A 0.632 M aqueous solution of a monoprotic base has a pH of 11.53.What is the ionization constant of this base? A)1.8 × 10^-5 B)2.1 × 10^-3 C)1.6 × 10^-23 D)5.0 × 10^-12 E)5.3 × 10^-3

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The pH of the solution is 11.53, which means the pOH is 14 - 11.53 = 2.47. The concentration of OH- ions ([OH-]) can be calculated using the pOH: [OH-] = 10^-pOH = 10^-2.47 ≈ 3.388 x 10^-3 M. Since the base is monoprotic and partially ionizes, we can set up an expression for the ionization constant (Kb) of the base using the formula Kb = [OH-]^2 / ([Base]initial - [OH-]), where [Base]initial is the initial concentration of the base. Given that the initial concentration of the base is 0.632 M and the concentration of OH- produced is 3.388 x 10^-3 M, we find that Kb ≈ (3.388 x 10^-3)^2 / (0.632 - 3.388 x 10^-3) ≈ 1.8 x 10^-5. This matches choice A.

A 0632 M Aqueous Solution of a Monoprotic Base Has a a PH