Discussion Forum

The conductance  of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinised pt electrodes.  The distance between the electrodes  is 120 cm with an area of cross section of 1 cm². The conductance of this solution was found to be 5× 10^-7 S. The pH of the solution is 4. The value of limiting molar conductivity ( $\wedge^{0}_{m}$) of this weak monobasic acid in aqueous solution is Z× 10² S cm^-1 mol^-1. The value of Z is

Answer:

Explanation:

 pH = C $\alpha =10^{-4}$

 $\Rightarrow$   $\alpha  =\frac{10^{-4}}{.0015}$

 Also, onductance (G)=  $K(\frac{A}{I})$

 $\Rightarrow$   $K=G(\frac{I}{A})=5 \times 10^{-7}\times \frac{120}{1}=6\times 10^{-5}$

  $\Rightarrow$   $\wedge^{c}=\frac{k\times 1000}{C}$

  =  $\frac{6 \times 10^{-5}\times 1000}{.0015}$

  $\Rightarrow$   $\wedge^{\infty}=\frac{\wedge^{c}}{\alpha}=\frac{6\times10^{-5}\times1000}{.0015}\times\frac{0.0015}{10^{-4}}$

        =600=6 x10²  S cm^-1 mol^-1