Question

For titration of \[100\text{ }mL\] of \[0.1\] M \[S{{n}^{2+}}.\] \[12.5mL\] of \[0.20\] M \[{{X}^{4+}}\] solution (where X is metal) is required. If \[S{{n}^{2+}}\] is oxidized to \[S{{n}^{4+}}.\] \[{{X}^{4+}}\] is reduced to
A.\[{{X}^{4-}}\]
B.\[{{X}^{3+}}\]
C.\[{{X}^{2+}}\]
D.\[{{X}^{2-}}\]

Answer 1

Hint: We know that while performing any titration, it is important to know the amount of substance present. This titration is the same process that lets you identify the amount of substance present and to know its proportion.

Complete answer:
When we mix any other substance to an unknown amount of one substance, it is known as titration. We can quickly identify the unknown amount of one substance with the reaction taking place. The results gained are in the form of proportion in which the two substances mix. For example when we add potassium permanganate into a conical flask containing oxalic acid, it gets discharged and the solution remains colourless. After complete consumption of oxalic acid ions, the end point is indicated by a pink colour due to excess of unreacted potassium permanganate (pink in colour).
Here we have given; \[100\text{ }mL\] of \[0.1\] M \[S{{n}^{2+}}\] and \[12.5\] mL of \[0.20\text{ }{{X}^{4+}}.\]
The reaction is given by: \[S{{n}^{2+}}\to S{{n}^{4+}}+2{{e}^{-}}\] and \[{{X}^{4+}}+n{{e}^{-}}={{X}^{\left( 4-n \right)}}.\]
Thus, the number of equivalents is same:
\[{{M}_{1}}{{V}_{1}}{{n}_{1}}={{M}_{2}}{{V}_{2}}{{n}_{2}}\] where \[{{M}_{1}}{{V}_{1}}=\] number of moles and \[n=\]valence factor.
On further substituting the value in above equation we get;
\[0.20\times 12.5\times n=100\times 0.1\times 2\]
On further solving we get;
\[2.5\times n=20\]
\[\Rightarrow n=\dfrac{20}{2.5}=8\]
The complete reduced reaction for \[{{X}^{4+}}\] is given by: \[{{X}^{4+}}+8{{e}^{-}}\to {{X}^{4-}}\]
And hence the answer is option A.

Note:
Remember that the \[S{{n}^{2+}}\] is an oxidizing agent which works in acidic medium more strongly than alkaline medium. So, for quantitative analysis potassium permanganate is generally used in acidic medium only. Its oxidizing action can be represented by the following reaction in an acidic medium.