Question

Give that ,${H_2}{O_2}$ solution of 0.2 N is prepared and after 5 hours it is titrated with $KMn{O_4}$ solution. If the volume of the ${H_2}{O_2}$ solution taken is half of the volume of $KMn{O_4}$​ used (in acidic medium), find the molarity of $KMn{O_4}$ solution?
A. 0.4 M
B. 0.01 M
C. 0.02 M
D. 0.2 M

Answer 1

We can find molarity of $KMn{O_4}$ solution by first finding the concentration and number of moles of $KMn{O_4}$ solution. From the reaction of these two compounds, we can find how many moles of ${H_2}{O_2}$ reacts with how many moles of $KMn{O_4}$ solution.

Complete step by step answer:
In the question statement, it is mentioned that after 5 hours the ${H_2}{O_2}$ solution is titrated with $KMn{O_4}$ and the half-life of hydrogen peroxide is also 5 hours. The half-life of any species in any chemical reaction is the time when its concentration or strength reaches half of its initial concentration (at the starting of the chemical reaction). Thus, when hydrogen peroxide is titrated with $KMn{O_4}$ solution its concentration or its strength is half as that of the initial strength.
Hence, the strength of ${H_2}{O_2}$ now is half of 0.2 N i.e., 0.1 N.
Also, we know that normality is twice as its molarity, thus, we can write 0.1 N hydrogen peroxide as 0.05 M hydrogen peroxide.
When hydrogen peroxide is reacted with potassium permanganate solution, the following reaction takes place:
$2Mn{O_4}^ - + 6{H^ + } + 5{H_2}{O_2} \to 2M{n_2}^ + + 8{H_2}O + 5{O_2}$
It is clear from the above reaction that 2 moles of $KMn{O_4}$ reacts with 5 moles of ${H_2}{O_2}$.
From here we can find the number of moles of $KMn{O_4}$ that will react with 1 litre of 0.05 M ${H_2}{O_2}$ as:
$Concentration{{ }}\;of\;{{ }}{{{H}}_2}{O_2} \times \dfrac{{number{{ }}\;of{{ }}\;moles{{ }}\;of\;{{ KMn}}{{{O}}_4}}}{{number{{ }}\;of{{ }}\;moles{{ }}\;of\;{{ }}{{{H}}_2}{O_2}}}$ = $0.05 \times \dfrac{2}{5}$ = 0.02 moles.
Now, it is given in the question that the volume used of hydrogen peroxide is half to that of potassium permanganate solution. Thus, the volume used of potassium permanganate solution is 2 litres.
We can now find the molarity because molarity is the number of moles divided by the volume. So, the molarity of $KMn{O_4}$ is $\dfrac{{0.02}}{2}$ = 0.01 M

So, the correct answer is Option B.

Note: In the given the volume used of hydrogen peroxide is not given, we assumed it to be 1 litre and thus the volume used of $KMn{O_4}$ solution becomes twice as that of ${H_2}{O_2}$ solution as per the question.