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ExamsNEETChemistry

In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the Ka values for successive loss of protons from these acids differ by at least a factor of 10³ i.e., Ka1 > Ka2.

  1. H2X ⇌ H⁺ + HX; Ka1
  2. HX ⇌ H⁺ + X²; Ka2
  3. Ka1 > Ka2
  4. Ka1 = Ka2

Correct answer: Ka1 > Ka2

Solution

In polyprotic acids, the successive ionization constants (Ka values) decrease significantly because it becomes increasingly difficult to remove a proton from a negatively charged species. Hence, Ka1 is always greater than Ka2.

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