Question:

At 298 K, the ionization constant of CN {CN}^- is 2.08×106 2.08 \times 10^{-6} . What is the ionization constant of its conjugate acid? (Given Kw=1014 K_w = 10^{-14} )

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The ionization constant of a conjugate acid and base pair is related by the ionization constant of water.
Updated On: Mar 15, 2025
  • 2.08×108 2.08 \times 10^{-8}
  • 4.8×109 4.8 \times 10^{-9}
  • 4.8×1010 4.8 \times 10^{-10}
  • 2.08×107 2.08 \times 10^{-7}
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The Correct Option is C

Solution and Explanation

The ionization constant of the conjugate acid HA {HA} can be found using the relation: KaKb=Kw K_a \cdot K_b = K_w Where: - Ka K_a is the ionization constant of the conjugate acid, - Kb K_b is the ionization constant of CN {CN}^- , - Kw=1014 K_w = 10^{-14} is the ionization constant of water. Given that Kb=2.08×106 K_b = 2.08 \times 10^{-6} , we can solve for Ka K_a : Ka=KwKb=10142.08×106=4.8×1010 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 4.8×1010 4.8 \times 10^{-10}
Final Answer: 4.8×1010 4.8 \times 10^{-10}
 

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