Question

The ionization of a weak acid in aqueous solution can be represented as:

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 $H{A}_1$ and $H{A}_2$ 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.

Solutions of $H{A}_1$ and $H{A}_2$ will be isohydric when

• A. ${\alpha }_1={\alpha }_2$
• B. $c_1=c_2$
• C. $\frac{{\alpha }_1}{{\alpha }_2}=\frac{c_1}{c_2}$
• D. $\frac{{\alpha }_2}{{\alpha }_1}=\frac{c_1}{c_2}$

Answer 1

Answer: (D)

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 :
$K_{a_1}=\frac{\left(c_1^{\prime }{\alpha }_1+c_2^{\prime }{\alpha }_2\right){\alpha }_1}{\left(1-{\alpha }_1\right)}(iii)$
$K_{a_2}=\frac{\left(c_1^{\prime }{\alpha }_1+c_2^{\prime }{\alpha }_2\right){\alpha }_2}{\left(1-{\alpha }_2\right)}(iv)$
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)
$\frac{{\alpha }_1\left(1-{\alpha }_2\right)}{{\alpha }_2\left(1-{\alpha }_1\right)}=\frac{c_1{\alpha }_1^2\left(1-{\alpha }_2\right)}{c_2{\alpha }_2^2\left(1-{\alpha }_1\right)}$
$<br/>c_1{\alpha }_1=c_2{\alpha }_2$
The two solutions are isohydric if the concentration of $H^{+}$in both the solutions before mixing are same,