Question

A $100$ml of $N{a_2}{S_2}{O_3}$ solution is divided into two equal parts A and B. A part requires $12.5$ ml of $0.2$ M ${I_2}$ solution(acidic medium) and part B is diluted x times and $50$ml of diluted solution requires $5$ ml of $0.8$ M ${I_2}$ solution in basic medium. What is the value of x?

Answer 1

Hint: We will use the law of equivalence in this question.It states that whenever two substances react, the equivalents of one will be equal to the equivalent of others. We will equate the equivalent for both the reaction, find the millimoles for both the reaction and consequently find then equate it to find the value of x.

Complete Step by step answer:
We know for an equivalent concept that equivalent to one entity is equal to another entity. So applying this in the given reaction
That is equivalents of A=equivalents of B
That is moles$ \times $ n factor(A)=moles $ \times $n factor(B)
So for the first part A, the reaction is as follows
$2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI$
We will equate the milliequivalents of $N{a_2}{S_2}{O_3}$ and ${I_2}$
$\therefore {M_{eq}}(N{a_2}{S_2}{O_3}) = {M_{eq}}({I_2})$
Mole $ \times $n factor = mole$ \times $n factor
n factor of $N{a_2}{S_2}{O_3}$ = $\dfrac{{6 - 2}}{2} = \dfrac{4}{2} = 2$
n factor of $N{a_2}{S_4}{O_6}$= $\dfrac{{12 - 2}}{4} = \dfrac{{10}}{4} = 2.5$
So the difference in n factor = $2.5 - 2 - 0.5$
Now substituting in the given equation,
 $M \times V \times 1 = M \times V \times 2$
Let ${M_1}$ be the required milliequivalents for part A
${M_1} \times 50 \times {10^{ - 3}} = 12.5 \times {10^{ - 3}} \times 0.5 \times 2$
$\Rightarrow {M_1} = 0.825$
Therefore millimoles of $N{a_2}{S_2}{O_3} = 0.825$
For part B
Coming to the second part the reactions involved are as follows
${S_2}O_3^{2 - } + {I_2} \to 2SO_4^{2 - } + 2I$
Equivalents of ${S_2}O_3^{2 - }$= Equivalents of ${I_2}$
${M_2} \times 8 = 5 \times {10^{ - 3}} \times 0.8 \times 2$
$\Rightarrow {M_2} = 1$
Therefore millimoles= $1$
Now initial concentration $ \times x$=final concentration
We know that concentration=$\dfrac{M}{V}$
$\therefore \dfrac{{0.825}}{{50}} \times x = \dfrac{1}{{50}}$
$ \Rightarrow x = 1.21$
Therefore the value of x is $1.21$

Note: The n factor calculated here for the given reaction is the change in the oxidation of the different reactants calculated individually. Also, the molarity here is the moles calculated is molarity times volume. It is to be meticulously calculated for finding the milliequivalents.