Animated Solution for Chemistry - Ionic Equilibrium: 0.01 moles of a weak acid $\text{HA}(K_a=2.0\times 10^{-6})$ is dissolved in $1.0\text{ L}$ of $0.1\text{ M HCl}$ solution. The degree of dissociation of HA is $\cdots \times 10^{-5}$ (Round off to the nearest integer). [Neglect volume change on adding HA. Assume degree of dissociation $\ll 1$]

A Tale of Two Acids

Imagine a beaker filled with $1.0\text{ L}$ of water. Into this beaker, we introduce two different characters: $0.01\text{ moles}$ of a weak acid, $\text{HA}$, and $0.1\text{ moles}$ of a strong acid, $\text{HCl}$.

Because the volume is exactly $1.0\text{ L}$, their molarities are simply equal to their moles. So, we have a solution containing $0.01\text{ M HA}$ and $0.1\text{ M HCl}$.

Now, $\text{HCl}$ is a strong acid. It doesn't hesitate; it completely shatters into ${\text{H}}^{+}$ and ${\text{Cl}}^{-}$ ions. Instantly, our beaker is flooded with $0.1\text{ M}$ of ${\text{H}}^{+}$ ions.

David vs. Goliath

Next, the weak acid $\text{HA}$ tries to dissociate:

But there's a problem. The solution is already packed with ${\text{H}}^{+}$ ions from the $\text{HCl}$. According to Le Chatelier's Principle, this massive presence of ${\text{H}}^{+}$ pushes the equilibrium of $\text{HA}$ heavily to the left. This phenomenon is known as the Common Ion Effect. The weak acid is suppressed, barely able to release any of its own ${\text{H}}^{+}$ ions.

Approximations Save the Day

Let's set up our equilibrium (ICE) table. If $\alpha$ is the degree of dissociation for $\text{HA}$, the change in concentration is $0.01\alpha$.

At equilibrium, the concentrations will be: - $[\text{HA}]=0.01-0.01\alpha =0.01(1-\alpha )$ - $[{\text{H}}^{+}]=0.1+0.01\alpha$ - $[{\text{A}}^{-}]=0.01\alpha$

Here is where we use a powerful approximation. The question explicitly tells us to assume $\alpha \ll 1$. Furthermore, the common ion effect makes $\alpha$ even smaller than it would be in pure water.

Because $\alpha$ is so tiny: - $1-\alpha \approx 1$, meaning $[\text{HA}]\approx 0.01\text{ M}$ - $0.1+0.01\alpha \approx 0.1$, meaning $[{\text{H}}^{+}]\approx 0.1\text{ M}$

Now, we plug these beautifully simplified values into the equilibrium constant expression for $\text{HA}$:

Substitute the known values:

Notice how the $0.01$ in the numerator and denominator perfectly cancel each other out! We are left with a very simple linear equation:

Solving for $\alpha$:

The question asks for the answer in the format $x\times 10^{-5}$. Comparing our result, we can clearly see that $x=2$.