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 H2SO4 + 10 mL 0.01 M Ca(OH)2 C. 10 mL 0.1 M H2SO4 + 10 mL 0.1 M KOH Choose the correct answer from the options given below:

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q56 – JEE-MAIN 2026

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Home Chemistry EquilibriumQuestion #425

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

10 mL 0.01 {M H}_{2} {SO}_{4} + 10 mL 0.01 {M Ca(OH)}_{2}

10 mL 0.1 {M H}_{2} {SO}_{4} + 10 mL 0.1 M KOH

Choose the correct answer from the options given below:

Clue 1 of 3: Identify Reactants

Determine if each mixture involves a strong acid, strong base, or a combination, and if it's a neutralization reaction.

🧠 Full Solution Path

Step 1: Calculate Moles of

H^{+}

and

O H^{-}

for each mixture✦ Active

For each mixture, calculate the initial millimoles (mmol) of $H^{+}$ and $O H^{-}$ ions. Remember that $C a (O H)_{2}$ provides $2 mol O H^{-}$ per mole, and $H_{2} S O_{4}$ provides $2 mol H^{+}$ per mole.

C a (O H)_{2}

10 mL \times 0.2 M = 2 {mmol Ca(OH)}_{2} \Rightarrow 4 {mmol OH}^{-}

H C l

25 mL \times 0.1 M = 2.5 mmol HCl \Rightarrow 2.5 {mmol H}^{+}

H_{2} S O_{4}

10 mL \times 0.01 M = 0.1 {mmol H}_{2} {SO}_{4} \Rightarrow 0.2 {mmol H}^{+}

C a (O H)_{2}

10 mL \times 0.01 M = 0.1 {mmol Ca(OH)}_{2} \Rightarrow 0.2 {mmol OH}^{-}

H_{2} S O_{4}

10 mL \times 0.1 M = 1 {mmol H}_{2} {SO}_{4} \Rightarrow 2 {mmol H}^{+}

K O H

10 mL \times 0.1 M = 1 mmol KOH \Rightarrow 1 {mmol OH}^{+}

Step 2: Determine Excess Ions and their Concentration○ Expand

Subtract the limiting reactant's moles from the excess reactant's moles to find the remaining ions. Then, divide by the total volume to get the concentration.

A. Excess

O H^{-} = 4 - 2.5 = 1.5 mmol

. Total volume

= 10 + 25 = 35 mL

[O H^{-}] = \frac{1.5 mmol}{35 mL} \approx 0.04286 M

H^{+}

and

O H^{-}

are equal (

0.2 mmol

each). Complete neutralization. The solution is neutral. C. Excess

H^{+} = 2 - 1 = 1 mmol

. Total volume

= 10 + 10 = 20 mL

[H^{+}] = \frac{1 mmol}{20 mL} = 0.05 M

💡 Teacher's Secret Hint

Remember to use the total volume after mixing for concentration calculations.

Step 3: Calculate pH and Arrange in Increasing Order○ Expand

Calculate the pH for each mixture using the concentrations found. For basic solutions, calculate pOH first, then $p H = 14 - p O H$ . For neutral solutions, $p H = 7$ .

p O H = - \log (0.04286) \approx 1.368 \Rightarrow p H = 14 - 1.368 = 12.632

p H = 7

p H = - \log (0.05) = 1.301

Arranging in increasing order of pH:

C (1.301) < B (7) < A (12.632)

💡 Teacher's Secret Hint

A strong acid-strong base neutralization results in a neutral solution (pH 7) if the moles of $H^{+}$ and $O H^{-}$ are exactly equal.

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Identify Reactants

🧠 Full Solution Path