Question

Specific conductance of 0.1M HA is $3.75\times {10}^{-4}oh{m}^{-1}c{m}^{-1}$. If ${\lambda }^{\infty }$ of HA is $250oh{m}^{-1}c{m}^{2}mo{l}^{-1}$, then dissociation constant Ka of HA is

• A. $1\times {10}^{-5}$
• B. $2.25\times {10}^{-4}$
• C. $2.25\times {10}^{-5}$
• D. $2.25\times {10}^{-13}$

Answer 1

The correct option is C $2.25\times {10}^{-5}$

${\lambda }_m=\frac{1000\times k}{m}=3.75\alpha =\frac{{\lambda }_m}{{\lambda }_m^r}=1.5\times {10}^{-2}\therefore k_a=c{\alpha }^2=2.25\times {10}^{-5}$