Question

Conductivity of an aqueous solution of 0.1 M HX (a weak mono-protic acid) is $5\times {10}^{4}S{m}^1$.Given: ${\wedge }_m^{\infty }\left[H^{+}\right]=0.04S{m}^{2}mo{l}^{-1};{\wedge }_m^{\infty }\left[X^{-}\right]=0.01S{m}^{2}mo{l}^{-1}$
Find $p{K}_a\left[HX\right].$

Answer 1

${\Lambda }_m^{\infty }={\Lambda }_m^{\infty }\left[H^{+}\right]+{\Lambda }_m^{\infty }\left[X^{-}\right]=0.04+0.01=0.05S{m}^2/mol$
Let $\alpha$ be the degree of dissociation.
$HX\rightleftharpoons H^{+}+X^{-}$
$0.1(1-\alpha )0.1\alpha 0.1\alpha$
Hence, the molar conductuivity is
${\wedge }_m=\frac{K}{M}=\frac{5\times {10}^{-4}S{m}^{-1}}{\frac{0.1mol}{{10}^{-3}{m}^3}}=0.05S{m}^2/mol$
The relationship between the degree of dissociation and molar conductivity is
$\alpha =\frac{{\wedge }_m}{{\wedge }_m^{\infty }}=\frac{0.05}{0.05}=1$
The expression for the dissociation constant is
$K_a=\frac{0.1{\alpha }^2}{1-\alpha }={10}^{-9}$
$pKa=-logKa=-log{10}^9=9$