Question

Enthalpy of neutralisation of acetic acid by $NaOH$ is $-50.6kJmo{l}^{-1}$. Calculate $\Delta H$ for ionisation of $C{H}_{3}COOH$. Given, the heat of neutralisation of a strong acid with a strong base is $-55.9kJmo{l}^{-1}$.

• A. $\Delta H=+5.3kJmo{l}^{-1}$
• B. $\Delta H=-5.3kJmo{l}^{-1}$
• C. $\Delta H=+10.6kJmo{l}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$\Delta H=+5.3kJmo{l}^{-1}$

The neutralisation of a strong acid by a strong base is represented by:

$H^{+}\left(aq\right)+\stackrel{\ominus }{O}H\left(aq\right)⟶H_{2}O\left(l\right)$ ; $\Delta H_1=-55.9kJ$ ......(i)

We have to calculate the enthalpy change for the ionization of acetic acid.

$C{H}_{3}COOH⟶C{H}_{3}CO{O}^{\ominus }+H^{\oplus };\Delta H=?$

Given:

the heat of neutralisation of a strong acid with a strong base

$C{H}_{3}COOH+\stackrel{\ominus }{O}H⟶C{H}_{3}CO{O}^{\ominus }+H_{2}O;\Delta H=-50.6$ ------ (ii)

Subtract enthalphy change for reaction (i) from the enthalpy change for reaction (ii) to get the enthalpy change for the ionization of acetic acid.

$\Delta H=\Delta H_2-\Delta H_1$ $=-50.6-(-55.9)$ $=5.3kJmo{l}^{-1}$

Hence, the enthalpy change for the ionisation of acetic acid is $=5.3kJmo{l}^{-1}$