At 25^$\circ$C, the base ionization constant for NH₃ is 1.8 * 10^-5. Determine the percentage ionization of a 0.150 M solution of ammonia at 25 ^$\circ$C.

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The percentage ionization of a weak base can be calculated using the following equation:% ionization = ( [H+] / [B] ) * 100%where [H+] is the molar concentration of hydrogen ions and [B] is the molar concentration of the base.For NH3, the base ionization constant (Kb) is given as:Kb = [NH4+][OH-] / [NH3]At 25 °C, Kb = 1.8 * 10^-5. Since [OH-] = [H+], we can substitute [H+] for [OH-] in the Kb expression:Kb = [H+]^2 / [NH3]Solving for [H+], we get:[H+] = sqrt(Kb * [NH3])Plugging in the values of Kb and [NH3] = 0.150 M, we get:[H+] = sqrt(1.8 * 10^-5 * 0.150 M) = 5.48 * 10^-4 MNow we can calculate the percentage ionization:% ionization = ( [H+] / [B] ) * 100%% ionization = ( 5.48 * 10^-4 M / 0.150 M ) * 100% = 0.365%Therefore, the percentage ionization of a 0.150 M solution of ammonia at 25 °C is approximately 0.365%, which is closest to choice C (1.1%).

At 25 ∘\circ∘ C, the Base Ionization Constant for NH3 Is 1

At 25^$\circ$C, the Base Ionization Constant for NH₃ Is 1