Question

A $0.46g$ sample of $A{{s}_{2}}{{O}_{3}}$ required $25.0mL$ of $KMn{{O}_{4}}$ solution for its titration. The molarity of $KMn{{O}_{4}}$ solution is:
A. $0.016$
B. $0.074$
C. $0.032$
D. $0.128$

Answer 1

To find the required answer of the given question, remember that in the process of titration, the milli equivalents of the two different compounds used will remain the same. First find out the molecular weight of the first compound and then, use the direct formula of milli equivalent to find out the molarity of the given compound.

Complete step by step answer:
Given that,
Mass of a sample of $A{{s}_{2}}{{O}_{3}}$ is $0.46g$.
The sample further required $25mL$ of $KMn{{O}_{4}}$ solution for the titration process.
We have to find out the molarity of the $KMn{{O}_{4}}$ solution.
So, the redox changes that will occur in the following reaction of the given sample are as follows:
$A{{s}_{2}}{{O}_{3}}\to 2As{{O}_{4}}^{3-}+4{{e}^{-}}$

We can see that there is a charge change of four electrons. So, the n-factor of $A{{s}_{2}}{{O}_{3}}$ will be $4$.
Similarly, for $KMn{{O}_{4}}$ solution, the reaction will be:
$5{{e}^{-}}+MnO_{4}^{-}\to M{{n}^{2+}}$

Thus, the charge change or the change in the electrons here is of five electrons. So, the n-factor of $KMn{{O}_{4}}$ solution will be $5$.

Now, we have to find out the equivalent weight of $A{{s}_{2}}{{O}_{3}}$by using the formula:
$E\text{q}\text{. Wt}\text{.}=\dfrac{\text{Mol}\text{. wt}\text{.}}{\text{n-factor}}$

So, now the equivalent weight of $A{{s}_{2}}{{O}_{3}}$ by using the above formula will be,
Equivalent weight of $A{{s}_{2}}{{O}_{3}}$ will be $\dfrac{Mol.\text{ wt}}{n-factor}=\dfrac{2\times 75+3\times 16}{4}=\dfrac{198}{4}$

As we know, in titration the milli equivalents of two compounds that react remain the same. So, the milli equivalent of $A{{s}_{2}}{{O}_{3}}$ will be equal to that of $KMn{{O}_{4}}$ solution.
The formula of milli equivalent is given by;
$milli\text{ eq}\text{.}=\dfrac{\text{mass given}}{\text{eq}\text{. wt}\text{.}}\times {{10}^{3}}$ or it is also given by
$milli\text{ eq}\text{.= Normality}\times \text{Volume}$

So, milli equivalent of $A{{s}_{2}}{{O}_{3}}$ will be equal to $=\dfrac{0.46}{\dfrac{198}{4}}\times {{10}^{3}}=\dfrac{0.46\times 4}{198}\times {{10}^{3}}$.
And, the milli equivalent of $KMn{{O}_{4}}$ will be equal to $=\text{Volume given}\times \text{Normality}=25\times N$.

Now, from the above inference we can say that,
The milli equivalent of $A{{s}_{2}}{{O}_{3}}$ will be equal to that of $KMn{{O}_{4}}$ solution.
So, $\dfrac{0.46\times 4}{198}\times {{10}^{3}}=25\times N$

Then,
$N=\dfrac{0.46\times 4}{198\times 25}\times {{10}^{3}}=0.037N$
And, we know normality equals the product of molarity and the n-factor of that compound.
So, the molarity of $KMn{{O}_{4}}$ will be $normality\times n-factor=\dfrac{0.037}{5}=0.074M$.
So, the correct answer is “Option B”.

Note: Normality and molarity are two important and commonly used expressions in chemistry. They are used to indicate the quantitative measurement of a substance. Molarity is also used in the calculation of pH. Normality is used for more advanced calculations in establishing a one-to-one relationship between acids and bases.