volumetric analysis

[tolga]

Q21.
A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus:
a   the pipette used to deliver the aliquot of sodium carbonate solution
b   the flask containing the aliquot
c   the burette.
A21.
a   higher
b   unchanged
c   lower

i am having trouble with this question and would like to know the reason and logic behind these answers and will be thankful if somebody could help.

[emb_23]

I did these errors in class today and those answers from everything ive done and been taught are wrong! Im not sure if anyone agrees with me but this is what I would say

a) Rinsing the Pipette used to deliver the aliquot with H2O
        By doing this it will result in the following
           - Diluting the Na2CO3
           - n(Na2CO3) in aliquot will be decreased
           - Therefore the n(HCl) required will decrease = deacrease the titre
           - The n(HCl) calculated will decrease
           - Concentration of Na2CO3 is decreased as
                             C(Na2CO3)= n/V     ~n is decreased

b)  Rinsing the flask containing the aliquot with H2O
       No change as the amount of mole present in the aliquot is not being changed thus the same amount of HCl will be required.

c) Rinsing the Burette with H2O
         By doing this it wil result in the following
           - dilute the HCl = lower concentration
           - Increase the titre of HCl
           - Increase the n(HCl) calculated
           - Increase the n(Na2CO3) calculated
           - Increase the concentration of Na2CO3 ~ C(Na2CO3) Calculated
          
I listed all the effects of rinsing the equipment with water just to systematically tell you what would happen i hope this helps sorry it doesnt match your answers

[tolga]

For part a i believe you are wrong because if there is less mole of na2co3 doesn't that mean that is it less concentrated and the average titre for HCL is lower as well because it is easier to reach the end point in titration with na2co3, because  HCL it's much more concentrated that the Na2C03 solution.

[chem-nerd]

you start with what you believe to be a known volume and known concentration of Na₂CO₃ and thus known number of mole. However, you actually have less n(Na₂CO₃)

you need to use less n(HCl) to neutralise this

you calculate HCl concentration to be higher because you are now using a smaller V(HCl) with an incorrectly assumed (from mole ratio) n(HCl).

[iffets12345]

What chem-nerd said. You are presuming that you have N amount of Na2CO3. You then have a titre of HCL that is less than usual because your Na2CO3 was diluted in the pipette from the water, meaning you were working with less mol than you thought. You have a small titre but you calculated your mol to be larger than in actuality because of its being a standard solution. If you have a number of mol and you divide it by a number that is smaller than the original volume it will of course, increase. That's math.

[Nomvalt]

i think this question has been answered in another thread. try having a good look around the chemistry forum if you still don't understand.

[Nomvalt]

found it
http://vcenotes.com/forum/index.php/topic,21242.0.html
good luck, hope that helps!