In a titration experiment,10 mL of an FeCl2 solution consumed 25 mL of a standard K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the FeCl2 solution is closest to [Given: molecular weight ofK2Cr2O7 = 294 g mol-1]

Question

In a titration experiment,10 mL of an FeCl2 solution consumed 25 mL of a standard K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the FeCl2 solution is closest to [Given: molecular weight ofK2Cr2O7 = 294 g mol-1] (A) 0.25 N (B) 0.50 N (C) 0.10 N (D) 0.04 N

Tardigrade Admin · Accepted Answer

In titration at equivalence point, number of equivalents of both reactants must be same.

∴ N_{1} V_{1} = N_{2} V_{2}

N

(normality) of

F e C l_{2}

= \frac{( N \times V ) of K _{2} C r _{2} O _{7}}{V of F e C l _{2}}

N (normality) of K_{2} C r_{2} O_{7}

= \frac{Mass \times 100}{Equivalent mass \times Volume of solution prepared}

Equivalent mass

= \frac{Molar mass}{n factor}

= \frac{294}{6}

= 49 g /

equi.

(K_{2} C r_{2} O_{7}

as oxidising agent is reduced to

2 C r^{3 +})

.
Oxidation state of

2

Cr atoms change from +

6

to +

3

.
Total change is

3 \times

2=6

, (∴ n - factor = 6)

∴ N (of K_{2} C r_{2} O_{7}) = \frac{1.225 \times 1000}{49 \times 250} = 0.1 N

∴ N (of F e C l_{2}) = \frac{0 , 1 \times 25}{10}

= 0.25 N