Question

The conductance of a $0.0015M$ 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 $120cm$ 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 $\left({\Lambda }_m^o\right)$ 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

Answer: 6

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