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Chemistry · Ionic Equilibrium

JEE Main 2026 — 2 April, Evening Shift — Question 49

The first and second ionization constants of a weak dibasic acid H2 A\mathrm{H}_{2} \mathrm{~A} are 8.1×10−88.1 \times 10^{-8} and 1.0×10−131.0 \times 10^{-13} respectively. 0.1 mol of H2 A\mathrm{H}_{2} \mathrm{~A} was dissolved in 1L of 0.1 M HCl solution. The concentration of HA−\mathrm{HA}^{-} in the resultant solution is :

  1. Option A:

    0.1 M

  2. Option B:

    9.53×10−6M9.53 \times 10^{-6} \mathrm{M}

  3. Option C:

    8.1×10−8M8.1 \times 10^{-8} \mathrm{M}

    Correct
  4. Option D:

    1.0×10−13M1.0 \times 10^{-13} \mathrm{M}

Answer: C

Step-by-step solution

HCl0.1M→H+0.1 m+Cl−0.1 m\underset{0.1 \mathrm{M}}{\mathrm{HCl}} \rightarrow \underset{0.1 \mathrm{~m}}{\mathrm{H}^{+}}+\underset{0.1 \mathrm{~m}}{\mathrm{Cl}^{-}} H2 A0.1−x⇌H+X+0.1+HA−X\underset{0.1-\mathrm{x}}{\mathrm{H}_{2} \mathrm{~A}} \rightleftharpoons \underset{\mathrm{X}+0.1}{\mathrm{H}^{+}}+\underset{\mathrm{X}}{\mathrm{HA}^{-}} (Due to common ion effect x is very less) Ka1=[H+][HA−][H2 A]\mathrm{Ka}_{1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HA}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{~A}\right]} Ka1=0.1×[HA−]0.1\mathrm{Ka}_{1}=\frac{0.1 \times\left[\mathrm{HA}^{-}\right]}{0.1} [HA−]=Ka1=8.1×10−8\left[\mathrm{HA}^{-}\right]=\mathrm{Ka}_{1}=8.1 \times 10^{-8}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Solutions with mixture of acids or bases

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