Chemistry - Electrochemistry Question with Solution | TestHub

Concepts : Ostwald's Dilution Law derivation.

Solution:

1. $K_a=\frac{C{\alpha }^2}{1-\alpha }$

2. Substitute $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$:

$$K_a=\frac{C{\left(\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}\right)}^2}{1-\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}}$$

$K_a=\frac{C{\Lambda }_m^2}{({\Lambda }_m^{\circ }{)}^{}\left({\Lambda }_m^{\circ }-{\Lambda }_m\right)}$

3. $K_a{\Lambda }_m^{\circ }({\Lambda }_m^{\circ }-{\Lambda }_m)=C{\Lambda }_m^2$

4. Rearranging:

$$\frac{1}{{\Lambda }_m}=\frac{1}{{\Lambda }_m^{\circ }}+\frac{C{\Lambda }_m}{K_a({\Lambda }_m^{\circ }{)}^2}$$

5. Plotting $\frac{1}{{\Lambda }_m}$ (y-axis) versus $C{\Lambda }_m$ (x-axis) gives a slope of $\frac{1}{K_a({\Lambda }_m^{\circ }{)}^2}$.

6. Conversely, the plot of $C{\Lambda }_m$ versus $\frac{1}{{\Lambda }_m}$ (reversing axes) gives a slope of $K_a({\Lambda }_m^{\circ }{)}^2$.

Correct Answer: Linear with slope $K_a({\Lambda }_m^{\circ }{)}^2$.