Question

In the reaction of $KMn{O_4}$ with ${H_2}{C_2}{0_4}$ , 20 mL of 0.2 M $KMn{O_4}$ is not equivalent to (in acidic medium):
A) 120 mL of 0.25 M ${H_2}{C_2}{0_4}$
B)100 mL of 0.1 M ${H_2}{C_2}{0_4}$
C)500 mL of 0.2 M ${H_2}{C_2}{0_4}$
D)10 mL of 1M ${H_2}{C_2}{0_4}$

Answer 1

To solve this problem, we must first find the oxidation state of Mn in $KMn{O_4}$ . Then we must find the manganese-based products formed by the reaction between $KMn{O_4}$ and oxalic acid $({H_2}{C_2}{0_4})$ . Finally, we will find the oxidation state of Mn in the corresponding product and compare the two values to find out the equivalent weight.

Complete step by step answer:
The reaction between $KMn{O_4}$ and oxalic acid takes place in an acidic medium. To create this acidic environment, sulphuric acid is used. Hence, this reaction can be represented as:
$2KMn{O_4} + 5{H_2}{C_2}{O_4}^ + 3{H_2}S{O_4} \to 2MnS{O_4}\,\, + {K_2}S{O_4} + 10C{O_{2}} + 8{H_{2}}O.$
Now, from this reaction, it is clear that 2 moles of $KMn{O_4}$ reacts with 5 moles of ${H_2}{C_2}{0_4}$ . Therefore 1 mole of $KMn{O_4}$ reacts with $\dfrac{5}{2}$ moles of ${H_2}{C_2}{0_4}$ .
Before we proceed with the question, let us first calculate the oxidation states of manganese in potassium permanganate.
Let O. S. of Mn in $KMn{O_4}$ = x. We know that the oxidation states of Potassium = $K = + 1$ ; and that of oxygen = $O = - 2$ . Also, the net charge on the compound is zero. Hence, the oxidation state of potassium permanganate can be represented as follows:

O.S. = O.S.\left( K \right) + O.S.\left( {Mn} \right) + \left[ {O.S.\left( O \right)} \right] \times \left( 4 \right) \\\ \begin{array}{*{20}{l}} {0 = \left( { + 1} \right) + \left( x \right) + 4 \times \left( { - 2} \right)} \\\ {x = 8-1 = + 7} \end{array} \\\

Hence the oxidation state of Mn in $KMn{O_4}$ is +7.
The manganese-based product formed in the reaction is $MnS{O_4}$ . Let the oxidation state of Mn in $MnS{O_4}$ be y. we know that the oxidation state of $S{O_4}$ the molecule is $\left( { - 2} \right)$ . Also, the net charge on this compound is zero. Hence, the oxidation state of $MnS{O_4}$ being represented as:

O.S.(MnS{O_4}) = O.S.\left( {Mn} \right) + O.S.(S{O_4}) \\\ \begin{array}{*{20}{l}} {0 = y + \left( { - 2} \right)} \\\ {y = + 2} \end{array} \\\

Now, the change of oxidation state is, $+ 7 - 2 = 5$
The number of moles of. 20 mL of 0.2 M $KMn{O_4}$ is,

$$\dfrac{{20 \times 0.2}}{{1000}} \\\ = \dfrac{4}{{1000}}moles \\\$$

So as per reaction the number of moles of ${H_2}{C_2}{0_4}$ required,

$$= \dfrac{4}{{1000}}\, \times \dfrac{5}{2} \\\ = \dfrac{1}{{100}}moles \\\$$

Now, check the option for the same equivalent ${H_2}{C_2}{0_4}$ .
In the case of 120 mL of 0.25 M, ${H_2}{C_2}{0_4}$ the number of moles is

$$= \dfrac{{0.25}}{{1000}} \times 120 \\\ = \dfrac{3}{{100}}moles \\\$$

In the case of 500 mL of 0.2 M ${H_2}{C_2}{0_4}$ the number of moles,

$$= \dfrac{{0.2}}{{1000}} \times 500 \\\ = \dfrac{1}{{10}}moles \\\$$

So, option A and C are not equivalent with20 mL of 0.2 M $KMn{O_4}$
The correct options are A and C.

Note: $KMn{O_4}$ or potassium permanganate is a crystalline solid which is usually purplish in color. On the other hand, oxalic acid is an organic compound that is found in the form of a white crystalline solid. In this reaction $KMn{O_4}$ is an oxidizing reagent and oxalic acid $({H_2}{C_2}{0_4})$ is a reducing agent.