Animated Solution for Chemistry - Ionic Equilibrium: 20 mL of 0.1 M ${\text{H}}_2{\text{SO}}_4$ solution is added to 30 mL of 0.2 M ${\text{NH}}_4\text{OH}$ solution. The pH of the resultant mixture is [${\text{pK}}_b$ of ${\text{NH}}_4\text{OH}=4.7$]

Mixing the Potions

Imagine you are standing in a laboratory with two flasks in your hands.

In one hand, you hold $20\text{ mL}$ of a $0.1\text{ M}$ solution of sulfuric acid (${\text{H}}_2{\text{SO}}_4$). In the other, you have $30\text{ mL}$ of a $0.2\text{ M}$ solution of ammonium hydroxide (${\text{NH}}_4\text{OH}$).

You pour them both into a single beaker. What exactly happens in that mixture? To find out, we first need to know exactly how many particles of each reactant we just threw into the arena.

We calculate the initial millimoles by simply multiplying the volume by the molarity.

We have $2\text{ mmol}$ of the strong acid and $6\text{ mmol}$ of the weak base. Let the chemical clash begin!

Stoichiometry in Action

Before we jump to conclusions, we must consult the balanced chemical equation. It is the ultimate rulebook for any reaction.

Notice the stoichiometry here. One mole of sulfuric acid requires exactly two moles of ammonium hydroxide to fully neutralize.

Since we only have $2\text{ mmol}$ of acid, it will consume exactly $4\text{ mmol}$ of the base ($2\times 2=4$).

The acid is our limiting reagent. It gets completely wiped out!

Birth of a Buffer

When the dust settles, what is left in our beaker?

The acid is gone ($0\text{ mmol}$). The base was in excess, so we have some left over: $6-4=2\text{ mmol}$ of unreacted ${\text{NH}}_4\text{OH}$. And we have newly formed salt: $2\text{ mmol}$ of $({\text{NH}}_4{)}_2{\text{SO}}_4$.

Take a close look at this mixture. We have a weak base and its conjugate salt coexisting in the same solution.

This is the classic, textbook definition of a Basic Buffer!

Henderson-Hasselbalch

To find the pH of a buffer, we summon the mighty Henderson-Hasselbalch equation. Since this is a basic buffer, we first calculate the pOH.

We are given the ${\text{pK}}_b$ of ammonium hydroxide as $4.7$. We also know we have $2\text{ mmol}$ of salt and $2\text{ mmol}$ of base. Let's substitute these values.

Since the logarithm of $1$ is exactly $0$, the equation simplifies beautifully.

A Trap to Avoid

We have our pOH, but the question specifically asks for the pH of the resultant mixture.

This is where many students make a heartbreaking silly mistake and lose marks. Always remember the fundamental relationship at room temperature:

Let's find our final answer.

The pH of our buffer solution is $9.3$.

Salt vs. Conjugate Ion

There is a fascinating catch in this problem that separates good students from great ones.

Strictly speaking, the Henderson-Hasselbalch equation uses the concentration of the conjugate acid ion, not just the salt molecule.

Because each molecule of $({\text{NH}}_4{)}_2{\text{SO}}_4$ releases two ${\text{NH}}_4^{+}$ ions, the actual millimoles of the conjugate acid is $4\text{ mmol}$.

If we used $4\text{ mmol}$ in our equation, the pH would calculate to $9.0$.

However, many standard textbook solutions and exam keys (including this specific JEE official answer) use the salt concentration directly as a simplified convention.

Always be aware of this dual convention. If $9.0$ was not an option, or if the key prefers the simplified formula, you must adapt and conquer!