Question

Two mole equimolar mixture of $N{a_2}{C_2}{O_4}$ and ${H_2}{C_2}{O_4}$ required ${V_1}L$ of $0.1\,M$ $KMn{O_4}$ in acidic medium for complete oxidation. The same amount of the mixture required ${V_2}L$ of $0.2\,M$ $NaOH$ for neutralization. The ratio of ${V_1}\& {V_2}$ is.
A) $1:2$
B) $2:1$.
C) $4:5$
D) $5:4$

Answer 1

We know that the number of equivalents per mole of an ion equals the charge on the ion
And this leads to the definition of milliequivalent. Equivalent per liter is a unit of concentration $mEq/L.$
Mathematically, equivalent is represented as,
${\text{mEq = }}\dfrac{{\left( {{\text{Mass}}} \right)\left( {{\text{Valence}}} \right)}}{{{\text{Molecular}}\,{\text{weight}}}}$
Example: To find the milliequivalent of potassium in $750\,ml$ of solution contains $58.65\,mg/L$ of potassium ion and the valence of potassium is $1$.
Using the above equation,
${\text{mEq}} = \dfrac{{\left( {58.65} \right)\left( 1 \right)}}{{39.1}} = 1.5\,{\text{mEq}}$
The milliequivalent can also be calculated using the formula,
${\text{mEq = Concentration \times volume \times number of ions}}$

Complete step by step answer:
As we know that titration is the process of the addition of a solution of known concentration and volume with other solutions of unknown concentration until the reaction attains neutralization. To find the normality of the acid and base titration we can use the relation.
${M_1}{V_1} = {M_1}{V_2}$
Where,
The molarity of the acidic solution is ${M_1}$.
The volume of the acidic solution is ${V_1}$.
The molarity of the basic solution is ${M_2}$.
The volume of the basic is ${V_2}$.
In case one:
Two mole equimolar mixture of $N{a_2}{C_2}{O_4}$ and ${H_2}{C_2}{O_4}$ required ${V_1}L$ of $0.1\,M$ $KMn{O_4}$ in acidic medium.
As the mixture is equimolar, one mole of each $N{a_2}{C_2}{O_4}$ and ${H_2}{C_2}{O_4}$ are present. Thus,
$Eq\,of\,N{a_2}{C_2}{O_4} + Eq\,of\,{H_2}{C_2}{O_4} = Eq\,of\,KMn{O_4}$
$1 \times 2 + 1 \times 2 = {V_1} \times 0.1\,M \times 5$
${V_1} = 8L$
In case two:
Two mole equimolar mixture of $N{a_2}{C_2}{O_4}$ and ${H_2}{C_2}{O_4}$ required ${V_2}L$ of $0.2\,M\,NaOH$ in acidic medium.
As the mixture is equimolar, one mole of each $N{a_2}{C_2}{O_4}$ and ${H_2}{C_2}{O_4}$ are present. Thus,
$Eq\,of\,N{a_2}{C_2}{O_4} + Eq\,of\,{H_2}{C_2}{O_4} = Eq\,of\,NaOH$
$1 \times 1 + 1 \times 1 = {V_2} \times 0.2\,M \times 1$
${V_2} = 10L$
${V_1}:{V_2} = 8:10 = 4:5$
Thus, The ratio of ${V_1}\& {V_2}$ is$4:5$.
Hence,
Therefore, the correct option is A.

Note:
We must remember that the normality used in precipitation reaction is used to find the number of ions which are precipitated in a reaction.
It is also used in redox reactions to determine the number of electrons that a reducing or oxidizing agent can accept or give.
We can calculate normality from molarity using the formula,
${\text{N = Molarity X Basicity = Molarity X Acidity}}$