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

Hint: 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.