Question

Ionic conductance of $H^{+}$ and ${SO}_4^{2-}$ are $x$ and $yS{cm}^2{mol}^{-1}$. Hence, equivalent conductivity of $H_2{SO}_4$ is:

• A. $2x+\frac{y}{2}$
• B. $x+\frac{y}{2}$
• C. $\frac{x}{2}+y$
• D. $\frac{x}{2}+\frac{y}{2}$

Answer 1

The correct option is A $2x+\frac{y}{2}$

Equivalent conductance can be defined as the conductance produce by an equivalent of ion present in the solution.

For $H^{+}$, no.of $e^{-}$ change $=1$

For ${SO}_4^{2-}$, no.of $e^{-}$ change $=2$

Thus equivalent conductance of $H_2{SO}_4$

$H_2{SO}_4$ $\rightleftharpoons 2H^{+}+{SO}_4^{2-}$

Eg Conductance $=\frac{2\left[Ionicconductanceof{H}^{+}\right]}{nf}+\frac{IonicConductanceof{SO}_4^{2-}}{nf}$

Eg. Conductance $=2\times \frac{x}{1}s{cm}^2+\frac{y}{2}s{cm}^2$

Eg. Conductance $=2x+\frac{y}{2}$