posted by Chemistry on .
What volume of 0.250M KOH is required to titrate 2.30 x 10^3 mol of the weak acid H2C2O4 ?
I know the answer is 18.4, but can you show the process?
The answer is 18.4 WHAT?
H2C2O4 + 2KOH ==> K2C2O4 + 2H2O
mols H2C2O4 = 2.30E3
mols KOH needed = twice that. Look at the coefficients. 1 mol H2C2O4 = 2 mols KOH.
M KOH = moles KOH/L KOH. Therefore,
L KOH = moles KOH/M KOH. I get 18,400 L. I wonder if you made a typo in the problem with mols 2.30E-3. If so the answer would be 0.0184L or 18.4 mL.