JEE Main 2025 — Chemistry Physical Chemistry

2025integermedium

The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right)$ is found to be 5. Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6. The new concentration of the diluted weak acid is given as $x \times 10^{-4} \mathrm{M}$ . The value of x is ________ (nearest integer)

Official previous-year question

Held on 4 Apr 2025 · Verified 6 Jul 2026.

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Solution

$\begin{aligned}
& \quad \mathrm{HX}{(\mathrm{aq)}} \rightleftharpoons \mathrm{H}{(\mathrm{aq})}^{+}+\mathrm{X}{(\mathrm{aq})}^{-} \quad \mathrm{K}{\mathrm{a}}=4 \times 10^{-10} \

& 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified } \

& \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \

& \mathrm{~K}_{\mathrm{a}}=0.01 \alpha^2=10^{-8}
\end{aligned}$

On dilution let final concentration of $\mathrm{HX}=\mathrm{c} \mathrm{M}$

$\begin{aligned}
& \mathrm{Hx}{(\mathrm{an})} \rightleftharpoons \mathrm{H}{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{an})}^{-} \

& \mathrm{C}(1-\alpha) \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \

& \Rightarrow \mathrm{C} \alpha=10^{-6}...(1) \

& \quad \frac{\mathrm{C} \alpha^{2}}{1-\alpha}=\mathrm{K}_{\mathrm{a}}=10^{-8}...(2) \

& \Rightarrow \frac{10^{-6} \alpha}{1-\alpha}=10^{-8}
\end{aligned}$

Data given is inconsistent & contradictory. This should be bonus.

Your note

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Solution

$\begin{aligned}
& \quad \mathrm{HX}{(\mathrm{aq)}} \rightleftharpoons \mathrm{H}{(\mathrm{aq})}^{+}+\mathrm{X}{(\mathrm{aq})}^{-} \quad \mathrm{K}{\mathrm{a}}=4 \times 10^{-10} \

& 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text { Not justified } \

& \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \

& \mathrm{~K}_{\mathrm{a}}=0.01 \alpha^2=10^{-8}
\end{aligned}$

On dilution let final concentration of $\mathrm{HX}=\mathrm{c} \mathrm{M}$

$\begin{aligned}
& \mathrm{Hx}{(\mathrm{an})} \rightleftharpoons \mathrm{H}{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{an})}^{-} \

& \mathrm{C}(1-\alpha) \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \

& \Rightarrow \mathrm{C} \alpha=10^{-6}...(1) \

& \quad \frac{\mathrm{C} \alpha^{2}}{1-\alpha}=\mathrm{K}_{\mathrm{a}}=10^{-8}...(2) \

& \Rightarrow \frac{10^{-6} \alpha}{1-\alpha}=10^{-8}
\end{aligned}$

Data given is inconsistent & contradictory. This should be bonus.

JEE Main 2025 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2025 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry