Values of dissociation constant, \(K_a\) are given as follows :
Acid \(K_a\) HCN \(6.2\times10^{-10}\) HF \(7.2\times10^{-4}\) \(\mathrm{HNO_2}\) \(4.0\times10^{-4}\)
Correct order of increasing base strength of the base \(\mathrm{CN^-}\), \(\mathrm{F^-}\) and \(\mathrm{NO_2^-}\) will be :
| Acid | \(K_a\) |
| HCN | \(6.2\times10^{-10}\) |
| HF | \(7.2\times10^{-4}\) |
| \(\mathrm{HNO_2}\) | \(4.0\times10^{-4}\) |
Show Hint
- \(\mathrm{F^-} \lt \mathrm{CN^-} \lt \mathrm{NO_2^-}\)
- \(\mathrm{NO_2^-} \lt \mathrm{CN^-} \lt \mathrm{F^-}\)
- \(\mathrm{F^-} \lt \mathrm{NO_2^-} \lt \mathrm{CN^-}\)
- \(\mathrm{NO_2^-} \lt \mathrm{F^-} \lt \mathrm{CN^-}\)
The Correct Option is C
Solution and Explanation
\(K_a\): \(\mathrm{HF}\) (\(7.2\times10^{-4}\)) \(\gt\) \(\mathrm{HNO_2}\) (\(4.0\times10^{-4}\)) \(\gt\) \(\mathrm{HCN}\) (\(6.2\times10^{-10}\)).
Hence base strength: \(\mathrm{F^-} \lt \mathrm{NO_2^-} \lt \mathrm{CN^-}\).
Answer: (3) \(\mathrm{F^-} \lt \mathrm{NO_2^-} \lt \mathrm{CN^-}\)
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