[toadesque]

It's asking me to calculate the normality of K2C2O4.

I did a lab and this is the information I have

Mass of K2C2O4 H2O in grams: 1.8029
Volume of solution: .1 L (i don't think it's anything other than water)
Volume of k2C2O4 in mL: 10.00

Is this the correct formula to use for normality?

Normality (N) = Molarity (M) × Equivalent (N/M)

I don't have the molarity. Would I just do moles of H2O divided by .1 liters? Also I have no idea how to get the equivalent or whatever that is.

My lab book says: Normality x Volume = # of equivalents but I don't have the normality so how do I do it?

I'm pretty sure I left out some information necessary to get it but maybe not. There are a bunch of balanced equations in my lab book that may be of use but I'm not sure which one would be of a use here.

[toadesque]

Haha okay I got help from another guy and he said this:

K2C2O4 molar mass = 166.2 g/mol
1.8029 g / 166.2 g/mol = 0.01085 mol
0.01085 mol / 0.100 L = 0.1085 M
There are 2 equivalents per mole (K2C2O4 + 2H+ --> H2C2O4 + 2K+)
Normality = 2x0.1085 = 0.217 N

The thing I had trouble understanding was how to get the equivalent. He says there are 2. How did he know that? Like I said before, there were a bunch of balanced equations in my lab manual but I wasn't sure which one to use or if they were necessary to use at all.

[AWK]

Fortunately, for oxalic acid and its salt the neutralization equivalent as well as the redox equivalent are the same (+2)

C₂O₄^2- + 2H⁺ = H₂C₂O₄
H₂C₂O₄ + 2OH^- = C₂O₄^2- + 2H₂O
C₂O₄^2- - 2e^- = 2CO₂

[Borek]

Unfortunately, oxalic acid is also used for precipitation of calcium, then its equivalent is 1

toadesque: in general, before calculating normality you must know the reaction.