[_cheers]

Im tryin to get the w/w% of calcium in skim milk via potentiometry.

my sample was 0.1010g in 250mL. 20 mL aliquot was taken and response measured at 203 mv

calibration graph of all the standards gave me m=22.1, b=261.1

to get the conc of [Ca2+]:

E=m*log[Ca2+] + b

203=22.1*log[Ca2+] + 261.1

[Ca2+] = 0.00235M

to find w/w%:

(0.00235M x (0.02L) x 40.08 g/mol Ca2+) / 0.1010g x100% =1.86% 

skim milk has ~40% so...help?

ty

[Borek]

skim milk has ~40% so...help?

Skim milk has 40% calcium? That's the same content as in solid, anhydrous calcium carbonate...

[_cheers]

Well Im not sure thats why I'm asking if this is right. I read on some site that skim milk has ~40%.  Is the calculation right?

[Borek]

Your calculation of concentration looks right (assuming m and b are correct values), no idea what else you did.

However, your question and data given doesn't make much sense. What is 0.1010g? Mass of calcium in 250mL of milk? If so, you don't need potentiometry, you can just divide calcium mass by solution mass to get the answer.

[Schrödinger]

(0.00235M x (0.02L) x 40.08 g/mol Ca2+) / 0.1010g x100% =1.86% 

I assume that all other data upto this point are correct, and your calculations can also be trusted. With this equation, however, there seems to be a problem. You are required to find out the w/w % of Ca in the main solution (from which the aliquot was taken).

But 0.00235 M x 0.02 L = moles of calcium in the aliquot, not the parent solution. Since the concentration of the aliquot and the parent solution are the same, you should use :
0.00235 M x 0.250L = mole of calcium in the 250 mL sample

I've not checked the numericals...