20.5 mL of oxalic acid,H2C2O4 were titrated with 0.800 M solution of lithium hydroxide. It took 40.0 mL of the base to reach the equivalence point. What is the molarity of the oxalic acid?
H2C2O4 + 2LiOH ==> LiC2O4 + 2H2O
moles oxalic acid = M x L.
moles LiOH = 2x moles oxalic acid (from the equation).
M LiOH = moles LiOH/L LiOH.
To determine the molarity of the oxalic acid (H2C2O4) solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between oxalic acid and lithium hydroxide (LiOH). In this reaction, each mol of oxalic acid reacts with 2 mol of lithium hydroxide in a 1:2 ratio.
The balanced chemical equation for the reaction is as follows:
H2C2O4 + 2LiOH -> Li2C2O4 + 2H2O
Given:
Volume of oxalic acid solution (V1) = 20.5 mL = 0.0205 L
Concentration of lithium hydroxide (C2) = 0.800 M
Volume of lithium hydroxide solution used (V2) = 40.0 mL = 0.0400 L
First, we need to determine the number of moles of lithium hydroxide used in the titration:
moles of LiOH (n2) = C2 * V2
Next, using the stoichiometric ratio from the balanced equation, we can find the moles of oxalic acid reacted:
moles of H2C2O4 (n1) = n2 * 1/2
Now, we can calculate the molarity of the oxalic acid solution (C1) by dividing the moles of oxalic acid by its volume:
C1 = n1 / V1
Let's substitute the given values into the equations:
moles of LiOH (n2) = 0.800 M * 0.0400 L = 0.0320 mol
moles of H2C2O4 (n1) = 0.0320 mol * 1/2 = 0.0160 mol
C1 = 0.0160 mol / 0.0205 L ≈ 0.780 M
Therefore, the molarity of the oxalic acid solution is approximately 0.780 M.