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

Hint: 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.