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  4. The heat of neutralization of NaOH with HCl is 57.3kJ and with HCN is 12.1kJ . The heat of ionization of HCN is

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Q. The heat of neutralization of $NaOH$ with $HCl$ is $57.3kJ$ and with $HCN$ is $12.1kJ$ . The heat of ionization of $HCN$ is

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Solution:

The heat of neutralization of $NaOH$ with $HCl$ is $57.3kJ$ and with $HCN$ is $12.1kJ$ .
Heat of ionisation $+$ Heat of neutralisation of $NaOH$ with $HCN=$ Heat of neutralisation of $NaOH$ with $HCl$ .
Heat of ionisation $+12.1KJ=57.3kJ$
$\therefore $ Heat of ionisation $=57.3-12.1=45.2 \, kJ$
$$