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Types of Redox Reactions

A mixture of ...

Question

A mixture of $H_{2} C_{2} O_{4}$ (oxalic acid) and $N a H C_{2} O_{4}$ weighing $2.02 g$ was dissolved in water and the solution made upto $1$ L. $10$ mL of the solution required $3.0$ mL of $0.1 N$ sodium hydroxide solution for complete neutralisation. In another experiment, $10.0$ mL of the same solution in hot dilute sulphuric acid medium required $4.0$ mL of $0.1$ N potassium permanganate solution for complete reaction. Calculate the amount of $H_{2} C_{2} O_{4}$ in the mixture

A

0.45

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B

0.9

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C

1.1

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D

1.8

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Solution

The correct option is C $1.1$
Let $x$ be the mass of oxalic acid in $2.02 g$ of the mixture. We will have
Molarity of oxalic acid $= \frac{n}{V} = \frac{(x / 90 g m o l^{- 1})}{1 L} = \frac{x}{90 g} m o l L^{- 1}$
Molarity of $N a H C_{2} O_{4} = \frac{(2.02 g - x) / 112 g m o l^{- 1}}{1 L} = \frac{(2.02 g - x)}{(112 g)} m o l L^{- 1}$
Amount of oxalic acid in $10 m L$ solution $= V M = (\frac{10}{1000} L) (\frac{x}{90 g} m o l L^{- 1}) = (\frac{10}{1000}) (\frac{x}{90 g}) m o l$
Amount of $N a H C_{2} O_{4}$ in $10 m L$ solution $= (\frac{10}{1000}) (\frac{2.02 g - x}{112 g}) m o l$
Now from the neutralization reactions
$H_{2} C_{2} O_{4} + 2 N a O H \to N a_{2} C_{2} O_{4} + 2 H_{2} O$ and $N a H C_{2} O + N a O H \to N a_{2} C_{2} O_{4} + H_{2} O$
We find that $1 m o l H_{2} C_{2} O_{4} \equiv 2 m o l$ and $1 m o l N a H C_{2} O_{4} \equiv 1 m o l N a O H$
Hence,
Amount of $N a O H$ equivalent to $H_{2} C_{2} O = 2 (\frac{10}{1000}) (\frac{x}{90 g}) m o l$
Amount of $N a O H$ equivalent to $N a H C_{2} O_{4} = (\frac{10}{1000}) (\frac{2.02 g - x}{112 g}) m o l$
Total amount will be $N a O H$ equivalent to $10 m L$ solution containing $H_{2} C_{2} O_{4}$ and $N a H C_{2} O_{4}$
$= [2 (\frac{10}{1000}) (\frac{x}{90 g}) + (\frac{10}{1000}) (\frac{2.02 g - x}{112 g})] m o l$
This amount will be equal to the amount of $N a O H$ present in $3.0 m L$ of $0.1 N N a O H$ solution. Hence,
$2 (\frac{10}{1000}) (\frac{x}{90 g}) + (\frac{10}{1000}) (\frac{2.02 g - x}{112 g}) = (\frac{3.0}{1000}) (0.1)$
Solving for $x$ , we get $x = 0.9 g$
Hence, in the original mixture, we have
Mass of oxalic acid $= 0.9 g$ and Mass of $N a H C_{2} O_{4} = 2.02 g - 0.9 g = 1.12 g$ .

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