JEE Main · 2025 · Shift-ImediumIEQ-082

A weak acid HA has degree of dissociation x. Which option gives the correct expression of pH - pKa?

Ionic Equilibrium · Class 11 · JEE Main Previous Year Question

Question

A weak acid HA has degree of dissociation xx. Which option gives the correct expression of pHpKa\text{pH} - \text{pK}_a?

Options
  1. a

    log(1+2x)\log(1+2x)

  2. b

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

  3. c

    00

  4. d

    log(x1x)\log\left(\frac{x}{1-x}\right)

Correct Answerd

log(x1x)\log\left(\frac{x}{1-x}\right)

Detailed Solution

Henderson-Hasselbalch: pH=pKa+log[\ceA][\ceHA]\text{pH} = \text{pK}_a + \log\frac{[\ce{A^-}]}{[\ce{HA}]}

With degree of dissociation xx: [\ceA]=ax[\ce{A^-}] = ax, [\ceHA]=a(1x)[\ce{HA}] = a(1-x)

pHpKa=logaxa(1x)=logx1x\text{pH} - \text{pK}_a = \log\frac{ax}{a(1-x)} = \log\frac{x}{1-x}

Answer: log(x/(1x))\log(x/(1-x)) → Option (d)

Verification: When pH = pKa: log(x/(1x))=0x=0.5\log(x/(1-x)) = 0 \Rightarrow x = 0.5 (50% dissociation) ✓

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Ionic Equilibrium) inside The Crucible, our adaptive practice platform.

A weak acid HA has degree of dissociation x. Which option gives the correct expression of pH - pKa? (JEE Main 2025) | Canvas Classes