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-?
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)
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.)
To calculate the excess concentration of C2O42- in the solution, you need to understand the stoichiometry of the reaction between CaC2O4 and (NH4)2C2O4.
From the balanced equation:
CaC2O4*H2O → Ca2+ + C2O42-
You can see that 1 mole of CaC2O4 produces 1 mole of Ca2+ and 1 mole of C2O42-, so the mole ratio of CaC2O4 to C2O42- is 1:1.
In your solution, you have 0.06028 moles of Ca2+ from the unknown [Ca2+] added to the reaction mixture. Therefore, the number of moles of C2O42- is also 0.06028 moles.
Now, you need to determine the excess concentration of C2O42- in the solution, considering that you started with 0.15 M (NH4)2C2O4. This means that for every 1 mole of (NH4)2C2O4, you have 0.15 moles of C2O42-.
Since the mole ratio of (NH4)2C2O4 to C2O42- is 1:1, you can calculate the initial number of moles of C2O42- present in the 25 mL of 0.15 M (NH4)2C2O4 as follows:
moles of C2O42- = concentration of (NH4)2C2O4 * volume of (NH4)2C2O4
= 0.15 M * 0.025 L
= 0.00375 moles
Now, subtract the moles of C2O42- used for Ca2+ from the initial moles of C2O42- to determine the excess concentration:
Excess moles of C2O42- = initial moles of C2O42- - moles of C2O42- used for Ca2+
= 0.00375 - 0.06028
= -0.05653 moles (negative because all C2O42- was consumed)
Since the mole ratio between CaC2O4 and C2O42- is 1:1, the negative excess moles of C2O42- indicate that there was an excess of CaC2O4 in the solution, and all C2O42- reacted with the Ca2+. Therefore, the excess concentration of C2O42- is 0 M.