Animated Solution for Chemistry - Ionic Equilibrium: The pH of a $0.1$ molar solution of the acid HQ is $3$. The value of the ionisation constant, $K_a$ of the acid is

The journey to mastering ionic equilibrium often begins with understanding the delicate dance of weak acids in water. When a weak acid dissolves, it doesn't completely break apart; instead, it establishes a dynamic equilibrium. In this problem, we are tasked with finding the ionization constant ($K_a$) of a weak acid, HQ, given its concentration and pH. Let's dive into the mechanics of this equilibrium and see how a simple approximation can make our lives much easier.

Every chemistry problem is a puzzle waiting to be solved, and the first step is to lay out the pieces. We are given a $0.1\text{ M}$ solution of a weak acid, HQ. This is our initial concentration, $C$. We are also given the pH of the solution, which is $3$.

The pH is a powerful piece of information because it acts as a direct window into the equilibrium state of the solution. By definition, pH is the negative logarithm of the hydrogen ion concentration. Therefore, we can easily reverse-engineer this to find the concentration of $H^{+}$ ions:

Substituting our given pH value:

This tells us exactly how many hydrogen ions are floating around in the solution once equilibrium has been reached.

Now, let's visualize what's happening at the molecular level. The weak acid HQ partially dissociates in water according to the following reversible reaction:

To keep track of the changing concentrations, we use an ICE (Initial, Change, Equilibrium) table. Initially, before any dissociation occurs, the concentration of HQ is $C$ ($0.1\text{ M}$), and the concentrations of the products, $H^{+}$ and $Q^{-}$, are zero.

As the system moves towards equilibrium, a certain amount of HQ, let's call it $x$, dissociates. This means the concentration of HQ decreases by $x$, while the concentrations of $H^{+}$ and $Q^{-}$ each increase by $x$.

At equilibrium, the concentrations are: $[HQ]=C-x$ $[H^{+}]=x$ * $[Q^{-}]=x$

From our earlier pH calculation, we already know the equilibrium concentration of $H^{+}$. This means we have found our $x$!

Because the stoichiometry of the dissociation is 1:1, the concentration of $Q^{-}$ is also equal to $x$:

Now we need to find the equilibrium concentration of the undissociated acid, HQ. According to our ICE table, this is $C-x$.

Here is where we can use a classic chemist's trick: approximation. The value $10^{-3}$ ($0.001$) is very small compared to $0.1$. If you subtract $0.001$ from $0.1$, you get $0.099$, which is practically still $0.1$.

In ionic equilibrium, if the degree of dissociation is less than $5\%$, we can safely ignore $x$ in the denominator. Here, the dissociation is only $1\%$, so the approximation is perfectly valid.

Don't rush through this. Always verify if the approximation is valid, as it saves immense calculation time during competitive exams like JEE.

The ionization constant, $K_a$, is the ratio of the product of the equilibrium concentrations of the products to the equilibrium concentration of the reactant.

Let's substitute the values we've found into this expression:

Now, it's just a matter of simple algebra. The numerator becomes $10^{-6}$, and the denominator is $10^{-1}$.

And there we have it! The ionization constant of the acid HQ is $1\times 10^{-5}$.

Ostwald's Dilution Law

While the ICE table method is fundamental and foolproof, you can also arrive at the answer using Ostwald's Dilution Law. For a weak acid where the degree of dissociation is small, the law provides a direct relationship:

If we square both sides, we get:

Substituting our values:

Both paths lead to the same elegant destination. Mastering both the foundational ICE table and the shortcut formulas will give you a significant edge in tackling ionic equilibrium problems.