The equilibrium hydrolysis constant () for the anilinium ion () is measured to be at . Calculate the — Ionic Equilibrium Chemistry Question
Question
The equilibrium hydrolysis constant ($K_h$) for the anilinium ion ($C_6H_5NH_3^+$) is measured to be $2.5 \times 10^{-5} \text{ M}$ at $25^\circ C$. Calculate the basic dissociation constant ($K_b$) for the weak base aniline ($C_6H_5NH_2$) at the same temperature. If $K_b = y \times 10^{-10}$, find the exact numerical value of $y$. ($K_w = 1.0 \times 10^{-14}$).
Answer: 4
💡 Solution & Explanation
The anilinium ion represents the conjugate acid of the weak base aniline. The relationship between the hydrolysis constant (which acts as $K_a$ for the conjugate acid) and the base dissociation constant is $K_h \times K_b = K_w$. Rearranging gives $K_b = \frac{K_w}{K_h} = \frac{1.0 \times 10^{-14}}{2.5 \times 10^{-5}} = 0.4 \times 10^{-9} = 4 \times 10^{-10}$. Thus, $y = 4$.
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