JEE Main 2025 — Ionic Equilibrium Question with Solution

An aqueous solution of HCl initially has a pH of 1.0, meaning the hydrogen ion concentration is $[{\text{H}}^{+}]=10^{-1}\text{M}$.

When we dilute the solution by adding an equal volume of water, the concentration of hydrogen ions will be halved:

$[{\text{H}}^{+}{]}_{\text{new}}=\frac{10^{-1}}{2}=5\times 10^{-2}\text{M}$

To find the new pH, we use the pH formula:

$\text{pH}=-\log [{\text{H}}^{+}]$

Substituting the new hydrogen ion concentration gives:

$\text{pH}=-\log (5\times 10^{-2})$

Using the logarithm property $\log (a\times b)=\log a+\log b$:

$\text{pH}=-(\log 5+\log 10^{-2})=-\log 5+2$

Given $\log 2=0.30$, we need $\log 5$, which can be calculated as:

$\log 10=\log (2\times 5)=\log 2+\log 5\Rightarrow \log 5=\log 10-\log 2=1-0.30=0.70$

Thus:

$\text{pH}=-(0.70)+2=1.3$

Therefore, the pH of the solution increases to 1.3.