Question

The enthalpy of neutralization of a weak acid in $1M$ solution with a strong base is $-56.1kJmo{l}^{-1}$. If enthalpy of ionization of the acid is $1.5kJmo{l}^{-1}$ and enthalpy of neutralization of the strong acid with a strong base is $-57.3kJ$ equiv${}^{-1}$, what is the $\%$ ionization of the weak acid in molar solution (assume the acid to be monobasic)?

Answer 1

Answer: 20

$HA+aq\to H_{\left(aq\right)}^{+}+A_{\left(aq\right)}^{-}\Delta H=xkJ$ mol${}^{-1}$
$H_{\left(aq\right)}^{+}+O{H}_{\left(aq\right)}^{-}\to H_2{O}_{\left(\ell \right)}\Delta H=-57.3kJ$ mol${}^{-1}$
Hence, $HA+O{H}_{\left(aq\right)}^{-}\to H_2{O}_{\left(\ell \right)}+A_{\left(aq\right)}^{-}$
$\Delta H=x-57.3$
But $\Delta H=x-57.3=-56.1$ (given),
$x=1.2kJ,mo{l}^{-1}$
if no self ionization of $HA$ occurs at all,
$\Delta H$(ionization) $=1.5kJmo{l}^{-1}$
Hence, $\%$ ionization in $1M$ solution
$=\frac{\left(1.5-1.2\right)}{1.5}\times 100=20$