Question

$k_1$ and $k_2$ are specific conductance of the solutions $A$ and $B$ in the same conductivity cell. If equal volumes of solutions $A$ and $B$ are mixed, what will be the resistance of the mixture using the same conductivity cell, whose cell constant is $x$? (Assume there Is no change in the degree of dissociation on mixing).

• A. $R=\frac{k_1+k_2}{2x}$
• B. $R=\frac{2(k_1+k_2)}{x}$
• C. $R=2x\left(k_1+k_2\right)$
• D. $R=\frac{2x}{k_1+k_2}$

Answer 1

The correct option is D $R=\frac{2x}{k_1+k_2}$

$\text{specific conductance}=\text{conductance}\times \text{cell constant}$

Let $R_1$ and $R_2$ be the resistance of mixture A and B respectively. So, the conductance of mixture A and B are $\frac{1}{R_1}$ and $\frac{1}{R_2}$ respectively.

For mixture A we have, $R_1=\frac{x}{k_1}$ and for mixture B, $R_2=\frac{x}{k_2}$

Since, equal volumes of the solutions are mixed therefore, both the solutions get doubly diluted and their specific conductance decreases by half (as on dilution specific conductance decreases).

Thus, we have $R_1=\frac{x}{k_1/2}=\frac{2x}{k_1}$ and $R_2=\frac{x}{k_2/2}=\frac{2x}{k_2}$

Resistance of mixture (R) can be found by using,

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{k_1}{2x}+\frac{k_2}{2x}$

$\frac{1}{R}=\frac{k_1+k_2}{2x}$

Therefore, $R=\frac{2x}{k_1+k_2}$