[homer5677]

Hello everyone,
I'm new here but am having a few trouble understanding normality and what it is, and how it can be related to molarity. Particularly the normality of a 0.005M iodine solution that is being titrated for sulphur dioxide content in wine.
Thanks for any replies!

[curiouscat]

Do you know what equivalent weight means?

[homer5677]

Do you know what equivalent weight means?

I think I have a rough idea. Isn't that just how much of 1 ion is needed to equal 1 mole of a halogen ion such as cl-

Eg. Ca2+
1 eqv = to 1/2 Ca2+? - I think??

How does this help to find the normality or is it kind of the same thing?

[magician4]

"Normality" is a concept quite irritating to people used to stoechiometric calculations based on mol/L and that like.

Nevertheless, it is a simplification for people who don't want to bother with calculations of what "on equivalent of iodine" would be like for a given reaction  (like: reaction with sulphur dioxide): x mL * 1 N (iodine) always equals findings of n = x mL* 1 mol/L (SO₂)
obviously, the equivalent factors usually belonging to a certain reaction here are included into the concentration of for example the iodine-solution, making concentration * eq-factor  become "normality"

hence, this is kind of a weighting-factor  exclusively for a well defined reaction (and might be completely different if the same substance would be used in a different reaction)

  "N" without explicit declaration of what reaction this "N" is belonging to is meaningless!


examples:
let's consider 2 reactions involving sulphuric acid
(a) formation of BaSO₄ :     1 Ba²⁺ + 1 SO₄^2-  BaSO₄ ( )
(b) formation of Ag₂SO₄:   2 Ag⁺ + 1 SO₄^2-   Ag₂SO₄ ( )

both substances are (next to) insoluble in water

let's say I wanted to measure the content of Ba²⁺ and Ag⁺ by their reaction with sulphuric acid by titration

with a 1 mol/L sulphuric acid i would detect n = x mL * 1 mol/L  barium ions , hence the sulphuric acid is 1 N in this case
on the other hand, with a 1 mol/L sulphuric acid i would detect n = 2 * x mL * 1 mol/L  silver ions ("catching two birds with one stone") , hence the sulphuric acid is 2 N in that case

... even if the very concentration in mol/L in both cases is identical


so for your very reaction of iodine with sulfurdioxide, looking at the stoechiometric factors involved:

1 SO₂ + 1 I₂ +  2 H₂O  1 SO₄^2- + 4 H⁺ + 2 I^-
you see that a 1 mol/L iodine-solution would be 1 N for this reaction


regards

Ingo