Question

$150mL$ of $0.5N$ nitric acid solution at ${25.35}^{o}C$ was mixed with $150mL$ of $0.5N$ sodium hydroxide solution at the same temperature. The final temperature was recorded to be ${28.77}^{o}C$. Calculate the heat of neutralization of nitric acid with sodium hydroxide.

Answer 1

Total mass of solution$=150+150=300g$
$Q=$Total heat produced$=300\times (28.77-25.35)cal$
$=300\times 3.42=1026cal$
Heat of neutralisation $=\frac{Q}{150}\times 1000\times \frac{1}{0.5}$
$=\frac{1026}{150}\times 1000\times \frac{1}{0.5}=13.68kcal$
since, heat is liberated, heat of neutralisation should be negative.
So, heat of neutralisation $=-13.68kcal$.