Animated Solution for Chemistry - Ionic Equilibrium: The pH of a solution obtained by mixing $50\text{ mL}$ of $1\text{ M}$ HCl and $30\text{ mL}$ of $1\text{ M}$ NaOH is $x\times 10^{-4}$. The value of $x$ is ......... (Nearest integer) [$\log 2.5=0.3979$]

Calculating the pH of a Mixture

Imagine you are standing in a chemistry lab with two beakers in front of you. In the left beaker, you have a strong, corrosive acid: $50\text{ mL}$ of $1\text{ M}$ $\text{HCl}$. In the right beaker, you have a strong base: $30\text{ mL}$ of $1\text{ M}$ $\text{NaOH}$. What happens when these two chemical titans clash in a single container? This classic neutralization problem is a favorite in JEE because it tests your ability to track ions, manage volumes, and handle logarithms without breaking a sweat.

Whenever a strong acid and a strong base are mixed, a neutralization reaction occurs. The $H^{+}$ ions from the acid seek out the $O{H}^{-}$ ions from the base to form neutral water molecules. To figure out who wins this battle, we first need to calculate the "troops" each side brings to the field. In chemistry, we measure these troops in milliequivalents.

Milliequivalents are simply the product of the normality (which is molarity $\times$ n-factor) and the volume in milliliters. Since both $\text{HCl}$ and $\text{NaOH}$ have an n-factor of $1$, their normality is equal to their molarity.

Let's calculate the milliequivalents for both: For the acid ($\text{HCl}$):

For the base ($\text{NaOH}$):

Look closely at the numbers. The acid brings $50$ milliequivalents to the battle, while the base only brings $30$. This means the base is the limiting reagent. It will be completely consumed, neutralizing exactly $30$ milliequivalents of the acid.

But what about the rest? The excess acid remains victorious in the solution. Let's find out exactly how much is left:

Finding the New Concentration

Here is where many students make a silly mistake. They take the $20$ milliequivalents and immediately try to find the pH. But remember, pH depends on concentration, not just the raw amount of ions.

When we poured the two beakers together, the total volume of the liquid increased. We must account for this new, larger volume to find the true concentration of the remaining $H^{+}$ ions.

Now, we divide the excess milliequivalents by the total volume:

To make the upcoming logarithm calculation easier, let's write this in scientific notation:

We are in the endgame now. The definition of pH is the negative logarithm (base 10) of the hydrogen ion concentration.

Using the properties of logarithms ($\log (a\times b)=\log a+\log b$), we can expand this:

The problem kindly provides the value of $\log 2.5$ as $0.3979$. Let's substitute that in:

We have our pH! But wait, the question has a specific formatting trap. It asks for the answer in the form of $x\times 10^{-4}$. Let's convert our decimal to match this format:

Therefore, the integer value of $x$ is $6021$.

Always remember to read the final line of the question carefully. A brilliant derivation can be ruined by a formatting error at the very last second. Keep your units straight, always divide by the total volume, and you'll master these equilibrium problems every time.