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Question: One gram of \(N{a_3}As{O_4}\) is boiled with excess of solid \(KI\) in the presence of strong \(HCl\...

One gram of Na3AsO4N{a_3}As{O_4} is boiled with excess of solid KIKI in the presence of strong HClHCl . The iodine evolved is absorbed in the KIKI solution and titrated against the 0.2N0.2N hypo solution. Assuming the reaction to be:
AsO43−+2H++2I−→AsO33−+H2O+I2AsO_4^{3 - } + 2{H^ + } + 2{I^ - } \to AsO_3^{3 - } + {H_2}O + {I_2}
The volume of thiosulphate hypo consumed is (As=75)(As = 75) :
A. 48.1ml48.1ml
B. 38.4ml38.4ml
C. 24.7ml24.7ml
D. 30.3ml30.3ml

Explanation

Solution

This question is based upon redox reactions of physical chemistry. Before attempting this question one must have prior knowledge of normality, n-factor and all the other terms used in this topic.

Complete step by step answer:
We will solve this question by the method of n-factor.
N factor considered as a conversion factor which is calculated by dividing molar mass of substance to get equivalent mass of the substance. N factor depends on the nature of substance which changes from one condition to another condition.
The balanced chemical equation is:
Na3AsO4+2HCl+2KI→Na3AsO3+2KCl+H2O+I2N{a_3}AsO_4^{} + 2HCl + 2KI \to N{a_3}AsO_3^{} + 2KCl + {H_2}O + {I_2}
N factor is calculated by the difference in oxidation number;
N factor of Na3AsO4=5−3⇒2N{a_3}As{O_4} = 5 - 3 \Rightarrow 2
Equivalent weight of Na3AsO4N{a_3}As{O_4} is calculated using the formula-
Equivalent weight = atomic massn−factor{\text{Equivalent weight = }}\dfrac{{{\text{atomic mass}}}}{{n - factor}}
Equivalent weight of Na3AsO4=2082⇒104N{a_3}As{O_4} = \dfrac{{208}}{2} \Rightarrow 104
Equivalent weight is defined as the mass of one equivalent that is the mass of a given substance which will either combine with or displace a fixed quantity of another substance. Equivalent weight is also known as gram equivalent.
Normality is defined as the ratio of mass of a substance and equivalent weight.
Generally, normality is explained as the number of gram or mole equivalents of solute present in one litre of a solution. When we say equivalent, it is the number of moles of reactive units in a compound.
Normality of Na3AsO4=1104N{a_3}As{O_4} = \dfrac{1}{{104}}
Equation of normality:
 N1V1(Thiosulphate)=N2V2(Na3AsO4) ⇒0.2×V1=1104×1000 ⇒V1=48.1ml  \ {N_1}{V_1}(Thiosulphate) = {N_2}{V_2}(N{a_3}As{O_4}) \\\ \Rightarrow 0.2 \times {V_1} = \dfrac{1}{{104}} \times 1000 \\\ \Rightarrow {V_1} = 48.1ml \\\ \
So, the correct answer is Option A.

Note: Normality is defined as the number of equivalent weights (or simply equivalents) of solute dissolved per liter of solution. It is the gram equivalent weight of a solute per liter of solution.