Calculate volume of $0.4$ $M N a O H$ required to react with following mixture.
$K H C_{2} O_{4} (1 m o l e) + H_{2} C_{2} O_{4} (2 m o l)$

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Solution

$K H C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ contain one and $2$ acidic hydrogens respectively. SO the mixture of $1$ mole $K H C_{2} O_{4}$ and $2$ moles of $H_{2} C_{2} O_{4}$ contains total of $5$ mols of acidic hydrogens.
now, $1000 m l$ $0.4 M$ $N a O H = 0.4$ mole of $N a O H$
Required volume of $0.4 M$ $N a O H = \frac{5}{0.4} \times 1000 m l$
$= 12.5 \times 10^{3} m l = 12.5 l i t$ .

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