Dilution of burettes and pipettes

[jamess]

This is really confusing me, but what are the 'steps' when;

1) a pipette is rinsed with water, instead of a base
2) a burette is rinsed with water, instead of an acid

I know that when a pipette is rinsed with water;
the base is diluted, therefore the [base] decreases, therefore [acid] will be falsely high/overestimated     BUT WHY?

and when an acid is diluted in a burette, the [acid] increases, therefore [base] is falsely low/underestimated     BUT WHY??

thanks

[nacho]

1 - 2 [same stuff]) when you rinse the pipette with water, it will contain droplets of that water thus diluting the base (as you've mentioned)
    So, when we put in the base, up to say, 20mL, it may actually only be 18mL of base +2mL of water. Then, our base is also less concentrated as it has been diluted.
    If our base is less concentrated, it will take more of it to neutralise the acid, correct?
Picture it like this:
We have 20mL 5.0M of HCl in a reaction flask, and 40mL of 5.0M NaOH in the burette. We will need 20mL of this to neutralise as it is equally concentrated, yes? (and also because they react in a 1:1 ratio, for 1 mol of HCl we need 1 mol of NaOH, i.e.    HCl(aq) + NaOH(aq) ----> + NaCl (aq)
If you follow me up to here, ask yourself, if we have less concentrated base, let's say 2.5M NaOH, THEN how much will it take to neutralise the acid?
Twice as much, correct? Why? Because there is half the no of mols in this less concentrated NaOH than there was before.

So, using the previous figures [ except we don't know the concentration of acid], if I put the 5.0M NaOH in the burette, and i previously washed it with water, it will be diluted, and have a new concentration of, say, 4.8M.
Therefore, when we titrate this against the acid, it will take more volume of this diluted base with 4.8M, than it would with the original un-diluted base of 5.0M.

Because it takes more volume, we will have a higher mean titre. Let's say, our mean titre with diluted base is 25.0mL and whilst is should have been 22.0mL if it werent diluted
bring your attention to the formula n = c x V
using stoich calculations, we now need to determine the no. of mol of NaOH needed to neutralise HCl. (When we have n(NaOH) we can say that this = n(HCl) (because they react in 1:1 ratio)
can you see that when we use n = 5.0mol multiplied by 0.025L = 0.125mol     ... answer 1 (the answer with diluted/error NaOH)
it will be higher than               n = 5.0mol multiplied by 0.022L ? = 0.11mol     ... answer 2 (the answer that we should have got, had we washed properly)
This is the fundamental step. So, when we determine this higher mol value for n(NaOH) we will also have a higher mol value for n(HCl)
At this stage,
we will have
n(HCl) and v(HCl) [this is the amount in the titration flask]
To figure out the concentration of HCl, we transpose our original formula, so that it is c = n/v
Now, look at this.
answer 1 will yield:
c = n/v = 0.125/0.02 = 6.25M

answer 2 will yield:
c = n/v = 0.11/0.02 = 5.5M

In comparison you can see that the answer we receive with dilute NaOH is higher than undilute NaOH.

So, a little flowchart should explain this (sorry for my unnecessarily lengthy explanation)

Washed burette with water ---> Solution which is added will be diluted slightly ----> Diluted solution will have less concentration than it should ----> Therefore more solution is needed to titrate against the solution in the titration flask----> more solution needed = higher mean titre---> high mean titre = higher no. of mols determined for solution in burette---> therefore higher no. of moles determined for solution in titration flask ----> when we use c = n/v we will have yielded a higher concentration than we should have

[jamess]

ohhh man, thanks heaps

[jamess]

would also be able to do one of those llittle flow charts for dilution of a pipette? cheers

[nacho]

Apply the same logic.
Assuming your going from pipette to titration flask,
in this titration flask you'd have a less concentrated solution. Slightly different at this stage, but it would require LESS solution from the burette to neutralise it, compared to if it weren't diluted.
Try drawing it out, and you should see it

[jamess]

hey so i summarised it;
is this true
and is there anything else that needs to be included in it?

Pipette and Burette Dilution
•   Same logic is applied to pipette and burettes, only need to be concerned about, whether there is an acid or base in there
•   Base, (let’s say it’s in a pipette)
1.   Base will be diluted
2.   [Base] decreases
3.   Need less v(acid) as more v(base) is used
4.   [acid] increases, because c=n/v
5.   [acid] will be overestimated
•   Acid, (let’s say it’s in a burette)
1.   Acid is diluted
2.   [acid] increases
3.   Need more v(base) as less v(acid) is used
4.   [base] decreases, because c=n/v
5.   [base] will be underestimated