Question

The conductance of a $0.0015\text{M}$ aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of $Pt$ electrodes. The distance between the electrodes is $120\text{cm}$ with an area of cross section of $1{\text{cm}}^2$ . The conductance of this solution was found to be $5\times {10}^{-7}\text{S}$. The $pH$ of the solution is $4$. The value of limiting molar conductivity $\left({\wedge }_m^0\right)$ of this weak monobasic acid in aqueous solution is $Z\times {10}^2{\text{S cm}}^{-1}{\text{mol}}^{-1}$ . The value of $Z$ is?

Answer 1

Let’s assume a weak monobasic acid $HA$.
It dissociates as per the equation:
$HA\left(aq\right)\rightleftharpoons H^{+}\left(aq\right)+A^{-}\left(aq\right)$
From this equation:
$[H^{+}{]}_{equilibrium}=C\times \alpha$
And as pH is given to be equal to 4
$\therefore$
$\left[H^{+}\right]={10}^{-4}M=C\times \alpha$
From the above $\alpha$ can be calculated as:
$\alpha =\frac{\left[H^{+}\right]}{\left[HA\right]}$
= $\frac{{10}^{-4}}{0.0015}$
= $\frac{1}{15}$
Also,
$\kappa =G\times \frac{l}{A}$
Putting values:
$\kappa =\frac{5\times {10}^{-7}\times 120}{1}$
= $6\times {10}^{-5}Sc{m}^{-1}$
And,
${\lambda }_M=\frac{\kappa \times 1000}{M}$
Putting values:
= $\frac{6\times {10}^{-5}\times {10}^3}{15\times {10}^{-4}}$
= $40Sc{m}^{2}mol-1$
Now, from the expression,
$\alpha =\frac{{\lambda }_M}{{\lambda }_M^o}$
Putting values:
$\frac{1}{15}=\frac{40}{Z\times {10}^2}$
$\therefore$
Z = $\frac{40\times 15}{100}$
= $6$
Hence, the correct answer is 6.