\(20 \mathrm{~mL}\) of \(\mathrm{Fe}^{2+}\) solution of certain concentration has completely reacted with…
\(20 \mathrm{~mL}\) of \(\mathrm{Fe}^{2+}\) solution of certain concentration has completely reacted with \(20 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) in acidic medium. If \(20 \mathrm{~mL}\) of same \(\mathrm{Fe}^{2+}\) solution has reacted completely with \(20 \mathrm{~mL}\) of \(\mathrm{KMnO}_4\) solution in acid medium, the molarity of \(\mathrm{KMnO}_4\) solution is
\(0.01 \mathrm{M}\)
\(0.12 \mathrm{M}\)
\(0.10 \mathrm{M}\)
\(0.012 \mathrm{M}\)
Solution
The reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)
(i.e. \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\)) in acidic medium occurs as follows :
\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}+6 \mathrm{Fe}^{3+}\)
The reaction between \(\mathrm{Fe}^{2+}\) and \(\mathrm{MnO}_4^{-}\)
(i.e. \(\mathrm{KMnO}_4\)) in acidic medium occurs as follows :
\(\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 \mathrm{Fe}^{2+} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}+5 \mathrm{Fe}^{3+}\)
\(\therefore \mathrm{Cr}_2 \mathrm{O}_7^{2-}\) react with \(6 \mathrm{Fe}^{2+} n_1\) and \(\mathrm{MnO}_4^{-}\)react with \(5 \mathrm{Fe}^{2+}\left(n_2\right)\) in a balanced equation.
Gram equivalent of \(\mathrm{Fe}^{2+}\)
\(=\) gram equivalent of \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\)
\(=\) gram equivalent of \(\mathrm{KMnO}_4\)
Therefore, on comparing gram equivalent of \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) and \(\mathrm{KMnO}_4\), we have
\(n_1=6, n_2=5\)
Conc. of \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=0.01 \mathrm{M}\left(M_1\right)\)
Volume of \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=20 \mathrm{~mL}\left(V_1\right)\)
react with \(\mathrm{Fe}^{2+}\) ions \(=6 \mathrm{~mol}\left(n_1\right)\)
Conc. of \(\mathrm{KMnO}_4=\) to find (say \(M_2\))
Volume of \(\mathrm{KMnO}_4=20 \mathrm{~mL}\left(V_2\right)\)
react with \(\mathrm{Fe}^{2+}\) ions \(=5 \mathrm{~mol}\left(n_2\right)\)
Thus, \(\quad M_1 \times V_1 \times n_1=M_2 \times V_2 \times n_2\)
or, \(0.01 \times 20 \times 6=M_2 \times 20 \times 5\)
or \(M_2=\frac{0.01 \times 6}{5}=0.012 \mathrm{M}\)
Hence, option (d) is the correct answer.