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Common Ion Effect

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Question

A sample containing only $N a_{2} C_{2} O_{4}$ and $K H C_{2} O_{4}$ required three times the volume of $0.1 N$ $K M n O_{4}$ for titration as compared to that of $0.1 N$ base (same size sample in each case). Percentage of $K H C_{2} O_{4}$ in the mixture is:

A

65.64

%

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B

34.36

%

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C

30.28

%

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D

69.72

%

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Solution

The correct option is A $65.64$ %
The dissociation equations are as shown below:
$N a_{2} C_{2} O_{4} \to 2 N a^{+} + {C_{2} O_{4}}^{2 -}$
$K H C_{2} O_{4} \to K^{+} + H^{+} {C_{2} O_{4}}^{2 -}$
Let $x$ moles of $N a_{2} C_{2} O_{4}$ and $y$ moles of $K H C_{2} O_{4}$ be present.
Therefore, $(x + y)$ moles of ${C_{2} O_{4}}^{2 -}$ ions are present.
This corresponds to $2 (x + y)$ equivalents.
$y$ moles of hydrogen ions are present.
This corresponds to $y$ equivalents.
Let $V$ ml be the volume of $0.1 N$ base required.
$\frac{0.1 \times V}{1000} = y$

$V = 10000 y$

The volume of $0.1 N K M n O_{4}$ is $3 V$ .

$∴ \frac{0.1 \times 3 V}{1000} = 2 (x + y)$

$\frac{0.1 \times 3 \times 10000 y}{1000} = 2 (x + y)$

$3 y = 2 x + 2 y$

$y = 2 x$

The amount of $N a_{2} C_{2} O_{4} = 134 x$ g
The amount of $K H C_{2} O_{4} = y$ mol $= 2 x$ mol $= 2 \times 128 x$ g $= 256 x$ g
The percentage of $K H C_{2} O_{4}$ in the mixture is $\frac{256 x}{256 x + 134 x} \times 100 = 65.64 %$

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