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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?

  • Chemistry -

    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.

  • Chemistry -


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