1. Use Ostwald’s dilution law for a weak monobasic acid

For a weak acid $\mathrm{HX}$, $\alpha = \frac{\Lambda_m}{\Lambda_m^0}$ where $\alpha$ is the degree of dissociation.

Also, $K_a = \frac{c\alpha^2}{1-\alpha}$ Substitute $\alpha = \dfrac{\Lambda_m}{\Lambda_m^0}$: $K_a = \frac{c\left(\dfrac{\Lambda_m}{\Lambda_m^0}\right)^2}{1-\dfrac{\Lambda_m}{\Lambda_m^0}}$

2. Simplify the expression

$K_a = \frac{c\Lambda_m^2}{(\Lambda_m^0)^2\left(1-\dfrac{\Lambda_m}{\Lambda_m^0}\right)}$ Since $1-\frac{\Lambda_m}{\Lambda_m^0} = \frac{\Lambda_m^0-\Lambda_m}{\Lambda_m^0}$ we get $K_a = \frac{c\Lambda_m^2}{\Lambda_m^0(\Lambda_m^0-\Lambda_m)}$

3. Rearrange into linear form

$K_a\Lambda_m^0(\Lambda_m^0-\Lambda_m)=c\Lambda_m^2$ Divide by $K_a\Lambda_m^0\Lambda_m$: $\frac{\Lambda_m^0-\Lambda_m}{\Lambda_m}=\frac{c\Lambda_m}{K_a\Lambda_m^0}$

So, $\frac{\Lambda_m^0}{\Lambda_m}-1 = \frac{c\Lambda_m}{K_a\Lambda_m^0}$ $\frac{\Lambda_m^0}{\Lambda_m} = 1 + \frac{c\Lambda_m}{K_a\Lambda_m^0}$ Now divide by $\Lambda_m^0$: $\frac{1}{\Lambda_m} = \frac{1}{\Lambda_m^0} + \frac{c\Lambda_m}{K_a(\Lambda_m^0)^2}$

4. Compare with straight-line form

If we plot $y = \frac{1}{\Lambda_m}$ against $x = c\Lambda_m,$ then $y = P + Sx$ where

• intercept: $P = \frac{1}{\Lambda_m^0}$
• slope: $S = \frac{1}{K_a(\Lambda_m^0)^2}$

5. Find the ratio $P/S$

$\frac{P}{S} = \frac{\dfrac{1}{\Lambda_m^0}}{\dfrac{1}{K_a(\Lambda_m^0)^2}} = K_a\Lambda_m^0$

Hence, $\boxed{\frac{P}{S} = K_a\Lambda_m^0}$

6. Option check

• A: $K_a\Lambda_m^0$ ✅
• B: $K_a\Lambda_m^0/2$ ❌
• C: $2K_a\Lambda_m^0$ ❌
• D: $1/(K_a\Lambda_m^0)$ ❌

Therefore, the correct option is A.