Performing a gravimetric analysis of Ca2+.
25 mL of unknown [Ca2+} added to 25 mL of 0.15 (NH4)2C2O4 and 75 mL of HCl. Urea also added.
H2NCONH2 + 3H2O --heat→ CO2 + 2NH4 + 2OH
CaC2O4*H2O --> Ca2+ + C2O42-
[Ca2+]unknown=0.06028 M <--0.0015433 moles of Ca2+/0.025 L
If 0.15 (NH4)2C2O4, how do I calculate for the excess concentration of C2O42-?
6 answers
HCl is also 0.15 M
Why do you want to calculate excess oxalate line? The only part you're interested in is the part the combined with calcium and you have weighed that. The excess oxalate simply stays in solution and I don't know that you need to worry about it.
I am somewhat concerned about why HCl was added.
In the analysis, some Ca2+ is not precipitated as CaC2O4, because there is always some Ca2+(aq) in equilibrium with the solid CaC2O4. I need to use the known Ksp= 2.3x10^-9 and the concentration of the excess oxalate ion to then use to find the concentration of Ca2+ in equilibrium with the solid CaC2O4*H2O
Pretaining to HCl....
H2C2O4(aq) + 2NH3(aq) --> 2NH4(aq) + C2O4(aq)
H2C2O4(aq) + 2NH3(aq) --> 2NH4(aq) + C2O4(aq)
OK You are raising the pH slowly in order to ppt larger crystals and the decomposition of the urea takes care of the excess HCl added. (You can't ppt CaC2O4 in HCl solution.)
2 answers