Question

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of $1c{m}^2$. The conductance of this solution was found to be $5\times {10}^{-7}S$. The pH of the solution is 4. The value of limiting molar conductivity $\left({\Lambda }_m^{\circ }\right)$ of this weak monobasic acid in aqueous solution is $Z\times {10}^{2}Sc{m}^{-1}mo{l}^{-1}$. The value of Z is___

Answer 1

$C=0.0015M$ $l=120cm$
$G=5\times {10}^{-7}S$ $a=1c{m}^2$
$G=\kappa \times \frac{a}{l}5\times {10}^{-7}=\kappa \times \frac{1}{120}\kappa =6\times {10}^{-5}Sc{m}^{-1}{\Lambda }_m^c=\frac{\kappa \times 1000}{M}=\frac{6\times {10}^{-5}\times 1000}{0.0015}pH=4$
$\left[H^{+}\right]={10}^{-4}=C\alpha =0.0015\alpha \alpha =\frac{{10}^{-4}}{0.0015}\alpha =\frac{{\Lambda }_m^c}{{\Lambda }_m^o}\Rightarrow \frac{{10}^{-4}}{0.0015}=\frac{\frac{6\times {10}^{-5}\times 1000}{0.0015}}{{\Lambda }_m^o}{\Lambda }_m^o=6\times {10}^{2}Sc{m}^{2}mol{e}^{-1}$