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NVAMediumJEE 2025Ionic Equilibria & pH

JEE Chemistry 2025 Question with Solution

The pH of a 0.01 M0.01 \, \text{M} weak acid HX\mathrm{HX} (Ka=4×10−10)\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right) is found to be 55 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 66 . The new concentration of the diluted weak acid is given as x×10−4 M\mathrm{x} \times 10^{-4} \, \text{M}. The value of xx is _____ (nearest integer).

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: Initial concentration of weak acid HX\mathrm{HX} is 0.01 M0.01 \, \text{M}, Ka=4×10−10\mathrm{K_a}=4 \times 10^{-10}, initial pH=5\text{pH}=5, and after dilution pH=6\text{pH}=6.

Find: The value of xx in the new concentration x×10−4 Mx \times 10^{-4} \, \text{M}.

From pH=6\text{pH}=6, the hydronium ion concentration is

[H3O+]=10−6 M[\mathrm{H_3O^+}] = 10^{-6} \, \text{M}

For the weak acid equilibrium,

Ka=[H3O+][X−][HX]\mathrm{K_a} = \frac{[\mathrm{H_3O^+}][\mathrm{X^-}]}{[\mathrm{HX}]}

Using [H3O+]=[X−]=10−6 M[\mathrm{H_3O^+}] = [\mathrm{X^-}] = 10^{-6} \, \text{M} after dilution,

4×10−10=(10−6)2[HX]new4 \times 10^{-10} = \frac{(10^{-6})^2}{[\mathrm{HX}]_{\text{new}}}

Now solve for the new acid concentration:

[HX]new=(10−6)24×10−10=2.5×10−3 M[\mathrm{HX}]_{\text{new}} = \frac{(10^{-6})^2}{4 \times 10^{-10}} = 2.5 \times 10^{-3} \, \text{M}

Write this in the form x×10−4 Mx \times 10^{-4} \, \text{M}:

2.5×10−3=25×10−42.5 \times 10^{-3} = 25 \times 10^{-4}

Therefore, the value of xx is 2525.

The solution notes a consistency check for the initial data; however, using the solution's equilibrium working, the required diluted concentration corresponds to x=25x=25.

Consistency Check and Calculation

Given: Weak acid HX\mathrm{HX} with Ka=4×10−10\mathrm{K_a}=4 \times 10^{-10}. Initial concentration is 0.01 M0.01 \, \text{M} and initial pH=5\text{pH}=5. After dilution, pH=6\text{pH}=6.

Find: The nearest integer value of xx in x×10−4 Mx \times 10^{-4} \, \text{M}.

For the initial solution, pH=5\text{pH}=5 gives

[H3O+]=10−5 M[\mathrm{H_3O^+}] = 10^{-5} \, \text{M}

the solution remarks on checking consistency of this data with the stated Ka\mathrm{K_a} and concentration.

For the diluted solution, pH=6\text{pH}=6 gives

[H3O+]=10−6 M[\mathrm{H_3O^+}] = 10^{-6} \, \text{M}

For a weak acid,

HX⇌H++X−\mathrm{HX} \rightleftharpoons \mathrm{H^+} + \mathrm{X^-}

and

Ka=[H+][X−][HX]new\mathrm{K_a} = \frac{[\mathrm{H^+}][\mathrm{X^-}]}{[\mathrm{HX}]_{\text{new}}}

Using the standard weak-acid approximation in the extracted solution, [H+]=[X−]=10−6 M[\mathrm{H^+}] = [\mathrm{X^-}] = 10^{-6} \, \text{M}. Hence,

4×10−10=(10−6)(10−6)[HX]new4 \times 10^{-10} = \frac{(10^{-6})(10^{-6})}{[\mathrm{HX}]_{\text{new}}}

So,

[HX]new=10−124×10−10=2.5×10−3 M[\mathrm{HX}]_{\text{new}} = \frac{10^{-12}}{4 \times 10^{-10}} = 2.5 \times 10^{-3} \, \text{M}

Now compare with x×10−4 Mx \times 10^{-4} \, \text{M}:

2.5×10−3 M=25×10−4 M2.5 \times 10^{-3} \, \text{M} = 25 \times 10^{-4} \, \text{M}

Therefore, the required value is 2525.

Common mistakes

  • Using the initial 0.01 M0.01 \, \text{M} concentration directly after dilution is incorrect because dilution changes the equilibrium concentration of HX\mathrm{HX}. Recalculate using the new pH=6\text{pH}=6 condition and the acid dissociation expression.

  • Forgetting to convert pH to hydronium ion concentration causes an incorrect setup. From pH=6\text{pH}=6, first write [H3O+]=10−6 M[\mathrm{H_3O^+}] = 10^{-6} \, \text{M}, then substitute into Ka\mathrm{K_a}.

  • Writing the final concentration as 2.5×10−32.5 \times 10^{-3} and stopping there misses the required form. The question asks for xx in x×10−4 Mx \times 10^{-4} \, \text{M}, so convert carefully to get 25×10−4 M25 \times 10^{-4} \, \text{M}.

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