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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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} $.

Answer

$ 2.08 \times 10^{-8} $

Answer

$ 4.8 \times 10^{-9} $

Answer

$ 4.8 \times 10^{-10} $

Answer

$ 2.08 \times 10^{-7} $

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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At 298 K, the ionization constant of CN− {CN}^- CN− is 2.08×10−6 2.08 \times 10^{-6} 2.08×10−6. What is the ionization constant of its conjugate acid? (Given Kw=10−14 K_w = 10^{-14} Kw​=10−14)