Question

At ${25}^{0}C$ the equivalent conductance of butanoic acid at infinite dilution is 386.6 $oh{m}^{-1}c{m}^{2}e{q}^{-1}$. If the ionization constant is $1.4\times {10}^{-5}$ , calculate equivalent conductance of 0.05 N butanoic acid solution at ${25}^{o}C$ ?

• A. 3.87
• B. 6.46
• C. 6.94
• D. 4.38

Answer 1

The correct option is B 6.46

$K=1.4\times {10}^{-5}$

$HA\to H^{+}+A^{-}$

$C(a-\alpha )$ $C\alpha$ $C\alpha$

$1.4\times {10}^{-5}=\frac{(c\alpha {)}^2}{c(1-\alpha )}$

$1.4\times {10}^{-5}=\frac{0.05\times {\alpha }^2}{1-\alpha }$

$\therefore 3571.43{\alpha }^2+\alpha -1=0$

$\therefore \alpha =\frac{-1\pm \sqrt{1+4\times 3571.43}}{2\times 3571.43}$

$=0.0165$

$\alpha =\frac{{\lambda }_e^{c}q}{{\lambda }_e^{0}q}=0.016$

$0.0165=\frac{\lambda }{386.6}$

$\therefore \lambda =6.46$