Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The ionization of a weak acid in aqueous solution can be represented as:
image
Dissociation constant of acid $= K _{ a }=\frac{\left[ H ^{+}\right]\left[ A ^{-}\right]}{[ HA ]}$
$=\frac{( c \alpha)( c \alpha)}{ c (1-\alpha)} $
$Ka =\frac{ c \alpha^{2}}{1-\alpha}$
Two solutions are said to be isohydric when the amount of dissociation of each solute is unchanged by mixing the solutions. Let two weak acids $HA _{1}$ and $HA _{2}$ are taken of concentration $c _{1}$ and $c _{2}$ respectively and their degrees of dissociation are $\alpha_{1}$ and $\alpha_{2}$ (before mixing) respectively. After mixing the two acids, let their degrees of dissociation remain unchanged but their concentration change.
image
Solutions of $HA _{1}$ and $HA _{2}$ will be isohydric when

Equilibrium

Solution:

Before mixing: $K _{ a _{1}}=\frac{ c _{1} \alpha_{1}^{2}}{\left(1-\alpha_{1}\right)}\,\,\, (i)$
$K _{ a _{2}}=\frac{ c _{2} \alpha_{2}^{2}}{\left(1-\alpha_{2}\right)}\,\,\,(ii)$
After mixing :
$K _{ a _{1}}=\frac{\left( c _{1}' \alpha_{1}+ c _{2}' \alpha_{2}\right) \alpha_{1}}{\left(1-\alpha_{1}\right)} \,\,\,(iii)$
$K _{ a _{2}}=\frac{\left( c _{1}' \alpha_{1}+ c _{2}' \alpha_{2}\right) \alpha_{2}}{\left(1-\alpha_{2}\right)}\,\,\,(iv)$
Dividing eq. (i) by (ii) gives
$\frac{ K _{ a _{1}}}{ K _{ a _{2}}}=\frac{ c _{1} \alpha_{1}^{2}\left(1-\alpha_{2}\right)}{ c _{2} \alpha_{2}^{2}\left(1-\alpha_{1}\right)}\,\,\, (v)$
Dividing eq. (iii) and iv gives
$\frac{ K _{ a _{1}}}{ K _{ a _{2}}}=\frac{\alpha_{1}\left(1-\alpha_{2}\right)}{\alpha_{2}\left(1-\alpha_{1}\right)}\,\,\, (vi)$
Comparing eq. (v) and (vi)
$\frac{\alpha_{1}\left(1-\alpha_{2}\right)}{\alpha_{2}\left(1-\alpha_{1}\right)} =\frac{c_{1} \alpha_{1}^{2}\left(1-\alpha_{2}\right)}{c_{2} \alpha_{2}^{2}\left(1-\alpha_{1}\right)} $
$c_{1} \alpha_{1} =c_{2} \alpha_{2}$
The two solutions are isohydric if the concentration of $H ^{+}$in both the solutions before mixing are same,