Animated Solution for Chemistry - Ionic Equilibrium: An acidic buffer is obtained on mixing

To master buffer solutions, we must first understand their fundamental anatomy. An acidic buffer is a solution that resists changes in pH upon the addition of small amounts of acid or base.

Chemically, it is constructed from a very specific recipe: a weak acid paired with its conjugate base (usually provided in the form of a salt formed with a strong base). If a solution does not contain both of these components in significant amounts, it simply cannot function as a buffer.

When faced with a problem asking to identify a buffer, the most common mistake is to just look for a weak acid and a salt in the options and assume that's the answer. However, chemistry is dynamic! We must always evaluate the final composition of the mixture after any possible neutralization reactions have occurred.

Let's calculate the initial millimoles ($n=M\times V$) for each option to see what really happens in the beaker.

The Process of Elimination

Option (a): We mix $100\text{ mL}$ of $0.1\text{ M}$ ${\text{CH}}_3\text{COOH}$ with $100\text{ mL}$ of $0.1\text{ M}$ $\text{NaOH}$.

Calculating the millimoles, we get $10\text{ mmol}$ of the weak acid and $10\text{ mmol}$ of the strong base. Because they are present in exact equimolar amounts, they will completely neutralize each other. The final solution will only contain the salt (${\text{CH}}_3\text{COONa}$) and water. Since there is no weak acid left, this is not a buffer.

Option (b): Here, we have a mixture of $\text{HCl}$ (a strong acid) and $\text{NaCl}$ (a neutral salt).

There is absolutely no weak acid present in this system. Strong acids dissociate completely and cannot establish the equilibrium required for buffer action. Thus, this is not a buffer.

Option (c): We mix $100\text{ mL}$ of $0.1\text{ M}$ ${\text{CH}}_3\text{COOH}$ ($10\text{ mmol}$) with $200\text{ mL}$ of $0.1\text{ M}$ $\text{NaOH}$ ($20\text{ mmol}$).

In this scenario, the strong base is in excess. The $10\text{ mmol}$ of weak acid will be completely consumed by $10\text{ mmol}$ of the base, leaving behind $10\text{ mmol}$ of unreacted $\text{NaOH}$ along with the newly formed salt. A mixture of a strong base and a salt is highly alkaline and is not a buffer.

Option D Unveiled

Option (d): We mix $100\text{ mL}$ of $0.1\text{ M}$ $\text{HCl}$ ($10\text{ mmol}$) with $200\text{ mL}$ of $0.1\text{ M}$ ${\text{CH}}_3\text{COONa}$ ($20\text{ mmol}$).

At first glance, you might think, "Wait, there is a strong acid here, how can this be a buffer?" This is the classic hidden buffer trap! Let's write down the reaction:

Initially, we have $20\text{ mmol}$ of the salt and $10\text{ mmol}$ of the strong acid. Here, $\text{HCl}$ acts as the limiting reagent. It will react completely, stealing sodium ions and forcing the acetate ions to accept protons, thereby generating the weak acid.

After the reaction is complete, the $10\text{ mmol}$ of $\text{HCl}$ will have converted $10\text{ mmol}$ of ${\text{CH}}_3\text{COONa}$ into $10\text{ mmol}$ of ${\text{CH}}_3\text{COOH}$.

Let's take an inventory of our final solution. We have: 1. $10\text{ mmol}$ of unreacted ${\text{CH}}_3\text{COONa}$ (the salt). 2. $10\text{ mmol}$ of newly generated ${\text{CH}}_3\text{COOH}$ (the weak acid).

We have successfully created a solution containing both a weak acid and its conjugate base! This perfectly satisfies the condition for an acidic buffer. Therefore, option (d) is the correct answer.