Solveeit Logo
Login

Question

Question: Oxalic acid \({H_2}{C_2}{O_4}\) is present in many plants and vegetables. If 24.0 mL of 0.01M \(KMn{...

Oxalic acid H2C2O4{H_2}{C_2}{O_4} is present in many plants and vegetables. If 24.0 mL of 0.01M KMnO4KMn{O_4} solution is needed to titrate 1.0 g of H2C2O4{H_2}{C_2}{O_4} to equivalence point, the mass percentage (as nearest integer) of H2C2O4{H_2}{C_2}{O_4} in the sample is:

Explanation

Solution

Equivalence point: point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution. At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.

Complete Step by step answer: Mass percentage is one way of representing the concentration of an element in a compound or a component in a mixture. Mass percentage is calculated as the mass of a component divided by the total mass of the mixture, multiplied by 100%.
Number of moles KMnO4KMn{O_4} = volume1000×molarity\dfrac{{volume}}{{1000}} \times molarity
Here, we are given with volume =24.0mL and molarity =0.01.
Number of moles KMnO4KMn{O_4}= 24.01000×0.01=0.00024mol.\dfrac{{24.0}}{{1000}} \times 0.01 = 0.00024mol.
2 moles of KMnO4KMn{O_4} = 5 moles of H2C2O4{H_2}{C_2}{O_4}
So 1 moles ofKMnO4KMn{O_4} =52\dfrac{5}{2} moles of H2C2O4{H_2}{C_2}{O_4}and therefore 0.00024 moles ofKMnO4KMn{O_4} = 52×0.00024=0.0006mol\dfrac{5}{2} \times 0.00024 = 0.0006mol of H2C2O4{H_2}{C_2}{O_4}.
Number of moles of H2C2O4{H_2}{C_2}{O_4} =52×0.00024=0.0006mol\dfrac{5}{2} \times 0.00024 = 0.0006mol
Mass of H2C2O4{H_2}{C_2}{O_4} = mass=molar  mass×molesmass = molar\;mass \times moles (Molar mass of oxalic acid = 90 g/mol. And
moles = 0.0006.) mass=0.0006×90=0.054gmass = 0.0006 \times 90 = 0.054g.
Mass percent is the mass of the element or solute divided by the mass of the compound or solute. The result is multiplied by 100 to give a percent.
The formula for the amount of an element in a compound is:
mass percent = (mass of element in 1 mole of compound / mass of 1 mole of compound) x 100
Mass percentage of H2C2O4{H_2}{C_2}{O_4}= 0.0541×100=5.4%\dfrac{{0.054}}{1} \times 100 = 5.4\%

The nearest integer is 5.

Note: The main difference between equivalence and endpoint is that the equivalence point is a point where the chemical reaction comes to an end while the endpoint is the point where the color change occurs in a system.