Question: write the balanced equations for the titration of oxalic acid (C2H2O4) with KOH.
Work so far: I know that the overall equation is COOH)2 + 2KOH ------.> (COOK)2 + 2H2O
but I am unsure how to proceed to the two step-wise equations
To write the balanced equations for the titration of oxalic acid (C2H2O4) with KOH, we need to consider the step-wise reactions that occur.
Step 1:
The reaction between oxalic acid (C2H2O4) and potassium hydroxide (KOH) can be considered as an acid-base reaction. Each oxalic acid molecule will react with two molecules of potassium hydroxide to produce potassium oxalate (COOK)2 and water (H2O). The balanced equation for step 1 is:
C2H2O4 + 2KOH -> (COOK)2 + 2H2O
Step 2:
In the second step, the remaining oxalic acid reacts with the excess potassium hydroxide. Since one mole of oxalic acid (C2H2O4) reacts with two moles of potassium hydroxide (KOH) in step 1, we subtract the moles of oxalic acid consumed in step 1 from the initial moles of oxalic acid to obtain the moles remaining. The balanced equation for step 2 is:
C2H2O4 + KOH -> KHC2O4 + H2O
It is important to note that the second step equation only occurs when there is an excess of potassium hydroxide present.
By combining these two steps, we have the overall reaction:
C2H2O4 + 3KOH -> (COOK)2 + KHC2O4 + 2H2O