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Chemistry · Ionic Equilibrium

JEE Main 2025 — 4 April, Morning Shift — Question 16

The pH of a 0.01 M weak acid HX(Ka=4×10−10)\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×10−4Mx \times 10^{-4} \mathrm{M}. The value of x is _____\_\_\_\_\_ (nearest integer)

Answer: 25

Numerical answer — enter this value.

Step-by-step solution

After dilution

pH=6[H+]=10−6[H+]=10−6=KaC10−6=4×10−10×C10−124×10−10=C0.25×10−2=C25×10−4=C∴x=25\begin{aligned} & \mathrm{pH}=6 \quad\left[\mathrm{H}^{+}\right]=10^{-6} \\& {\left[\mathrm{H}^{+}\right]=10^{-6}=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}} \\& 10^{-6}=\sqrt{4 \times 10^{-10} \times \mathrm{C}} \\& \frac{10^{-12}}{4 \times 10^{-10}}=\mathrm{C} \\& 0.25 \times 10^{-2}=\mathrm{C} \\& 25 \times 10^{-4}=\mathrm{C} \therefore \mathrm{x}=25 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Solutions containing one Acid or Base