Question

$50c{{m}^{3}}$ of 0.04M ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ in acidic medium oxidises a sample of ${{H}_{2}}S$ gas to sulphur. The volume of 0.03M $KMn{{O}_{4}}$ required to oxidise the same amount of ${{H}_{2}}S$ gas to sulphur, in acidic medium is:
[A] $80c{{m}^{3}}$
[B] $120c{{m}^{3}}$
[C] $60c{{m}^{3}}$
[D] $90c{{m}^{3}}$

Answer 1

Hint: To solve this, firstly equate the number of equivalents of potassium dichromate and hydrogen sulphide. Find out the weight of hydrogen sulphide from there. Then, equate the number of equivalents of hydrogen sulphide and potassium permanganate and use it to find out the required volume.

Complete step by step answer:
To solve this, firstly let us consider the $50c{{m}^{3}}$0.04M ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ sample. It oxidised hydrogen sulphide gas to sulphur.
We know that we can calculate the number of equivalents by dividing weight by equivalent weight. And to calculate equivalent weight, we have to divide molecular weight by the number of moles of equivalents per mole of that base or acid.
The molecular weight of hydrogen sulphide is 2 + 32 = 34 g/mol.
Therefore, equivalent weight of hydrogen sulphide is $\dfrac{34}{2}=17$
Here, the number of equivalents of hydrogen sulphide is equal to the number of equivalents of potassium dichromate.

Therefore, we can write that- $\dfrac{wt.of\text{ }{{\text{H}}_{2}}S}{Eq.wt\text{ of }{{\text{H}}_{2}}S}={{N}_{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}\times {{V}_{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}(in\text{ L)}$

We can write normality as the product of molarity and n-factor.
Therefore, $\dfrac{wt.of\text{ }{{\text{H}}_{2}}S}{Eq.wt\text{ of }{{\text{H}}_{2}}S}={{M}_{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}\times {{n}_{f}}\times {{V}_{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}}(in\text{ L)}$

Putting the value of volume which is 50mL i.e. 0.05L and n-factor is 6 and molarity is given to us which is 0.04and the calculated equivalent weight of hydrogen sulphide, we get-
     $\begin{align}
 & \dfrac{wt.of\text{ }{{\text{H}}_{2}}S}{17}=0.04M\times 6\times 0.05\text{L} \\
 & \text{or,}wt.of\text{ }{{\text{H}}_{2}}S=0.204 \\
\end{align}$

Now, we can find out the volume of potassium permanganate needed by equating the number of equivalents of hydrogen sulphide and number of equivalents of potassium permanganate.
The n-factor for $KMn{{O}_{4}}$ is 5 and the concentration is 0.03M.

Therefore, we can write that-
     $\begin{align}
  & \dfrac{wt.of\text{ }{{\text{H}}_{2}}S}{Eq.wt\text{ of }{{\text{H}}_{2}}S}={{M}_{{{K}_{2}}Mn{{O}_{4}}}}\times {{n}_{f}}\times {{V}_{{{K}_{2}}Mn{{O}_{4}}}}(in\text{ L)} \\
 & \text{Or, }\dfrac{0.204}{17}=0.03M\times 5\times V\text{L} \\
 & \text{or,V=0}\text{.08L=80L=80c}{{\text{m}}^{3}} \\
\end{align}$

As we can see from the above calculation that 80mL of 0.03 M $KMn{{O}_{4}}$ is required to oxidise the same amount of ${{H}_{2}}S$ gas in acidic medium.
So, the correct answer is “Option A”.

Note: We should not be confused between equivalent weight and number of equivalents. Equivalent weight is the gram molecular weight of the substance divided by the valency factor of the substance. Valency factor is basically the charge it carries whereas the number of equivalents is the weight/ mass of the compound divided by its equivalent weight.