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 $1\,cm^2$. The conductance of this solution was found to be $5\times10^{-7}S$. The $pH$ of the solution is $4$. The value of limiting molar conductivity $\left(\Lambda^{o}_{m}\right)$ of this weak monobasic acid in aqueous solution is $Z\times10^{2}S\,cm^{-1} mol^{-1}$. The value of $Z$ is

Answer 1

$\ell=$ distance between the electrodes $=120 cm$
$A =$ Area of cross-section of electrode $=1 cm ^{2}$
cell constant $=\frac{R}{A}=120 cm ^{-1}$
Conductance $=5 \times 10^{-7} S$
Conductivity $=$ Conductance $ \times$ cell constant
$=\left(5 \times 10^{-7}\right) \times(120) $
$=6 \times 10^{-5} S cm ^{-1} $
Molar conductance $=\frac{ K \times 1000}{ C }$
$=\frac{6 \times 10^{-5} \times 1000}{0.0015}=40$
$\left[ H ^{+}\right]=10^{-4}= C \alpha=0.0015 \alpha$
$\alpha=\frac{10^{-4}}{0.0015}$
$\alpha=\frac{\Lambda_{ m }^{ C }}{\Lambda_{ m }^{\infty}}$
$ \Rightarrow \Lambda_{ m }^{\infty} =\frac{\Lambda_{ m }^{ C }}{\alpha}=\frac{40 \times 0.0015}{10^{-4}} $
$=6 \times 10^{2} \,S \,cm ^{-1} \,mol ^{-1} $
$ z =6 $