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JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Values of dissociation constant, \(K_a\) are given as follows
| Acid | \(K_a\) |
| \(HCN\) | \(6.2\times 10^{-10}\) |
| \(HF\) | \(7.2\times 10^{-4}\) |
| \(HNO_2\) | \(4.0\times 10^{-4}\) |
- A \(F^- < CN^- < NO_2^-\)
- B \(NO_2^- < CN^- < F^-\)
- C \(F^- < NO_2^- < CN^-\)
- D \(NO_2^- < F^- < CN^-\)
Answer & Solution
Correct Answer
(C) \(F^- < NO_2^- < CN^-\)
Step-by-step Solution
Detailed explanation
Higher the value of \(K_a\) lower will be the value of \(pK_a\) \(i.e.\) higher will be the acidic nature. Further since \(CN^-, F^-\) and \(NO_2^-\) are conjugate base of the acids \(HCN, HF\) and \(HNO_2\) respectively hence the correct order of base strength will be…
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