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

$k=G\times \frac{\ell }{a}k=5\times {10}^{-7}\times \frac{120cm}{1c{m}^2}=6\times {10}^{-5}Sc{m}^{-1}{\Lambda }_m^c=\frac{k\times 1000}{C}=\frac{6\times {10}^{-5}\times 1000}{0.0015}pH=4,\left[H^{+}\right]={10}^{-4}=C\alpha \alpha =\frac{{10}^{-4}}{0.0015}\alpha =\frac{{\Lambda }_m^c}{{\Lambda }_m^{\circ }}\frac{{10}^{-4}}{0.0015}=\frac{6\times {10}^{-5}\times 1000}{0.0015\times {\Lambda }_m^{\circ }}{\Lambda }_m^{\circ }=6\times {10}^{2}Z=6$