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Molarity

A mixture of ...

Question

A mixture of $N a_{2} C_{2} O_{4}$ ( $A$ ) and $K H_{2} C_{2} O_{4} .2 H_{2} O$ ( $B$ ) required equal volumes of $0.1$ M $K M n O_{4}$ and $0.1$ M $N a O H$ separately. Molar ratio of $A$ and $B$ in the mixture is :

A

1 : 1

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B

1 : 5.5

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C

5.5 : 1

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D

3.1 : 1

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Solution

The correct option is C $5.5 : 1$
Let (A) $N a_{2} C_{2} O_{4}$ = x mol = x mol ${C_{2} O_{4}}^{2 -}$
(B) $K H C_{2} O_{4} . H_{2} C_{2} O_{4}$ = y mol = 3y mol $H^{+}$ and 2y mol ${C_{2} O_{4}}^{2 -}$
Total oxalate $= (x + 2 y)$
Let volume of $K M n O_{4}$ or $N a O H$ = V mL
$H^{+} + O H^{-} \to H_{2} O$
$5 {C_{2} O_{4}}^{2 -} + 2 {M n O_{4}}^{2} \to 2 M n^{2 +} + 10 C O_{2}$
$1 H^{+} = 1 O H^{-}$
3y mol $H^{+}$ = 3y mol $O H^{-}$
3y = $\frac{V \times 0.1}{1000}$ mol $N a O H$
y = $\frac{V}{30000}$
$5 {C_{2} O_{4}}^{2 -} = 2 {M n O_{4}}^{-}$
$∴$ (x+2y) mol $5 {C_{2} O_{4}}^{2 -}$ = $\frac{2 (x + 2 y)}{5}$ mol ${M n O_{4}}^{-}$
$\frac{2 (x + 2 y)}{5}$ = $\frac{V \times 0.1}{1000}$ mol $K M n O_{4}$
$\frac{x}{y} = 5.5 : 1$

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