Question

The specific conductance of an acid HA of 0.1 M concentration is $3.75\times {10}^{-4}{\Omega }^{-1}c{m}^{-1}$. If ${\lambda }^{\infty }\left(HA\right)=250{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$. The dissociation constant $K_a$ of HA is $y\times {10}^{-5}$. Find the value of y

Answer 1

${\lambda }^{0.1M}\left(HA\right)=\frac{k\times 1000}{C}=\frac{3.75\times {10}^{-4}\times 1000}{0.1}{\Omega }^{-1}c{m}^{2}mo{l}^{-1}=3.75{\Omega }^{-1}c{m}^{2}mo{l}^{-1}\alpha =\frac{{\lambda }^{0.1M}}{{\lambda }^{\infty }}=\frac{3.75}{250}=0.015Ka=C{\alpha }^2=0.1(0.015{)}^2=2.25\times {10}^{-5}$