JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Equilibrium

In mixture A the base is in excess: the hydroxide supplied is $10\times 0.2\times 2=4$ mmol against $25\times 0.1=2.5$ mmol of acid, leaving $1.5$ mmol of $\mathrm{O}{\mathrm{H}}^{-}$ in $35\mathrm{mL}$, a strongly basic solution of pH near $12.6$. In mixture B the acid, $10\times 0.01\times 2=0.2$ mmol, exactly cancels the base, $10\times 0.01\times 2=0.2$ mmol, giving a neutral solution of pH $7$. In mixture C the acid is in excess: $10\times 0.1\times 2=2$ mmol of ${\mathrm{H}}^{+}$ against $10\times 0.1=1$ mmol of $\mathrm{O}{\mathrm{H}}^{-}$ leaves $1$ mmol of ${\mathrm{H}}^{+}$ in $20\mathrm{mL}$, so $[{\mathrm{H}}^{+}]=0.05\mathrm{M}$ and the pH is about $1.3$. In order of increasing pH the arrangement is $\mathrm{C}<\mathrm{B}<\mathrm{A}$.