Question

What is the specific conductance of 0.1 M HA solution, if ${\lambda }^{\infty }\left(HA\right)=250{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ and the dissociation constant $K_a$ of HA is $2.25\times {10}^{-5}$? (Assume $\alpha <<1)$

• A. $2.5\times {10}^{-2}{\Omega }^{-1}c{m}^{-1}$
• B. $2.5\times {10}^{-4}{\Omega }^{-1}c{m}^{-1}$
• C. $3.75\times {10}^{-4}{\Omega }^{-1}c{m}^{-1}$
• D. $3.5\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}$

Answer 1

The correct option is C $3.75\times {10}^{-4}{\Omega }^{-1}c{m}^{-1}$

$HA\rightleftharpoons H^{+}+A^{-}$
$C00$
$C-C\alpha C\alpha C\alpha$
So,
$Ka=C{\alpha }^2$, assume $\alpha <<1$
$\Rightarrow 0.1\times {\alpha }^2=2.25\times {10}^{-5}$ $\Rightarrow \alpha =1.5\times {10}^{-2}$
So, $\alpha =\frac{{\lambda }^{0.1M}}{{\lambda }^{\infty }}\Rightarrow \frac{{\lambda }^{0.1M}}{250}=0.015$ $\Rightarrow {\lambda }^{0.1M}=3.75{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$
Hence, ${\lambda }^{0.1M}\left(HA\right)=\frac{k\times 1000}{C}$ $\Rightarrow 3.75=\frac{k\times 1000}{0.1}$$\Rightarrow k=3.75\times {10}^{-4}{\Omega }^{-1}c{m}^{-1}$