Animated Solution for Chemistry - Ionic Equilibrium: The ${\text{OH}}^{-}$ concentration in a mixture of $5.0\text{ mL}$ of $0.0504\text{ M}$ ${\text{NH}}_4\text{Cl}$ and $2\text{ mL}$ of $0.0210\text{ M}$ ${\text{NH}}_3$ solution is $x\times 10^{-6}\text{ M}$. The value of $x$ is ......... . (Nearest integer) [Given, $K_w=1\times 10^{-14}$ and $K_b=1.8\times 10^{-5}$]

The problem of finding the pH or hydroxide ion concentration of a buffer solution is a classic staple in physical chemistry. It tests not just your ability to plug numbers into a formula, but your fundamental understanding of chemical equilibrium.

In this thrilling journey, we will dissect a mixture of ammonia and ammonium chloride. We will bypass the tedious logarithmic calculations often taught in textbooks and uncover a much more elegant, time-saving approach. Let's dive in!

Imagine you are standing in a laboratory. In one hand, you hold a beaker containing $5.0\text{ mL}$ of a $0.0504\text{ M}$ solution of ammonium chloride (${\text{NH}}_4\text{Cl}$). In your other hand, you have a test tube with $2.0\text{ mL}$ of a $0.0210\text{ M}$ solution of ammonia (${\text{NH}}_3$).

You pour them both into a single flask. What exactly have you created?

Ammonia (${\text{NH}}_3$) is a classic weak base. It doesn't dissociate completely in water; instead, it establishes a delicate equilibrium. Ammonium chloride (${\text{NH}}_4\text{Cl}$), on the other hand, is a salt derived from this weak base and a strong acid ($\text{HCl}$). In solution, it dissociates completely to provide a massive influx of ammonium ions (${\text{NH}}_4^{+}$), which is the conjugate acid of ammonia.

Whenever you have a substantial amount of a weak base and its conjugate acid coexisting in the same solution, you have created a Basic Buffer. This solution has the remarkable ability to resist drastic changes in its pH when small amounts of strong acids or bases are added.

To find the hydroxide ion concentration, $[{\text{OH}}^{-}]$, of a basic buffer, most students immediately reach for the Henderson-Hasselbalch equation:

While this equation is perfectly valid, it forces you to calculate logarithms, which can be incredibly time-consuming and prone to silly mistakes during a high-pressure exam like JEE.

Instead, let's look at the raw equilibrium expression from which the Henderson-Hasselbalch equation is derived. For the dissociation of ammonia:

The equilibrium constant is:

If we rearrange this to solve directly for $[{\text{OH}}^{-}]$, we get a much friendlier equation:

Or, more generally:

Here is the beautiful part: concentration is simply moles divided by total volume ($C=n/V$). Since both the base and the salt are swimming in the exact same total volume of the mixture, the volume term completely cancels out!

This means we don't even need to calculate the final concentrations. We just need the raw number of moles (or millimoles). This is a massive time-saver!

Let's calculate the millimoles of our two components. Millimoles are simply molarity multiplied by volume in milliliters.

First, let's look at the salt, ammonium chloride.

Next, let's calculate the millimoles of the weak base, ammonia.

We now have the exact amounts of our buffer components. Notice how we completely ignored the total volume of $7.0\text{ mL}$. It is irrelevant to our ratio!

Now, we bring back our master equation and substitute the values we just found. The problem provides the base dissociation constant, $K_b=1.8\times 10^{-5}$.

At first glance, dividing $0.042$ by $0.252$ might look intimidating without a calculator. But this is a JEE question, and the numbers are almost always rigged to cancel out beautifully.

Look closely at the significant digits: $42$ and $252$. If you multiply $40$ by $6$, you get $240$. If you multiply $2$ by $6$, you get $12$. Add them together, and $240+12=252$!

Therefore, $0.042$ goes into $0.252$ exactly $6$ times. The ratio simplifies perfectly:

Now, substitute this back into our equation:

Dividing $1.8$ by $6$ is straightforward. It gives $0.3$.

To match the format requested by the question ($x\times 10^{-6}$), we simply shift the decimal point one place to the right, which decreases the exponent by one:

The question states that the hydroxide ion concentration is $x\times 10^{-6}\text{ M}$.

By comparing our elegantly derived result with the given format, it is crystal clear:

We arrived at the exact answer without ever touching a logarithm or calculating a final concentration. By understanding the physical reality of the buffer and manipulating the equilibrium expression, we turned a potentially tedious calculation into a smooth, satisfying logical deduction.