Question

The equivalent conductance of a weak monobasic acid at infinite dilution is $100c{m}^3$ $e{q}^{-1}$ and that of its $0.01$M solution is $5c{m}^2$ $e{q}^{-1}$ at ${25}^o$C. The dissociation constant $K_a$ of the acid is:

• A. $2.5\times {10}^{-4}$
• B. $5\times {10}^{-4}$
• C. $1.25\times {10}^{-5}$
• D. $2.5\times {10}^{-5}$

Answer 1

The correct option is D $2.5\times {10}^{-5}$

The degree of dissociation, $\alpha =\frac{\text{Equivalent conductance at given dilution}}{\text{Equivalent conductance at infinite dilution}}$

$=\frac{5c{m}^{2}e{q}^{-1}}{100c{m}^{2}e{q}^{-1}}=\frac{1}{20}$

For weak monobasic acid,

$K_a=\frac{C{\alpha }^2}{1-\alpha }$

$=C{\alpha }^2(\because 1>>\alpha )$

$=0.01\times {\left(\frac{1}{20}\right)}^2=2.5\times {10}^{-5}$