The conductivity of a weak acid HA of concentration 0.001 mol L$$^{-1}$$ is $$2.0 \times 10^{-5}$$ S cm$$^{-1}$$. If $$\Lambda_m^0(HA) = 190$$ S cm$$^2$$ mol$$^{-1}$$, the ionization constant (K$$_a$$) of HA is equal to _________ $$\times 10^{-6}$$ (Round off to the Nearest Integer)

We have been given the specific conductivity (also called specific conductance) of the weak monoprotic acid HA as

$$\kappa = 2.0 \times 10^{-5}\ {\rm S\;cm^{-1}}$$

and its concentration as

$$c = 0.001\ {\rm mol\;L^{-1}}.$$

First we calculate the molar conductivity at this concentration. The relation between specific conductivity $$\kappa$$ and molar conductivity $$\Lambda_m$$ is stated as

$$\Lambda_m = \frac{\kappa \times 1000}{c}$$

because there are 1000 cm$$^3$$ in 1 L. Substituting the given numbers, we get

$$\Lambda_m = \frac{2.0 \times 10^{-5}\ {\rm S\;cm^{-1}} \times 1000\ {\rm cm^{3}\,L^{-1}}}{0.001\ {\rm mol\;L^{-1}}}.$$

Multiplying inside the numerator,

$$2.0 \times 10^{-5} \times 1000 = 2.0 \times 10^{-2},$$

so

$$\Lambda_m = \frac{2.0 \times 10^{-2}\ {\rm S\;cm^{2}\,mol^{-1}}}{0.001}.$$

Dividing by 0.001 is equivalent to multiplying by 1000, hence

$$\Lambda_m = 2.0 \times 10^{-2} \times 1000= 20\ {\rm S\;cm^{2}\,mol^{-1}}.$$

Next, we obtain the degree of ionization (dissociation) $$\alpha$$ for the weak acid using

$$\alpha = \frac{\Lambda_m}{\Lambda_m^{0}},$$

where $$\Lambda_m^{0}$$ is the molar conductivity at infinite dilution (given). Substituting,

$$\alpha = \frac{20}{190}.$$

Carrying out the division,

$$\alpha = 0.105263 \approx 0.1053.$$

For a weak monoprotic acid HA ⇌ H$$^{+}$$ + A$$^{-}$$, the ionization (dissociation) constant is written as

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

We now substitute every known quantity step by step:

First, compute $$\alpha^{2}$$:

$$\alpha^{2} = (0.1053)^{2} = 0.011084.$$

Multiply by the initial concentration $$c=0.001$$ mol L$$^{-1}$$:

$$c\alpha^{2} = 0.001 \times 0.011084 = 1.1084 \times 10^{-5}.$$

Next, evaluate the denominator $$1-\alpha$$:

$$1 - \alpha = 1 - 0.1053 = 0.8947.$$

Finally, divide the numerator by the denominator to obtain $$K_a$$:

$$K_a = \frac{1.1084 \times 10^{-5}}{0.8947}.$$

Carrying out the division,

$$K_a = 1.2386 \times 10^{-5}.$$

To express $$K_a$$ in the requested form “$$\times 10^{-6}$$”, write

$$K_a = 12.386 \times 10^{-6}.$$

On rounding to the nearest integer, the numerical value becomes 12.

So, the answer is $$12$$.