Chemistry Q54 – JEE-MAIN 2025

An aqueous solution of HCl with pH

1.0

is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would (Given

\log 2 = 0.30

)

Clue 1 of 3: pH Definition

Recall the definition of pH and how it relates to the hydrogen ion concentration in a solution.

🥷

Ninja Strategy•Dilution Effect on pH

Diluting an acid solution always makes it less acidic, meaning its pH must increase.

🧠 Full Solution Path

Step 1: Calculate initial hydrogen ion concentration✦ Active

Given initial pH is $1.0$ . For a strong acid like HCl, the hydrogen ion concentration is related to pH by the formula $p H = - \log [H^{+}]$ . Therefore, the initial concentration of $H^{+}$ ions is:

[H^{+}]_{1} = 10^{- p H} = 10^{- 1.0} = 0.1 M

Step 2: Calculate hydrogen ion concentration after dilution○ Expand

The solution is diluted by adding an equal volume of water, which means the initial volume ( $V_{1}$ ) is doubled to $V_{2} = 2 V_{1}$ . Using the dilution formula $C_{1} V_{1} = C_{2} V_{2}$ , the new concentration of $H^{+}$ ions ( $[H^{+}]_{2}$ ) is:

[H^{+}]_{2} = \frac{[H^{+}]_{1} V_{1}}{V_{2}} = \frac{0.1 M \times V_{1}}{2 V_{1}} = \frac{0.1}{2} = 0.05 M

Step 3: Calculate the new pH○ Expand

The new pH is calculated using the diluted hydrogen ion concentration. Given $\log 2 = 0.30$ , we can find $\log 5 = \log (10 / 2) = \log 10 - \log 2 = 1 - 0.30 = 0.70$ .

p H_{2} = - \log [H^{+}]_{2} = - \log (0.05) = - \log (5 \times 10^{- 2})

p H_{2} = - (\log 5 + \log 10^{- 2}) = - (\log 5 - 2)

p H_{2} = 2 - \log 5 = 2 - 0.70 = 1.30

💡 Teacher's Secret Hint

Remember that diluting an acid solution increases its pH (makes it less acidic), so options suggesting a decrease or no change in pH can be immediately ruled out.

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Chemistry Question 54 – JEE-MAIN 2025

Clue 1 of 3: pH Definition

🧠 Full Solution Path