Question

The enthalpy of neutralization of a weak acid in $1 M$ solution with a strong base is $-56. 1kJ mol^{-1}$. If enthalpy of ionization of the acid is $1.5 kJ mol^{-1}$ and enthalpy of neutralization of the strong acid with a strong base is $-57.3 kJ$ equiv$^{-1}$, what is the $\%$ ionization of the weak acid in molar solution (assume the acid to be monobasic)?

Answer 1

$HA +aq \to H^{+}_{\left(aq\right)} +A^{-}_{\left(aq\right)} \Delta H = x kJ$ mol$^{-1}$
$H^{+}_{\left(aq\right)} +OH^{-}_{\left(aq\right)}\to H_{2}O_{\left(\ell\right)}\Delta H = -57.3 kJ$ mol$^{-1}$
Hence, $HA +OH^{-}_{\left(aq\right)}\to H_{2}O_{\left(\ell\right)}+A^{-}_{\left(aq\right)}$
$\Delta H =x -57.3$
But $\Delta H =x-57.3 = -56.1$ (given),
$x = 1.2\, kJ\ , mol^{-1}$
if no self ionization of $HA$ occurs at all,
$\Delta H$(ionization) $ = 1.5 \,kJ \,mol^{-1}$
Hence, $\%$ ionization in $1 M$ solution
$ = \frac{\left(1.5 -1.2\right)}{1.5}\times100 = 20$