1.984 g of a hydrate of K2CO3 is dissolved in 250. mLs of H2O. 10.0 mLs of this solution is tirade against 0.100 M HCl, producing the following results.

Question

Flask 1 - Initial 0, final 9.2
Flask 2 - Initial 9.2, final 18.3
Flask 3 - initial 18.3, final 27.8
Flask 4 - initial 27.8, final 36.9

Determine the formula of the hydrate.

For some reason my instructor doesn't like to teach us anything until the day after it's due. I really have no idea how to even start this question. Help would be much appreciated!! Thanks!

Answer 1

What indicator did you use?

Answer 2

I'm assuming the indicator was something in the acid range, say change pH about 4 to 5 which would mean CO3^2- was titrated all the way to H2O and CO2.
K2CO3.xH2O + 2HCl ==> H2O + CO2 + xH2O _ 2KCl

mols HCl. Subtract from each other and take the average. That's 9.2 for #1 and 9.1 for #2 etc.
mols HCl = M x L = ?
mols K2CO3.xH2O = 1/2 that (from the coefficients in the balanced equation)
Then since mol = grams/molar mass you can rearrange to molar mass = grams/mol
g of sample = 1.984 x (10/250) = 0.07936
So molar mass = about 0.0936/0.000461 = about 172 or so.
K2CO3 weighs 138.2 so 172 - 138.2 = about 34 which tells me there must be 2 H2O for 2*18 = 36.
Why is it not closer than that. Probably for a couple of reasons. First, you have a sample weighed to 4 places but you read the buret to only 2 places. Second is you can read another place on the buret if you had used above 10 and we like to use more like 35 or 40 mL. It improves the precision. Third, I don't know how pure the K2CO3.2H2O was.

Answer 3

Thank you very much! this makes much more sense now! I really appreciate your help.