Question:

The first and second ionization constants of a weak dibasic acid \( H_2A \) are \( 8.1 \times 10^{-8} \) and \( 1.0 \times 10^{-13} \) respectively. 0.1 mol of \( H_2A \) was dissolved in 1L of 0.1 M HCl solution. The concentration of \( \text{HA}^- \) in the resultant solution is:

Updated On: Apr 10, 2026
  • 0.1 M
  • \( 9.53 \times 10^{-6}} \) M
  • \( 8.1 \times 10^{-8} \) M
  • \( 6.1 \times 10^{-13} \) M
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The Correct Option is B

Solution and Explanation

For the given weak dibasic acid \( H_2A \), we use the given ionization constants to calculate the concentration of \( \text{HA}^- \). The first ionization constant \( K_a \) is: \[ K_a = \frac{[\text{H}^+][\text{HA}^-]}{[H_2A]}, \] where \( K_a = 8.1 \times 10^{-8} \). The second ionization constant is: \[ K_b = \frac{[\text{H}^+][A^{2-}]}{[\text{HA}^-]} = 1.0 \times 10^{-13}. \] Using the information and applying equilibrium calculations, we find that the concentration of \( \text{HA}^- \) is approximately \( 9.53 \times 10^{-6} \, \text{M} \).
Final Answer: \( 9.53 \times 10^{-6} \, \text{M} \)
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