Answer:
The heat of ionization of CH₃COOH is 0.5kcal.
Explanation:
- HCl is a strong acid and NaOH is a strong base. So, when NaOH reacts with CH₃COOH which is a weak acid, energy is only required to form CH₃COO⁻ +H⁺. And the reaction is given as,
[tex]CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O[/tex] (1)
[tex]E=T-\Delta H[/tex] (2)
Where,
E=heat of neutralization of the given reaction
T=total heat of neutralization of strong base and strong acid
ΔH=heat of ionization
From the question we have,
E=13.2kcal
T=13.7kcal
By substituting the value of E and T in equation (2) we get;
[tex]13.2kcal=13.7kcal-\Delta H[/tex]
[tex]\Delta H=13.7-13.2=0.5kcal[/tex]
Hence, the heat of ionization CH₃COOH is 0.5kcal.