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Types of Redox Reactions

0.1 N K2C2O4....

Question

$0.1 N K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O$ solution reacts completely with $20 ml of 0.05 M$ $K M n O_{4}$ solution in acidic medium. Another sample of the same solution of $K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O$ having the same volume is titrated with $\frac{1}{8} M N a O H$ solution. Then, the volume of $N a O H$ solution (in ml) required for titration is:

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Solution

Redox titration
$Eq. of K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} . 4 H_{2} O = Eq. of K M n O_{4}$
$\frac{0.1 \times V}{1000} = \frac{20 \times 0.05 \times 5}{1000}$
$V = 50 ml$
$n$ factor of $K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O$
For redox titration $= 8$
For acid base titration $= 6$
$∴$ for acid base titration normality of
$K_{2} C_{2} O_{4} .3 H_{2} C_{2} O_{4} .4 H_{2} O = \frac{0.1}{8} \times 6 N$
Eq. of acid = Eq. of base
$\frac{0.1 \times 50}{8 \times 1000} \times 6 = \frac{1}{8} \times \frac{V ml}{1000}$
$V ml = 30 ml$

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