Question

The $pH$ of pure water at ${25}^{o}C$ and ${60}^{o}C$ are $7$ and $6.5$ respectively. $HCl$ gas is passed through water at ${25}^{o}C$ till the resulting $1$ litre solution which aquires a $pH$ of $3$. Now $4\times {10}^{-3}$ mole of $NaCN$ are added into this solution. Also a fresh $0.1M$ $HCN$ solution has $pH$ $5.1936$. Now in the one part of solution obtained after addition of $NaCN$, one milli mole of $NaOH$ are added and in the second part of this solution $0.5$ miili mole of $HCl$ are added.

The heat of formation of water from $H^{+}$ and ${OH}^{-}$ is :

• A. $13.06kcal$
• B. $-13.06kcal$
• C. $-16.32kcal$
  
• D. $16.32kcal$

Answer 1

The correct option is B $-13.06kcal$

At ${25}^{o}C$, $pH=7$

$\therefore K_w={10}^{-14}$;

At ${60}^{o}C$, $pH=6.5$

$\therefore K_w={10}^{-13}$

$H_{2}O\rightleftharpoons H^{+}+{OH}^{-};\Delta H=?$

$2.303\log \frac{{10}^{-13}}{{10}^{-14}}=\frac{\Delta H}{R}\left[\frac{35}{333\times 298}\right]$

$\therefore \Delta H=13.06kcal$

$H^{+}+{OH}^{-}⟶H_{2}O;\Delta H=-13.06kcal$