The heat of neutralization of a strong base and a strong acid is $57 k J$ . The heat released when $0.5$ mole of $H N O_{3}$ solution is added to $0.20 m o l e s$ of $N a O H$ solution, is:

11.4 k J

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34.7 k J

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23.5 k J

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58.8 k J

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Solution

The correct option is A $11.4 k J$
$H^{+} + O H^{-} \to H_{2} O + H e a t$
One mole of $H^{+}$ and one mole of $O H^{-}$ give $57 k J$ . When $0.5 m o l e s$ of $H^{+}$ (from $H N O_{3})$ reacts with $0.2 m o l e$ of $O H^{-}$ (from $N a O H$ ), $0.2 m o l e$ of $H^{+}$ is neutralised by $0.2 m o l e$ of $O H^{-}$ and $0.3 m o l e$ of $H^{+}$ remains unreacted.
$H^{+} 0.2 m o l e + O H^{-} 0.2 m o l e \to H_{2} O$
Heat evolved will be $= \frac{57 \times 0.2}{1} = 11.4 k J$

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