In a titration experiment,10 , mL of an FeCl_ | vedclass

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(A) At the equivalence point,the number of equivalents of $FeCl_{2}$ equals the number of equivalents of $K_{2}Cr_{2}O_{7}$. $N_{1}V_{1} = N_{2}V_{2}$

Vedclass · Accepted Answer

$0.25$

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(A) At the equivalence point,the number of equivalents of $FeCl_{2}$ equals the number of equivalents of $K_{2}Cr_{2}O_{7}$. $N_{1}V_{1} = N_{2}V_{2}$ First,calculate the normality of the $K_{2}Cr_{2}O_{7}$ solution: $n$-factor of $K_{2}Cr_{2}O_{7} = 6$ (since $Cr_{2}O_{7}^{2-} + 14H^{+} + 6e^{-} ightarrow 2Cr^{3+} + 7H_{2}O$). Equivalent mass of $K_{2}Cr_{2}O_{7} = \frac{294}{6} = 49 \, g \, eq^{-1}$. Normality $(N) = \frac{	ext{mass}}{	ext{eq. mass} 	imes 	ext{volume in } L} = \frac{1.225}{49 	imes 0.250} = 0.1 \, N$. Now,using the titration formula $N_{FeCl_{2}} 	imes V_{FeCl_{2}} = N_{K_{2}Cr_{2}O_{7}} 	imes V_{K_{2}Cr_{2}O_{7}}$: $N_{FeCl_{2}} 	imes 10 \, mL = 0.1 \, N 	imes 25 \, mL$. $N_{FeCl_{2}} = \frac{0.1 	imes 25}{10} = 0.25 \, N$.

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