Chemistry

Problem

The conductivity of

20\text{ mL}

0.2\text{ N NaOH}

solution is

a\text{ S cm}^{-1}

. If

10\text{ mL}

0.4\text{ N HCl}

solution is added to this solution, the conductivity decreases by

b\text{ S cm}^{-1}

. The resultant solution has equivalent conductivity in

\text{S cm}^2\text{eq}^{-1}

is : a.

a-b

\frac{(a-b) \times 30}{4}

\frac{(a-b) \times 3 \times 10^4}{4}

\frac{(a-b) \times 4}{3}