100mL of oxalic acid (H2C2O4) requires 35mL of 0.04M KMnO4 to titrate it to the endpoint. calculate the molarity of the oxalic acid.
0.035
0.1
Well, let's put on our chemistry clown shoes and juggle some numbers!
Step 1: Find the number of moles of KMnO4 used.
Given that the volume of KMnO4 used is 35 mL and the molarity is 0.04 M, we can use the formula Molarity (M) = Moles (mol) / Volume (L) to find the number of moles:
0.04 M = Moles / 0.035 L
Let's solve for moles:
Moles = 0.04 M * 0.035 L
Moles = 0.0014 mol of KMnO4
Step 2: Find the number of moles of oxalic acid.
The balanced equation between KMnO4 and H2C2O4 is 5H2C2O4 + 2KMnO4 + 3H2SO4 = 10CO2 + 2MnSO4 + 8H2O.
We can see that the ratio between KMnO4 and H2C2O4 is 2:5.
So, for every 2 moles of KMnO4, we have 5 moles of H2C2O4.
Using this ratio, we can calculate the number of moles of H2C2O4:
(0.0014 mol of KMnO4) * (5 mol of H2C2O4 / 2 mol of KMnO4) = 0.0035 mol of H2C2O4
Step 3: Calculate the molarity of oxalic acid.
Now, to find the molarity of oxalic acid, we divide the moles of H2C2O4 by the volume in liters:
Molarity (M) = Moles (mol) / Volume (L)
The volume given is 100 mL, which is 0.1 L:
Molarity = 0.0035 mol / 0.1 L
Molarity = 0.035 M
So, the molarity of oxalic acid (H2C2O4) is 0.035 M.
To calculate the molarity of the oxalic acid (H2C2O4), we can use the concept of stoichiometry. The balanced chemical equation between oxalic acid and potassium permanganate (KMnO4) is:
5H2C2O4 + 2KMnO4 + 3H2SO4 → 10CO2 + 2MnSO4 + 8H2O + K2SO4
From the equation, we can see that 5 moles of oxalic acid react with 2 moles of potassium permanganate. This ratio can be used to determine the moles of oxalic acid present in the solution.
Given that 35 mL of 0.04M KMnO4 is required to titrate 100 mL of oxalic acid, we can calculate the moles of potassium permanganate used:
Moles of KMnO4 = volume (in L) × concentration = 0.035 L × 0.04 mol/L = 0.0014 mol
Now, using the stoichiometric ratio, we can determine the moles of oxalic acid:
Moles of H2C2O4 = (0.0014 mol KMnO4) × (5 mol H2C2O4/2 mol KMnO4) = 0.0035 mol
Finally, we can calculate the molarity of the oxalic acid solution:
Molarity = moles of solute/volume of solution (in L)
= 0.0035 mol/ 0.1 L
= 0.035 M
Therefore, the molarity of the oxalic acid solution is 0.035 M.
5C2O4^2- + 2MnO4^- ==> 2Mn^2+ + 10CO2
I know that doesn't balanced the equation and if you wish to add the H^+ on the left and H2O on the right I recommend it; however, the redox part is balanced and that is all that is needed to work the problem. Your prof may want you to completely balance the equation.
mols KMnO4 = M x L = ?
Using the coefficients in the equation above, convert mols KMnO4 to mols H2C2O4.
Then M H2C2O4 = mols H2C2O4/L H2C2O4. You know L and mols, solve for M.