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MCQEasyJEE 2025Ionic Equilibria & pH

JEE Chemistry 2025 Question with Solution

A weak acid HA has degree of dissociation xx. Which option gives the correct expression of pH−pKapH - pK_a?

  • A

    log⁡(1+2x)\log(1 + 2x)

  • B

    00

  • C

    log⁡(x1−x)\log\left(\frac{x}{1-x}\right)

  • D

    log⁡(1−xx)\log\left(\frac{1-x}{x}\right)

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: A weak acid HA has degree of dissociation xx and initial concentration CC.

Find: The correct expression for pH−pKapH - pK_a in terms of xx.

For the dissociation

HA⇌H++A−HA \rightleftharpoons H^+ + A^-

at equilibrium,

[H+]=Cx,[A−]=Cx,[HA]=C(1−x)[H^+] = Cx, \quad [A^-] = Cx, \quad [HA] = C(1-x)

Therefore,

Ka=[H+][A−][HA]=(Cx)(Cx)C(1−x)=Cx21−xK_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}

Now,

pKa=−log⁡Ka=−log⁡(Cx21−x)pK_a = -\log K_a = -\log\left(\frac{Cx^2}{1-x}\right)

and

pH=−log⁡[H+]=−log⁡(Cx)pH = -\log[H^+] = -\log(Cx)

So,

pH−pKa=(−log⁡C−log⁡x)−(−log⁡C−2log⁡x+log⁡(1−x))pH - pK_a = (-\log C - \log x) - (-\log C - 2\log x + \log(1-x)) pH−pKa=log⁡x−log⁡(1−x)=log⁡(x1−x)pH - pK_a = \log x - \log(1-x) = \log\left(\frac{x}{1-x}\right)

Therefore, the correct option is C.

The solution contains a discrepancy because one part marks option A, but the derivation shown leads to log⁡(x1−x)\log\left(\frac{x}{1-x}\right), which is option C.

Using Henderson-type relation

Given: Degree of dissociation of weak acid HA is xx.

Find: pH−pKapH - pK_a.

From equilibrium composition,

[A−][HA]=CxC(1−x)=x1−x\frac{[A^-]}{[HA]} = \frac{Cx}{C(1-x)} = \frac{x}{1-x}

Using the relation

pH=pKa+log⁡([A−][HA])pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right)

we get

pH−pKa=log⁡(x1−x)pH - pK_a = \log\left(\frac{x}{1-x}\right)

Hence, the correct option is C.

Common mistakes

  • Using the option label given on the page without checking the derivation. The worked equilibrium expression leads to log⁡(x1−x)\log\left(\frac{x}{1-x}\right), so the algebra must be trusted over the mislabeled option tag.

  • Reversing the ratio of undissociated acid and conjugate base. Since [A−]=Cx[A^-] = Cx and [HA]=C(1−x)[HA] = C(1-x), the correct ratio is x1−x\frac{x}{1-x}, not 1−xx\frac{1-x}{x}.

  • Forgetting that pH=−log⁡[H+]=−log⁡(Cx)pH = -\log[H^+] = -\log(Cx). If [H+][H^+] is written incorrectly, the final logarithmic expression will also be wrong.

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