JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The first and second ionization constants of a weak dibasic acid \(H_2 A\) are \(8.1 \times 10^{-8}\) and \(1.0 \times 10^{-13}\) respectively. \(0.1\) mol of \(H_2 A\) was dissolved in \(1\) L of \(0.1\) M HCl solution. The concentration of \(HA^-\) in the resultant solution is:
- A \(0.1\) M
- B \(9.53 \times 10^{-6}\) M
- C \(8.1 \times 10^{-8}\) M
- D \(1.0 \times 10^{-13}\) M
Answer & Solution
Correct Answer
(C) \(8.1 \times 10^{-8}\) M
Step-by-step Solution
Detailed explanation
Given: Weak dibasic acid \(H_2A\) with \(K_{a1} = 8.1 \times 10^{-8}\) and \(K_{a2} = 1.0 \times 10^{-13}\) Initial concentration: \([H_2A]_0 = 0.1\text{ M}\) The solution also contains \(0.1\text{ M HCl}\), which dissociates completely to give \([H^+]_0 = 0.1\text{ M}\). The…
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