40 ml of x M KMnO4 solution is required to react completely with 200 ml of 0.02 M oxalic acid solution in acidic medium. The value of x is
Answers
Answer: x = 0.04 M
Explanation:
To answer this question we must know about 3 things:
1. Normality (N) of a solution. The normality of a solution is the gram equivalent weight of a solute per liter of solution.
2. Change in Oxidation State/Number. The Charge present on a particular element when it’s is present in Ionic state.
3. Number of Gram Equivalents. The number of Gram Equivalents of a ion in a solution can be defined as the number of moles divided by the n-factor of the substance.
To calculate the change in Oxidation State
Oxidation Half Reaction: C2O42- → 2CO2 + 2e-
Reduction Half Reaction MnO41- + 8H+ + 5e- → Mn2+ + 4H2O
As we now know, the change in Oxidation State for the Carbon in Oxalic acid on conversion into Carbon di oxide is of 1 charge. For calculating the total change, we need to balance the total charge of oxidation and reduction halves, simply multiply the reduction half by 5 and the oxidation half by 5. Therefore n1 = 5 & n2 = 2
Using the formula : N1V1 = N2V2 and N = nM where n is the number of moles and M is the Molarity of the Solution.
M1 × n1V1 = M2 × n2V2
x × 5 × 40 = 0.02 × 2 × 200
x = 0.04 M
Therefore 0.04 M of KMnO4 solution is required.