Help with volumetric calculation?
I'm stuck at one point. I'll show you the task first:
For alkalimetric determinations, prepare 1 liter of 0.2 mol/L sodium hydroxide solution. For titer determination, it is titrated against oxalic acid.
a) Calculate the weight of oxalic acid dehydrate if the consumption of the standard solution is to be approximately 20 mL.
So what we are looking for is m(H2C2O4 ⋅ 2 H2O).
I have already calculated the amount of NaOH.
n(NaOH) = V (NaOH) ⋅ c shall (NaOH) = 0.004 mol.
How do I find out what the amount of oxalic acid dehydrate is?
My teacher calculated it with z*, but I couldn't keep up. :/
Does anyone know more?
(1 votes)
Loading...
Oxalic acid is a divalent acid which consumes twice the amount of NaOH:
H2C2O4 + 2 OH ̄ ⟶ C2O42 ̄ + 2 H2O
Now you want to use so much oxalic acid that you approx. V=20 ml of your c≅0.2 mol/l measure solution at titration. The quantity of NaOH is then n=cV=4 mmol, so you must half as much so approx. Use 2 mmol of oxalic acid. The molar mass is M=126.065 g/mol, so you want approx. m=nM≅262 mg weigh — but please weigh sharply.
The real concentration of your dimensional solution is then to be reduced as follows: If the consumption of NaOH V is the same as that of oxalic acid when the oxalic acid is weighed in, then the amount of NaOH was equal to cV, and the amount of oxalic acid m/M is the same.