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100 mL of 0.2...

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100 mL of 0.2 N $H N O_{3}$ at $25^{o} C$ was mixed with 100 mL of 0.2 N $N a O H$ for neutrilization at the same temperature. The final temperature was recorded to be $26.37^{o} C$ . Calculate the heat of neutrilization (in~ $k c a l$ ) of $H N O_{3}$ with $N a O H$ .

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Solution

$Q = m s Δ t = mass of sol. (due to H_{2} O) \times sp. heat (of H_{2} O) \times 1.37$
$= 200 \times 1 \times 1.37 = 274 c a l .$
$M_{e q}$ of acid = $M_{e q}$ of base $= 100 \times 0.2 = 20$

Heat of neutrilization $= \frac{Q \times 1000}{20} = \frac{274 \times 1000}{20} = 13.7 k c a l$

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