JEE Main 2025 Apr 7 Shift 1, Chemistry Q5: Types Concentration And Solubility of Solutions

Initial $[{\text{H}}^{+}]=10^{-1}$ M (from pH = 1).

On adding an equal volume of water, the total volume doubles and the concentration is halved:

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

The new pH is

$\text{pH}=-\log \left(\frac{10^{-1}}{2}\right)=1+\log 2=1+0.30=1.3$

$\therefore$ pH increases to 1.3 $\Rightarrow$ Option (2).