Question

Heat liberated in the neutralisation of $500ml$ of $1N$ $HCl$ and $500ml$ of $1N$ ${NH}_{4}OH$ is $-1.36K.Cal$. The heat of ionization of ${NH}_{4}OH$ is:

• A. $10.98kCals$
• B. $-12.34KCals$
• C. $-10.98KCals$
• D. $12.34KCals$

Answer 1

The correct option is A $10.98kCals$

Solution:- (A) $10.98kcal$

$NaOH+HCl⟶NaCl+H_{2}O\Delta H=-13.7kcal/eq.$

As we know that heat of neutralisation of a strong acid by a strong base is heat of fomation of water.

Therefore,

$H^{+}+{OH}^{-}⟶H_{2}O\Delta H=-13.7kcal$

$H_{2}O⟶H^{+}+{OH}^{-}\Delta H=13.7kcal.....\left(1\right)$

Given that heat liberated in the neutralisation of $500mL$ of $1NHCl$ and $500mL$ of $1NN{H}_{4}OH$ is $-1.36kcal$

As we know that,

$\text{gm eq.}=N\times V_{\left(\text{in L}\right)}$

$\therefore {\text{gm eq.}}_{HCl}={gmeq.}_{N{H}_{4}OH}=1\times \frac{500}{1000}=0.5$

Now,

Heat liberated in the neutralisation of $0.5$ gram equivalent of $HCl$ and $N{H}_{4}OH=-1.36kcal$

Heat liberated in the neutralisation of $1$ gram equivalent of $HCl$ and $N{H}_{4}OH=\frac{-1.36}{0.5}=-2.72kcal$

Therefore,

$H^{+}+N{H}_{4}OH⟶{N{H}_4}^{+}+H_{2}O\Delta H=-2.72kcal.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$H_{2}O+N{H}_{4}OH+H^{+}⟶H^{+}+{OH}^{-}+{N{H}_4}^{-}+H_{2}O\Delta H=13.7+(-2.72)$

$N{H}_{4}OH⟶{N{H}_4}^{+}+{OH}^{-}\Delta H=10.98kcal$

Hence the heat of ionisation of $N{H}_{4}OH$ is $10.98kcal$.