JEE Main 2025 Apr 4 Shift 1, Chemistry Q24: Ionization Of Acids And Bases

JEE Main2025Apr 4, Shift 1Chemistry

The pH of a 0.01 M weak acid HX ( $K_{a} = 4 \times 1 0^{- 10}$ ) 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 $x \times 1 0^{- 4}$ M. The value of $x$ is ________ (nearest integer)

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Solution

For a weak acid,

$K_{a} = \frac{[ H ^{+} ] ^{2}}{C}$

After dilution,

$pH = 6 \Rightarrow [H^{+}] = 1 0^{- 6} M$

Let the new concentration be $C = x \times 1 0^{- 4} M$ .

$∴ 4 \times 1 0^{- 10} = \frac{( 1 0 ^{- 6} ) ^{2}}{C}$

$\Rightarrow C = \frac{1 0 ^{- 12}}{4 \times 1 0 ^{- 10}} = 2.5 \times 1 0^{- 3} M$

$= 25 \times 1 0^{- 4} M$

$∴ x = 25$

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JEE Main 2025 Apr 4 Shift 1, Chemistry Q24: Ionization Of Acids And Bases

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