Question

A mixture of $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ and $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ required equal volumes of $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$and $\text{ 0}\text{.2 M NaOH }$ separately for complete titration. The mole ratio of $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$and $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$in the mixture is:
(A) $\text{ }\dfrac{2}{11}\text{ }$
(B) $\text{ }\dfrac{15}{2}\text{ }$
(C) $\text{ }\dfrac{5}{2}\text{ }$
(D) $\text{ }\dfrac{7}{2}\text{ }$

Answer 1

The $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$and $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$are reacting with the $\text{ 0}\text{.2 M NaOH }$ and the $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$.Assume that the volume of the $\text{KMn}{{\text{O}}_{\text{4}}}\text{ }$and $\text{NaOH }$ reacting to the mixture be $\text{ 1 L }$ and calculate the number of moles of the $\text{KMn}{{\text{O}}_{\text{4}}}\text{ }$and$\text{NaOH }$. The reaction is given as follows,
$\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ + 3NaOH }\to \text{ }{{\text{K}}^{\text{+}}}\text{ + 2 }{{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}\text{ + 3N}{{\text{a}}^{\text{+}}}\text{ + 3}{{\text{H}}_{\text{2}}}\text{O }$
$\text{ 2KMn}{{\text{O}}_{\text{4}}}\text{+ 5N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{+ 8}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\to \text{ }{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ + 2MnS}{{\text{O}}_{\text{4}}}$

Complete step by step solution:
We have provided with the mixture $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ and the $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$.the mixture of the oxalates requires the equal volume of the$\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$ and the $\text{ 0}\text{.2 M NaOH }$ solution for the separately complete titration of the mixture.
Let's assume that the mixture of the$\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$and the $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$requires $\text{ 1 L }$ of the $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$solution and $\text{ 0}\text{.2 M NaOH }$solution separately for the complete titration of the oxalate.
We know that the mixture requires the equal volume of the $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$and the $\text{ 0}\text{.2 M NaOH }$for neutralisation, thus if we find the volume of the $\text{ 0}\text{.2 M NaOH }$then we can directly get the volume of the$\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$.
The number of moles which corresponds to the $\text{ 1 L }$of the solution calculated as,
$\begin{aligned} & \text{ Molarity (M) = }\dfrac{\text{n}}{\text{V}} \\\ & \therefore \text{ n = M }\times \text{ V} \\\ \end{aligned}$
Thus, the number of moles of $\text{ 0}\text{.2 M NaOH }$equal to,
$\text{ }{{\text{n}}_{\text{NaOH}}}\text{ = 1 }\times \text{ 0}\text{.2 = 0}\text{.2 mole of NaOH }$
The reaction between the $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$and $\text{ 0}\text{.2 M NaOH }$is as follows,
$\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ + 3NaOH }\to \text{ }{{\text{K}}^{\text{+}}}\text{ + 2 }{{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}\text{ + 3N}{{\text{a}}^{\text{+}}}\text{ + 3}{{\text{H}}_{\text{2}}}\text{O }$
1 mole of the $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$requires the 3 moles of$\text{NaOH}$.
So, $\text{0}\text{.2 mole}$ of $\text{NaOH}$ will react with the,$\text{ }\dfrac{0.2}{3}\text{ = 0}\text{.0667 moles of KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$
Similarly, assume that the $\text{ 1 L }$of the $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$reacting with the sodium oxalate.$\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$, then the number of moles associated with the $\text{ 0}\text{.2 M KMn}{{\text{O}}_{\text{4}}}\text{ }$would-be,
$\text{ }{{\text{n}}_{\text{KMn}{{\text{O}}_{\text{4}}}}}\text{ = 1 }\times \text{ 0}\text{.2 = 0}\text{.2 mole of KMn}{{\text{O}}_{\text{4}}}$
Let's consider a reaction of $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$with$\text{KMn}{{\text{O}}_{\text{4}}}\text{ }$.
$\text{ 2KMn}{{\text{O}}_{\text{4}}}\text{+ 5N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{+ 8}{{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ }\to \text{ }{{\text{K}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}\text{ + 2MnS}{{\text{O}}_{\text{4}}}$
The 2 moles of the $\text{KMn}{{\text{O}}_{\text{4}}}\text{ }$reacts with the 5 moles of the$\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$. Therefore, the $\text{0}\text{.2 mole}$of the $\text{KMn}{{\text{O}}_{\text{4}}}\text{ }$will react with the, $\text{ }\dfrac{5}{2}\text{ }\times \text{0}\text{.2 = 0}\text{.5 moles of N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$
We have to find the mole ratio of $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ the mixture of$\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$. The mole ratio would be,
$\text{ }\dfrac{\text{mole of N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }}{\text{mole of KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }}\text{ = }\dfrac{0.5}{0.0667}\text{ = }\dfrac{15}{2}$
Thus, the mole ratio of the $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ and $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$is equal to the$\text{ }\dfrac{15}{2}\text{ }$.

Hence, (B) is the correct option.

Note: Note that, the mixture contains the total x moles of $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ and each gives the total of one ${{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}$ ion and y moles of $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ and each gives two moles of ${{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}$. Thus, the mixture contains a total of three moles of${{\text{C}}_{\text{2}}}\text{O}_{4}^{2-}$. The $\text{ N}{{\text{a}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ reacts with the potassium permanganate and $\text{ KH}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{.}{{\text{H}}_{\text{2}}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\text{ }$ reacts with the sodium hydroxide and give as neutralization reaction.