Animated Solution for Chemistry - Ionic Equilibrium: The $p{K}_a$ of a weak acid (HA) is $4.5$. The pOH of an aqueous buffered solution of HA in which $50\%$ of the acid ionised is

Imagine you are a chemist working on a delicate biological reaction, perhaps synthesizing a complex protein or studying the behavior of a sensitive enzyme. You know that even the slightest change in the acidity of your solution could completely destroy your hard work, denaturing the proteins and halting the reaction. How do you protect your reaction from the chaotic fluctuations of hydrogen ions? You use a buffer solution.

Buffers are the unsung heroes of chemistry and biology. They are special solutions that stubbornly resist changes in their pH, even when you throw strong acids or strong bases at them. They achieve this chemical superpower by containing a mixture of two specific components: a weak acid and its conjugate base (or a weak base and its conjugate acid).

In our specific problem, we are dealing with a weak acid, which we will generically call $HA$. When you dissolve this weak acid in water, it does not completely break apart like a strong acid (such as hydrochloric acid) would. Instead, it only partially dissociates to form hydrogen ions ($H^{+}$) and its conjugate base ($A^{-}$). This creates a dynamic equilibrium, a continuous back-and-forth dance of molecules, which looks like this:

The problem gives us a crucial piece of information right at the beginning: the $p{K}_a$ of this weak acid is $4.5$. But what exactly is $p{K}_a$? It is the negative base-10 logarithm of the acid dissociation constant ($K_a$). The $K_a$ is a quantitative measure of the acid's strength—how much it "wants" to give up its proton. A lower $p{K}_a$ means a stronger weak acid, while a higher $p{K}_a$ means a weaker one. But more importantly for our purposes, the $p{K}_a$ serves as the central anchor point for our buffer's pH.

Now, let's look at the most interesting and revealing part of the problem statement. It says that the acid in our buffered solution is 50% ionised.

What does this physically mean at the molecular level? Imagine you start with exactly 100 molecules of the weak acid $HA$ in your beaker. If the acid is 50% ionised, it means that exactly 50 of those molecules have successfully given up their protons to the surrounding water, transforming into the conjugate base $A^{-}$. The other 50 molecules remain completely intact as the original acid $HA$.

This creates a beautiful state of chemical symmetry. Because exactly half of the acid has dissociated, the concentration of the intact acid left in the solution is perfectly equal to the concentration of the newly formed conjugate base. Mathematically, we can write this elegant relationship as:

This specific state is known in chemistry as the half-equivalence point. It is the absolute sweet spot for any buffer system. Why? Because at this point, the buffer has its maximum capacity to neutralize both added acids and added bases. It has equal amounts of its defensive components ready to react: the $A^{-}$ is ready to neutralize any invading $H^{+}$ ions, and the $HA$ is ready to neutralize any invading $O{H}^{-}$ ions.

Henderson-Hasselbalch

To find the pH of an acidic buffer, we rely on one of the most famous and widely used equations in all of chemistry: the Henderson-Hasselbalch equation. This equation elegantly connects the pH of the solution to the intrinsic $p{K}_a$ of the acid and the ratio of the concentrations of the conjugate base and the weak acid.

The equation is written as:

Let's take a moment to appreciate where this equation comes from. It is simply a logarithmic rearrangement of the standard equilibrium constant expression for the dissociation of the weak acid ($K_a=\frac{[H^{+}][A^{-}]}{[HA]}$). By taking the negative logarithm of both sides, we arrive at this incredibly useful tool.

Now, let's substitute what we know into this master equation. We know from the problem that the $p{K}_a$ is $4.5$. We also established through our logical deduction that because the acid is 50% ionised, the concentrations $[HA]$ and $[A^{-}]$ are exactly equal.

When you divide any non-zero number by itself, the result is exactly $1$. Therefore, the ratio $\frac{[A^{-}]}{[HA]}$ simplifies perfectly to $1$. Our equation now looks like this:

Here is where the math simplifies beautifully. The logarithm of $1$ in any base (whether it's base 10 or the natural base $e$) is always $0$. Why? Because any number raised to the power of $0$ equals $1$ (i.e., $10^0=1$).

So, the entire logarithmic term vanishes into thin air, leaving us with a very simple addition:

This leads us to a fundamental and highly testable rule of buffer chemistry: At the half-equivalence point (when the acid is 50% ionised), the pH of the buffer is exactly equal to the $p{K}_a$ of the weak acid.

At this point in the calculation, you might be feeling quite triumphant. You've successfully navigated the chemistry, calculated the pH, and you see the number $4.5$ sitting right there in option (a). It is incredibly tempting to just tick that box, feel good about yourself, and move on to the next question.

But wait! Stop right there. This is where the examiner has set a classic, devious trap.

Read the question carefully one more time. It does not ask for the pH of the solution. It explicitly asks for the pOH of the solution. This is a very common pitfall in highly competitive exams like the JEE. The examiners are testing not only your conceptual knowledge of ionic equilibrium but also your presence of mind, your patience, and your attention to detail under pressure.

To find the pOH, we need to recall the cosmic balance between acidity and basicity in an aqueous solution. Water itself undergoes a very slight self-ionization, producing both $H^{+}$ and $O{H}^{-}$ ions. At standard room temperature ($25^{\circ }C$ or $298K$), the product of their concentrations is a constant known as the ionic product of water ($K_w$), which is $10^{-14}$.

When we take the negative logarithm of this relationship, we get the beautiful balancing equation:

This means that as a solution becomes more acidic (pH drops), it must simultaneously become less basic (pOH rises), and vice versa. They are locked in a chemical seesaw.

We already know our pH is $4.5$. Let's substitute that value into our balancing equation:

Now, it's just a simple matter of basic arithmetic. We subtract $4.5$ from $14$ to isolate the pOH:

And there we have it! The pOH of our buffered solution is $9.5$. This matches perfectly with option (c).

By staying alert, reading the question thoroughly, and following the chemistry through to the very end, we successfully navigated the examiner's trap and arrived at the correct answer. Always remember to double-check what the question is actually asking for before you finalize your choice!