Question

The ionization constant of a weak acid is $1.6\times {10}^{-5}$ and the molar conductivity at infinite dilution is $380\times {10}^{-4}$$S{m}^{2}mo{l}^{-1}$. If the cell constant is $0.01m^{-1}$, then conductance ($G$) of $0.01M$ acid solution is:

• A. $1.52\times {10}^{-5}S$
• B. $1.52\times {10}^{-4}S$
  
• C. $1.52\times {10}^{-3}S$
  
• D. $1.52S$

Answer 1

The correct option is D $1.52S$

By Ostwald dilution law,
$K_a=\frac{C{\alpha }^2}{1-\alpha }\Rightarrow 1.6\times {10}^{-5}$
For weak acid,
$\alpha <<1$
$K_a=C{\alpha }^2$

$\alpha =\sqrt{\frac{K_a}{C}}$

$\alpha =\sqrt{\frac{1.6\times {10}^{-5}}{0.01}}\Rightarrow 0.04$

​​​​​​$\alpha =\frac{\text{Molar conductivity at a given concentration}}{\text{Molar conductivity at a infinite dilution}}$

​​​​​​$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^0}$

${\Lambda }_m=\alpha \times {\Lambda }_m^0$

${\Lambda }_m=0.04\times 380\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

${\Lambda }_m=15.2\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

$1L={10}^{-3}{m}^3$
$1L^{-1}={10}^3{m}^{-3}$

$\text{Molar concentration,}C=0.01mol{L}^{-1}$
$\text{Molar concentration,}C=0.01\times {10}^{3}mol{m}^{-3}$

$\text{Specific conductance,}\kappa ={\Lambda }_m\times C$
$\kappa =15.2\times {10}^{-4}\times {10}^3\times {10}^{-2}$
$\kappa =15.2\times {10}^{-3}S{m}^{-1}$

We know,
$\text{Specific conductance,}\kappa =\text{Conductance}\times \text{Cell constant}$

$\text{Conductance}=\frac{\text{Specific conductance}}{\text{Cell constant}}$

$\text{Conductance}=\frac{15.2\times {10}^{-3}}{0.01}$

$\text{Conductance,}G=1.52S$