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Chemistry · EQUILIBRIUM
हिन्दीThe base dissociation constant \( K_b \) of aniline is \( 4.27 \times 10^{-10} \) at \( 298\text{ K} \). What is the value of the conjugate acid dissociation constant \( K_a \) for anilinium ion (\( \text{C}_6\text{H}_5\text{NH}_3^+ \)) at this temperature? (\( K_w = 1.0 \times 10^{-14} \))
Correct answer · Option 2
\( 2.34 \times 10^{-5} \)
Explanation
For a conjugate acid-base pair, \( K_a \times K_b = K_w \). Therefore, \( K_a = \frac{1.0 \times 10^{-14}}{4.27 \times 10^{-10}} \approx 2.34 \times 10^{-5} \).
