Arrange the following resultant mixtures in increasing order of their pH values A. 10 mL 0.2 M Ca ( OH ) 2 +25 mL 0.1 M HCl B. 10 mL 0.01 M H 2 SO 4 +10 mL 0.01 M Ca ( OH ) 2 C. 10 mL 0.1 M H 2 SO 4 +10 mL 0.1 M KOH Choose the correct answer from the options given below :

Chemistry · Ionic Equilibrium

JEE Main 2026 — 5 April, Morning Shift — Question 54

Arrange the following resultant mixtures in increasing order of their pH values A. $10 \mathrm{~mL} 0.2 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}+25 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{HCl}$ B. $10 \mathrm{~mL} 0.01 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL} 0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}$ C. $10 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}+10 \mathrm{~mL} 0.1 \mathrm{M} \mathrm{KOH}$

Choose the correct answer from the options given below :

Option A:
$\mathrm{B}<\mathrm{C}<\mathrm{A}$
Option B:
C $<$ A $<$ B
Option C:
C $<$ B $<$ A
Correct
Option D:
A $<$ C $<$ B

Answer: C

Step-by-step solution

(A) $\mathrm{H^+_{(aq)}} + \mathrm{OH^-_{(aq)}} \rightarrow \mathrm{H_2O}(\ell)$

$t=0 \quad 2.5\,\mathrm{mmol} \quad 4\,\mathrm{mmol}$

$t=\infty \quad 0 \quad 1.5\,\mathrm{mmol}$

In final solution, $\left[\mathrm{OH^-}\right] = \frac{1.5\,\mathrm{mmol}}{35\,\mathrm{mL}} = \frac{1.5}{35}$

$\mathrm{pOH} = -\log\frac{1.5}{35}$

$\mathrm{pH} = 14 + \log\left(\frac{1.5}{35}\right)$

(B) $\mathrm{H^+_{(aq)}} + \mathrm{OH^-_{(aq)}} \rightarrow \mathrm{H_2O}(\ell)$

$t=0 \quad 0.2\,\mathrm{mmol} \quad 0.2\,\mathrm{mmol}$

Final $\mathrm{pH}=7$

$t=0 \quad 2\,\mathrm{mmol} \quad 1\,\mathrm{mmol}$

$t=\infty \quad 1\,\mathrm{mmol} \quad 0$

Final $\mathrm{pH}<7$

So, order of pH values: (C) $<$ (B) $<$ (A)

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Chemistry
Chapter: Ionic Equilibrium
Topic: pH of solution containing implicit reaction