[gobuckskb9]

Here is my problem:
a 20 mL solution of 0.005 M Sn2+ in 1 M HCl was titration with 0.020 M Ce4+ to give Sn4+ and Ce3+. Calculate the potential (versus S.C.E.) at the following volumes of Ce4+: 0.1, 1.0, 5.0, 9.5, 10.0, 12.0 mL.

I can't get the equivalence point. Every time I do the calculation (M1)(V1)=(M2)(V2).
I get 5 mL and the answer is 10 mL. What am I doing wrong here?

[Borek]

Every time I do the calculation (M1)(V1)=(M2)(V2).

Every time you use wrong formula you get wrong result, nothing surprising about it.

http://www.titrations.info/titration-calculation

http://www.titrations.info/potentiometric-titration-equivalence-point-calculation

[gobuckskb9]

I am simply trying to find how many mL I have to add to get to the equivalence point, I am aware of the Nernst equation and how it should be used most of the time (this particular problem does not play out the way it should).

So when I am looking to get the equivalence point I know that I have 20 mL of 0.005 M Sn(2+), and I am adding 0.02 M Ce(4+)
so... the volume of Ce(4+) that I need to add should be:
(20)(0.005)=(x)(0.002)
x = 5.
but the book has x=10.
The reaction for this problem is:
2Ce(4+) + 2e(-) -->  2Ce(3+) 
Sn(4+) + 2e(-) -->  Sn(2+)

2Ce(4+) + Sn(2+) --> 2Ce(3+) + Sn(4+)
Is my answer supposed to change if the number of electrons transferred is 2?

[sjb]

I am simply trying to find how many mL I have to add to get to the equivalence point, I am aware of the Nernst equation and how it should be used most of the time (this particular problem does not play out the way it should).

So when I am looking to get the equivalence point I know that I have 20 mL of 0.005 M Sn(2+), and I am adding 0.02 M Ce(4+)
so... the volume of Ce(4+) that I need to add should be:
(20)(0.005)=(x)(0.002)
x = 5.
but the book has x=10.
The reaction for this problem is:
2Ce(4+) + 2e(-) -->  2Ce(3+) 
Sn(4+) + 2e(-) -->  Sn(2+)

2Ce(4+) + Sn(2+) --> 2Ce(3+) + Sn(4+)
Is my answer supposed to change if the number of electrons transferred is 2?

For each mole of Sn(II) you oxidise, you need two moles of Ce(IV) to reduce it. So 20 ml of 0.005 M Sn(II) is how many moles?

[Borek]

(20)(0.005)=(x)(0.002)

2Ce(4+) + Sn(2+) --> 2Ce(3+) + Sn(4+)

You are contradicting yourself here. Read how to calculate titration results - I have already posted the link.

You are sad proof of the fact that equation M₁V₁=M₂V₂ should be banned from classrooms.