Question

Excess of $KI$ and ${\text{dil}}{\text{.}}\;{H_2}S{O_4}$, were mixed in $50{\text{ml}}\,{H_2}{O_2}$. The ${I_2}$ liberated required $20ml$ of $0.1N\;N{a_2}{S_2}{O_3}$. Find out the strength of ${H_2}{O_2}$ in $g/li{t^{ - 1}}?$.
A.$0.17g/lit$
B. $0.34g/lit$
C.$0.68g/lit$
D.$1.36g/lit$

Answer 1

The volume strength ${H_2}{O_2}$ is simply the measure of oxygen at standard, temperature, and pressure given by one volume sample of hydrogen peroxide ${H_2}{O_2}$ on heating. If we write the chemical reaction between Iodide ${I_2}$ and sodium thiosulfate $N{a_2}{S_2}{O_3}$we can easily find the number of moles or equivalent ${I_2}$ which will help us in calculating the strength ${H_2}{O_2}$.

Formula Used:
Amount of ${H_2}{O_2} = \dfrac{{N \times E \times V}}{{1000}}$
Where $N$ represents normality, $E$is the equivalent mass, and $V$ is the volume.

Complete step-by-step answer: First, we will discuss the reactions used in the reaction. As the question says that excess of $KI$and ${\text{dil}}{\text{.}}\;{H_2}S{O_4}$, were mixed in $50{\text{ml}}\,{H_2}{O_2}$. So, first, we will write the reaction to understand the question properly. The reaction can be given ${H_2}{O_2} + 2KI + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O + {I_2}$. We can observe that the above reaction is balanced. Now according to the question, the amount of Iodine liberated is reacted with $N{a_2}{S_2}{O_3}$. So, the reaction can be written as $2N{a_2}{S_2}{O_3} + 2KI + {H_2}S{O_4} \to N{a_2}{S_4}{O_6} + 2NaI$. So, now we have the reactions required to solve the problem.
Now, we will calculate the equivalent mass of ${H_2}{O_2}$. To calculate equivalent mass we have ${M_{{H_2}{O_2}}} = 34g/mol,Basicity = 2$. So the equivalent mass is calculated as $E = \dfrac{{{M_{{H_2}{O_2}}}}}{{Basicity}} = \dfrac{{34}}{2} = 17$. Now, we can observe that $20ml\;0.1N\;N{a_2}{S_2}{O_3} = 20ml\;0.1N\;{I_2} = 20ml\;0.1N\;{H_2}{O_2}$. So, now we can calculate the amount of ${H_2}{O_2}$. We have $N = 0.1,V = 20ml,E = 17$. Now substitute in the formula, amount of ${H_2}{O_2} = \dfrac{{N \times E \times V}}{{1000}} = \dfrac{{0.1 \times 17 \times 20}}{{1000}} = 0.034g$. But we need the strength of ${H_2}{O_2}$ in $g/li{t^{ - 1}}.$ so, for that, we weight of ${H_2}{O_2} = 0.034,V = 50ml$. Therefore, concentration in $g/li{t^{ - 1}}$ is ${H_2}{O_2} = \dfrac{{0.034}}{{50}} \times 1000 = 0.68g/li{t^{ - 1}}$.

Therefore, the correct option is (C).

Note: Sodium thiosulfate is an inorganic compound that is available as a colorless solid. Sodium thiosulfate is also known as hypo solution. The solid is crystalline and dissolves in water. It is used for both film and photographic processing.