = 8.3875/42.7
=0.196M

2. How many cm3 of 0.48M H2SO4 is required to neutralise 26.0cm3 of 0.61M NaOH?

H2SO4 + 2NaOH Na2SO4 + 2H2O(l)
Mole ratio of H2SO4 to NaOH is 1:2
concentration = moles/volume
conc. x vol (H2SO4) = 2[conc. x vol. (NaOH)]
0.48v = 2(0.61 x 26)
vol(H2SO4) = 2(0.61 x 26)/0.48
= 66.1 cm3

3. To what volume must 400cm3 of 0.61M NaOH be diluted to yield a 0.50 solution

conc = mol/vol
0.61= mols/400
= 244 mol
0.5 = 244/vol
vol = 488cm3

hopfully this will also benefit mroe of you!
thanks

 

[Scanorama]

I find titration is very hard, the hardest part by far in chemistry...

 

[The Bograt]

can someone just confirm I have done the right steps here?

 

[CM_Tutor]

Originally posted by The Bograt
1. What is the molarity of a solution of H2SO4 is 42.7cm3 is required to exactly titrate 27.5cm3 of 0.61M NaOH
here is what i got:

H2SO4 + 2NaOH --> Na2SO4 + 2H2O(l)
Mole ratio of H2SO4 to NaOH = 1:2
therefore moles NaOH = 0.61 x 27.5
= 16.775 mol
therefore moles of H2SO4 needed = 16.775/2
= 8.3875 moles
conc = mol/vol
M Na2SO4 = 8.3875/42.7
=0.196M

The answer is correct (should really be 0.20 M, to 2 sig fig, as only 2 sig fig in qestion).
HOWEVER, your working is incorrect, as you have made two mistakes that cancel one another out. In determining the number of moles of NaOH, you have multiplied concentration, in mol/L, by volume in mL. As a result, you have an incorrect calculation, as n(NaOH) = 0.016775 mol. You should either include a 10^(-3) factor to correct volume into L, ie:

therefore moles NaOH = 0.61 x 27.5 x 10^(-3)
= 16.775 x 10^(-3) mol

or you should work in millimoles (mmol), ie:

therefore moles NaOH = 0.61 x 27.5
= 16.775 mmol

The reason that you have the correct answer is that in the last step, you have divided by volume, again in mL, and so the two 10^(-3) factors that you have missed would have cancelled.

2. How many cm3 of 0.48M H2SO4 is required to neutralise 26.0cm3 of 0.61M NaOH?

H2SO4 + 2NaOH Na2SO4 + 2H2O(l)
Mole ratio of H2SO4 to NaOH is 1:2
concentration = moles/volume
conc. x vol (H2SO4) = 2[conc. x vol. (NaOH)]
0.48v = 2(0.61 x 26)
vol(H2SO4) = 2(0.61 x 26)/0.48
= 66.1 cm3

This one is wrong, the correct answer is 17 cm3 (to 2 sig fig). The mistake lies in the statement that
"conc. x vol (H2SO4) = 2[conc. x vol. (NaOH)]".

You have said that n(H2SO4) : n(NaOH) = 1 : 2
Upon cross multiplying, this gives 2 * n(H2SO4) = n(NaOH)
and thus your statement should read 2 * conc. x vol (H2SO4) = [conc. x vol. (NaOH)].

You are also working in mL instead of L again.

3. To what volume must 400cm3 of 0.61M NaOH be diluted to yield a 0.50 solution

conc = mol/vol
0.61= mols/400
= 244 mol
0.5 = 244/vol
vol = 488cm3

This is the correct answer, although you again have used mL instead of L, and thus have chemical amount in mmol instead of mol. Note that this question could be done quicker using the dilution formula.

I have posted some approaches to mole calculations in another recent thread, if you want to go have a look.

Hope this helps.

 

[The Bograt]

Hope this helps.

It sure does thanks so much, I wondered why I still got a reasonable answer but didn't do it in litres.
I will go check out that thread
Thanks for the help