Question

What volume of $2N$ ${K_2}C{r_2}{O_7}$ solution is required to oxidize $0.81$ gm of ${H_2}S$ in acid medium?
A. $47.8ml$
B. $23.8ml$
C. $40ml$
D. $72ml$

Answer 1

As we know that the potassium dichromate is a chemical compound having the formula, ${K_2}C{r_2}{O_7}$. This is widely used as an oxidizing agent in industrial applications and used in various laboratories. And the hydrogen sulphide is an inorganic compound having the molecular formula, ${H_2}S$. It contains two sulphur and hydrogen bonds around the sulphur atom and the sulphur atom contains two lone pairs of electrons. And it exhibits dipole – dipole intermolecular forces.

Complete answer:
We have to know that the volume of potassium dichromate is not equal to$47.8ml$. Hence, option (A) is incorrect.
The given concentration of potassium dichromate is equal to $2N$ and weight of hydrogen sulphide is $0.81$. Here, the potassium dichromate oxidizes the hydrogen sulphide in the presence of an acidic medium. Let’s see the reaction,
${K_2}C{r_2}{O_7} + {H_2}S + {H_2}S{O_4} \to S + {K_2}S{O_4} + Cr{\left( {S{O_4}} \right)_3} + {H_2}O$
By the oxidation of ${H_2}S$, there is a formation of sulphur, potassium sulphate, chromium sulphate and water.
Here, equivalence of {K_2}C{r_2}{O_7}$$$$ = equivalence of ${H_2}S$
The equivalence can be find out by using the equation,
$equivalence = {n_f} \times no.of moles$
$= normality \times volume\left( L \right)$
$= {n_f} \times M \times V$
Hence, equivalence of {K_2}C{r_2}{O_7}$$$$ = 2 \times V
To find out the equivalence of ${H_2}S$, first we have to find out the number of moles of ${H_2}S$. So,
$n\left( {{H_2}S} \right) = \dfrac{W}{M} = \dfrac{{0.81}}{{34}}$
${n_f}$of ${H_2}S$ is equal to change in oxidation state of sulphur from ${H_2}S$ to sulphur. The oxidation number in sulphur is equal to $- 2$ and the oxidation state of sulphur is equal to zero. Hence, ${n_f}of\,{H_2}S = + 2$
Thus, equivalence of ${H_2}S = {n_f} \times no.of moles$
$= 2 \times \dfrac{{0.81}}{{34}}$
We know, equivalence of {K_2}C{r_2}{O_7}$$$$ = equivalence of ${H_2}S$
Therefore, $2 \times V = 2 \times \dfrac{{0.81}}{{34}}$
By simplification,
$V = 23.8 \times {10^{ - 3}}L = 23.8ml$
So, volume of $2N$ ${K_2}C{r_2}{O_7}$ solution is required to oxidize $0.81$ gm of ${H_2}S$ in acidic medium is equal to $23.8ml$. Hence, option (B) is correct.
The volume of $2N$ ${K_2}C{r_2}{O_7}$ to oxidize $0.81g$ of hydrogen sulphide is not equal to$40ml$. Hence, option (C) is incorrect.
The volume of $2N$ ${K_2}C{r_2}{O_7}$ solution is required to oxidize $0.81$ gm of ${H_2}S$ in acidic medium is not equal to $72ml$. Hence, option (D) is incorrect.

So, the correct answer is “Option B”.

Note:
We have to know that the required volume of potassium dichromate can be found out by using the terms, n-factor and number of moles etc. The n-factor of a compound is equal to the number of replaced hydrogen ions from the one mole of an acid in a chemical reaction. And the n-factor of an acid will not be the same as its basicity. And the number of moles of a compound can be found out by dividing given weight with its molecular mass.