Animated Solution for Chemistry - Ionic Equilibrium: A solution is prepared by mixing $0.01\text{ mol}$ each of ${\text{H}}_2{\text{CO}}_3$, ${\text{NaHCO}}_3$, ${\text{Na}}_2{\text{CO}}_3$, and $\text{NaOH}$ in $100\text{ mL}$ of water. pH of the resulting solution is ______. [Given : $p{K}_{a1}$ and $p{K}_{a2}$ of ${\text{H}}_2{\text{CO}}_3$ are $6.37$ and $10.32$, respectively ; $\log 2=0.30$]

Title: The Battle of the Beaker: Unraveling a Multi-Component Buffer System

Have you ever looked at a chemistry problem and felt like you were staring at a chaotic soup of molecules? Mixing four different compounds into a single beaker might seem like a recipe for disaster, but beneath the surface, there is a beautiful, logical sequence of events governed by the laws of chemical equilibrium. Let's dive into this fascinating problem and see how a strong base navigates a sea of weak acids and conjugate bases.

We start by taking inventory of our beaker. We are given $0.01\text{ mol}$ of four different substances: carbonic acid (${\text{H}}_2{\text{CO}}_3$), sodium bicarbonate (${\text{NaHCO}}_3$), sodium carbonate (${\text{Na}}_2{\text{CO}}_3$), and sodium hydroxide ($\text{NaOH}$).

To make our calculations smoother, let's convert these moles into millimoles (mmol). Since $1\text{ mol}=1000\text{ mmol}$, we have exactly $10\text{ mmol}$ of each component. They are all swimming in $100\text{ mL}$ of water.

Now, the real action begins. We have a strong base in the mix: $\text{NaOH}$. A strong base is like a highly reactive predator; it will immediately seek out the most acidic proton available to neutralize.

In our mixture, we have two potential acids: ${\text{H}}_2{\text{CO}}_3$ and ${\text{NaHCO}}_3$. Which one will $\text{NaOH}$ attack first? The answer lies in their acid strength. Carbonic acid is a stronger acid than the bicarbonate ion. Therefore, the $\text{NaOH}$ will preferentially react with ${\text{H}}_2{\text{CO}}_3$.

The reaction is a straightforward one-to-one neutralization:

Since we started with $10\text{ mmol}$ of ${\text{H}}_2{\text{CO}}_3$ and $10\text{ mmol}$ of $\text{NaOH}$, they will completely consume each other. The $\text{NaOH}$ is entirely neutralized, and the ${\text{H}}_2{\text{CO}}_3$ is completely converted into an additional $10\text{ mmol}$ of ${\text{NaHCO}}_3$.

With the reaction complete, let's look at what remains in our beaker.

The ${\text{H}}_2{\text{CO}}_3$ and $\text{NaOH}$ are gone. However, we must remember that we already had $10\text{ mmol}$ of ${\text{NaHCO}}_3$ from the very beginning. Adding the newly formed $10\text{ mmol}$, we now have a total of $20\text{ mmol}$ of ${\text{NaHCO}}_3$.

What about the ${\text{Na}}_2{\text{CO}}_3$? It was a spectator during this specific reaction, so its amount remains unchanged at $10\text{ mmol}$.

Take a close look at our final composition: we have a weak acid (${\text{HCO}}_3^{-}$ from ${\text{NaHCO}}_3$) and its conjugate base (${\text{CO}}_3^{2-}$ from ${\text{Na}}_2{\text{CO}}_3$). This is the textbook definition of a buffer solution!

To find the pH of a buffer, we rely on the elegant Henderson-Hasselbalch equation:

But wait, which $p{K}_a$ do we use? Carbonic acid is diprotic, meaning it has two $p{K}_a$ values. Since our buffer system consists of the equilibrium between ${\text{HCO}}_3^{-}$ and ${\text{CO}}_3^{2-}$, we must use the second dissociation constant, $p{K}_{a2}$, which is given as $10.32$.

Now, it's just a matter of plugging in the numbers. Because both the salt and the acid are in the same $100\text{ mL}$ volume, their concentration ratio is identical to their mole ratio.

Simplifying the fraction gives us $\frac{1}{2}$.

We are given that $\log 2=0.30$.

And there we have it! By carefully tracking the moles and understanding the hierarchy of acid-base reactions, we transformed a complex mixture into a simple buffer calculation. Always remember: in chemistry, as in life, it's crucial to take things one step at a time.