Question

Given, 2mole, equimolar mixture of $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ required ${{V}_{1}}L$ of 0.1M $KMn{{O}_{4}}$ in acidic medium for complete oxidation. The same amount of the mixture required ${{V}_{2}}L$ of 0.2M NaOH for neutralization. The ratio of ${{V}_{1}}$ and ${{V}_{2}}$ is:
(A) 1:2
(B) 2:1
(C) 4:5
(D) 5:4

Answer 1

Equimolar mixture indicates that $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ are in the same proportions that is one mole each. Neutralization of this mixture is carried out using 0.1M $KMn{{O}_{4}}$ and 0.2M NaOH. We need to find out the volume of titrants required and their ratios. Start by taking into account the equivalent of each species and then find the volume of titrant.

Complete answer:
- An equimolar mixture of $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ is titrated with 0.1M $KMn{{O}_{4}}$ in acidic medium.
- Therefore, one mole of $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ and one mole of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ is present.
- Now, 1eq. of 1M $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ + 1eq. of 1M ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ = 1eq. of 0.1M $KMn{{O}_{4}}$
- An equivalent is the product of the number of moles into the number of reacting species.
- For $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ and ${{H}_{2}}{{C}_{2}}{{O}_{4}}$, number of reacting species is 2 because two sodium and protons can be donated respectively. For $KMn{{O}_{4}}$, the number of reacting species is 5 because manganese will get converted from +7 to +2 oxidation state.
- Therefore, $1\times 2+1\times 2=0.1\times 5\times {{V}_{1}}$
- Therefore, ${{V}_{1}}=8L$
- Same amount of mixture was also titrated against 0.2M NaOH.
- Therefore, 1eq. of 1M $N{{a}_{2}}{{C}_{2}}{{O}_{4}}$ + 1eq. of 1M ${{H}_{2}}{{C}_{2}}{{O}_{4}}$ = 1eq. of 0.2M NaOH
- For NaOH, the number of reacting species is 1 only.
- Therefore, $1\times 2+1\times 2=0.2\times 1\times {{V}_{2}}$
- Therefore, ${{V}_{2}}=10L$
- Hence, the ratio of the two volumes, ${{V}_{1}}:{{V}_{2}}=8:10=4:5$

Therefore, the correct option is option (C).

Note:
Remember that one equivalent of a species is equal to the number of moles of that species into the number of reacting species. For calculating, the volume of titrant just equates to the number of equivalents of analyte and titrant.