Animated Solution for Chemistry - Electrochemistry: The conductivity of a weak acid HA of concentration $0.001{\text{ mol L}}^{-1}$ is $2.0\times 10^{-5}{\text{ S cm}}^{-1}$. If ${\Lambda }_m^{\circ }(\text{HA})=190{\text{ S cm}}^2{\text{ mol}}^{-1}$, the ionisation constant ($K_a$) of HA is equal to ............ $\times 10^{-6}$. (Round off to the nearest integer)

Imagine you are standing in a laboratory, holding a beaker filled with a weak acid, $\text{HA}$. The concentration is incredibly low, just $0.001\text{ M}$. You dip a conductivity cell into this solution and measure its conductivity, $\kappa$, which reads $2.0\times 10^{-5}{\text{ S cm}}^{-1}$.

The problem also provides the limiting molar conductivity, ${\Lambda }_m^{\circ }=190{\text{ S cm}}^2{\text{ mol}}^{-1}$. This value represents the theoretical maximum conductivity if the acid were completely dissociated into ions at infinite dilution. Our goal is to find the ionisation constant, $K_a$, which tells us exactly how weak this acid truly is.

To find $K_a$, we first need to know how much of the acid has actually dissociated. This is measured by the degree of dissociation, $\alpha$. But to find $\alpha$, we need the molar conductivity, ${\Lambda }_m$, at our specific concentration.

The bridge connecting our measured conductivity to molar conductivity is the formula:

Let's carefully substitute our known values into this equation. We plug in $\kappa =2.0\times 10^{-5}$ and $C=0.001$:

Calculating this is straightforward. The denominator $0.001$ is $10^{-3}$, so dividing by it is like multiplying by $10^3$. Combined with the $1000$ (which is another $10^3$), we get a factor of $10^6$. Multiplying $2.0\times 10^{-5}$ by $10^6$ gives us a clean, whole number:

Now that we have the molar conductivity, finding the degree of dissociation, $\alpha$, is simply a matter of comparing our current state to the theoretical maximum.

Substituting our calculated ${\Lambda }_m=20$ and the given ${\Lambda }_m^{\circ }=190$:

We will keep this as a fraction. Converting it to a decimal now would introduce rounding errors and make the upcoming algebra much messier.

Ostwald's Dilution Law

We are finally ready to calculate the ionisation constant, $K_a$. For a weak acid dissociating as $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$, Ostwald's dilution law gives us:

Let's substitute $C=0.001$ and $\alpha =\frac{2}{19}$:

Don't get intimidated by the fractions! Let's simplify the denominator first: $1-\frac{2}{19}=\frac{17}{19}$. In the numerator, squaring the fraction gives $\frac{4}{361}$.

When we divide these fractions, one of the $19$s in the denominator cancels out beautifully. We are left with:

Dividing $0.004$ by $323$ gives approximately $1.238\times 10^{-5}$. The question specifically asks for the value that multiplies $10^{-6}$, so we rewrite our answer:

Rounding 12.38 to the nearest integer gives us our final, triumphant answer: 12.