What is the pH of a 0.040 mol L^-1 aqueous Ba(OH)₂ solution? A)1.10 B)1.40 C)12.60 D)12.90

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Quizplus · Accepted Answer

Ba(OH)2 is a strong base and fully dissociates in water to form OH- ions. The concentration of OH- ions in the solution can be calculated using the molarity of the Ba(OH)2 solution: [OH-] = 2 x 0.040 mol/L = 0.080 mol/L Since the solution contains only Ba(OH)2 and L^-1 ions, the concentration of H+ ions can be calculated using the equation for the ion product of water: Kw = [H+][OH-] 1.0 x 10^-14 = [H+][0.080] [H+] = 1.25 x 10^-13 mol/L To find the pH, we take the negative logarithm of the H+ concentration: pH = -log[H+] = -log(1.25 x 10^-13) = 12.90 Therefore, the pH of the solution is 12.90, and the best choice is D.

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