posted by Mike .
A 23.596 g sample of aqueous waste leaving a fertilizer manufacturer contains ammonia. The sample is diluted with 79.504 g of water. A 13.803 g aliquot of this solution is then titrated with 0.1098 M HCl. It required 29.02 ml of the HCl solution to reac the methyl red endpoint. Calculate the weight percent NH3 in the aqueous waste.
HCl + NH3 ---> NH4^+ +Cl^-
I have to find the moles of hcl consumed , convert them to moles of NH3 and then convert that to grams of NH3 but i really don't know how to do each one of these steps. Help?
moles HCl = M x L = 0.1098M x 0.02902L = approximately 0.003
The NH3 + HCl equation is balanced and it is 1 mole NH3 to 1 mole HCl; therefore, moles NH3 = moles HCl.
grams NH3 = moles NH3 x molar mass NH3 = approximately 0.05 g NH3 in the titrated sample (the 13.803 g portion). You need to convert this to grams in the sample. You do that by
0.05 g in the titration part x (23.596 + 79.504)/(13.803 = approximately 0.4 g in the original sample. Then
%NH3 = (grams NH3/mass sample)*100 = ??
mass of the sample is 23.596
I got 1.6134% which doesnt make sense :s