A certain weak base B has a base-ionization constant K_b of 7.3 $\times$ 10^-4 at 25°C.If strong acid is added to a solution of B,at what pH will [B] = [BH⁺]? A) 8.5 B) 10.9 C) 7.0 D) 3.1 E) 5.5

Question

Quizplus · Accepted Answer

The equation for the ionization of the weak base B is: B + H2O ⇌ BH+ + OH- The base-ionization constant K_b is given by: K_b = [BH+][OH-]/[B] At equilibrium, the law of mass action can be used to relate the equilibrium concentrations of the species: K_b = [(x)(x)]/[B-x] where x represents the concentration of [BH+] and [OH-] (because they are equal in this case) and B-x represents the concentration of the remaining, un-ionized weak base. The problem asks us to find the pH at which [B] = [BH+], or in other words, when half of the weak base has been converted to its conjugate acid. At this point, the concentration of [BH+] = [OH-] = x, and the concentration of [B] = B-x. Substituting these values into the expression for K_b and solving for the pH yields: K_b = x2/(B-x) = 7.3 x 10-11 x2 = K_b(B-x) = 7.3 x 10-11(B-x) Since we are looking for the pH at which [B] = [BH+], we can set B-x = x and solve for x: x2 = 7.3 x 10-11(x) x = sqrt(7.3 x 10-11) = 2.7 x 10-6 M At equilibrium, the concentration of [OH-] = 2.7 x 10-6 M, and the pH is given by: pH = 14 - pOH = 14 + log([H3O+]) = 14 + log(Kw/[OH-]) = 14 + log(1.0 x 10-14/2.7 x 10-6) ≈ 10.9 Therefore, the best choice is B.

A Certain Weak Base B Has a Base-Ionization Constant Kb ×\times×

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