JEE Main 2026 — Ionic Equilibrium Question with Solution

For mixture A:
Millimoles of OH${}^{-}$ = $10\times 0.2\times 2=4$
Millimoles of H${}^{+}$ = $25\times 0.1\times 1=2.5$
Remaining OH${}^{-}$ = $4-2.5=1.5$ mmol
The solution is basic, so pH $>7$.

For mixture B:
Millimoles of H${}^{+}$ = $10\times 0.01\times 2=0.2$
Millimoles of OH${}^{-}$ = $10\times 0.01\times 2=0.2$
The solution is completely neutralized, so pH $=7$.

For mixture C:
Millimoles of H${}^{+}$ = $10\times 0.1\times 2=2$
Millimoles of OH${}^{-}$ = $10\times 0.1\times 1=1$
Remaining H${}^{+}$ = $2-1=1$ mmol
The solution is acidic, so pH $<7$.

Therefore, the increasing order of pH values is C $<$ B $<$ A.

Answer: $C<B<A$