In an acid-base titration, 33.12ml of 0.1177 M H2C2O4 was titrated with 0.2146 M NaOH to the end point. What volume of the NaOH solution was used?
H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O
mols H2C2O4 = M x L = ?
mols NaOH = 2x mols H2C2O4 (look at the coefficients)
M NaOH = mols NaOH/L NaOH
You know M and mols, solve for L NaoH.
To find the volume of the NaOH solution used in the acid-base titration, you need to use the concept of stoichiometry and the balanced equation of the reaction between H2C2O4 (oxalic acid) and NaOH (sodium hydroxide).
The balanced equation for the reaction between H2C2O4 and NaOH is:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, you can see that it takes 2 moles of NaOH to react with 1 mole of H2C2O4.
First, you need to determine the number of moles of H2C2O4 in the solution. To do this, use the equation:
moles = concentration × volume
moles of H2C2O4 = 0.1177 M × 33.12 ml
moles of H2C2O4 = 0.1177 mol/L × 0.03312 L
moles of H2C2O4 = 0.003890224 mol
Using the stoichiometry of the balanced equation, you can see that 2 moles of NaOH react with 1 mole of H2C2O4. Therefore, the number of moles of NaOH used in the reaction is also 0.003890224 mol.
Finally, you can find the volume of the NaOH solution used by dividing the moles of NaOH by its concentration:
volume of NaOH = moles of NaOH / concentration of NaOH
volume of NaOH = 0.003890224 mol / 0.2146 M
volume of NaOH = 0.018083176 L
To convert the volume to milliliters:
volume of NaOH = 0.018083176 L × 1000 ml/L
volume of NaOH = 18.083176 ml
Therefore, the volume of the NaOH solution used in the titration is 18.083176 ml.