JEE Main Chemistry question

The equilibrium constant for the reaction, HONO+CN⁻→HCN+ONO⁻ is 1×10⁻¹⁴. The magnitude of the equilibrium constant suggests that:

Answer: (B)

Solution

K=Ka(HONO)/Ka(HCN); since K≪1, Ka(HONO)≪Ka(HCN), meaning HCN is (per this given K) the stronger acid of the pair. (Options

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Question ID VK-136299