At 298 K, the ionization constant of \( {CN}^- \) is \( 2.08 \times 10^{-6} \). What is the ionization constant of its conjugate acid? (Given \( K_w = 10^{-14} \))
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- \( 2.08 \times 10^{-8} \)
- \( 4.8 \times 10^{-9} \)
- \( 4.8 \times 10^{-10} \)
- \( 2.08 \times 10^{-7} \)
The Correct Option is C
Solution and Explanation
The ionization constant of the conjugate acid \( {HA} \) can be found using the relation: \[ K_a \cdot K_b = K_w \] Where: - \( K_a \) is the ionization constant of the conjugate acid, - \( K_b \) is the ionization constant of \( {CN}^- \), - \( K_w = 10^{-14} \) is the ionization constant of water. Given that \( K_b = 2.08 \times 10^{-6} \), we can solve for \( K_a \): \[ K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{2.08 \times 10^{-6}} = 4.8 \times 10^{-10} \] Thus, the ionization constant of the conjugate acid is \( 4.8 \times 10^{-10} \).
Final Answer: \( 4.8 \times 10^{-10} \).
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