In order to standardize a KMnO4 solution, 0.3498 g Na2C2O4 was dissolved in 30 mL water and 15 mL 3.0 M H2SO4. The KMnO4 solution was added to the Na2C2O4 solution until a pale pink color persisted. The titration took 29.5 mL of KMnO4 solution. What is the concentration of the KMnO4 solution?

To determine the concentration of the KMnO4 solution, you can use the concept of stoichiometry. The balanced equation for the reaction between KMnO4 and Na2C2O4 is:

2 KMnO4 + 5 Na2C2O4 + 8 H2SO4 → 2 MnSO4 + 10 CO2 + 8 H2O + 5 Na2SO4 + K2SO4

From the equation, we can see that two moles of KMnO4 react with five moles of Na2C2O4. Therefore, we need to calculate the number of moles of Na2C2O4 used in the titration.

First, let's calculate the molar mass of Na2C2O4:

Na2C2O4: (2 * atomic mass of Na) + (2 * atomic mass of C) + (4 * atomic mass of O)
Na2C2O4: (2 * 22.99 g/mol) + (2 * 12.01 g/mol) + (4 * 16.00 g/mol)
Na2C2O4: 61.98 g/mol + 24.02 g/mol + 64.00 g/mol
Na2C2O4: 150.00 g/mol

Next, calculate the number of moles of Na2C2O4 used:

moles of Na2C2O4 = mass / molar mass
moles of Na2C2O4 = 0.3498 g / 150.00 g/mol
moles of Na2C2O4 = 0.00233 mol

Now, let's determine the number of moles of KMnO4 used:

According to the balanced equation, the stoichiometric ratio of Na2C2O4 to KMnO4 is 5:2.
So, the number of moles of KMnO4 used = (5/2) * moles of Na2C2O4
number of moles of KMnO4 used = (5/2) * 0.00233 mol
number of moles of KMnO4 used = 0.00583 mol

Since we know the volume of the KMnO4 solution used (29.5 mL) and assuming the density of the KMnO4 solution is approximately 1 g/mL, we can calculate the molarity (concentration) of the KMnO4 solution.

Molarity (M) = moles / volume (in liters)
Molarity (M) = 0.00583 mol / 0.0295 L
Molarity (M) = 0.198 M

Therefore, the concentration of the KMnO4 solution is approximately 0.198 M.