Find the pH of the a 0.500M solution of the weak base Ammonia (NH₃) . The k_b = 1.78*10^-5
NH₃ +H₂O --> NH₄⁺ + OH^-

Question

Find the pH of the a 0.500M solution of the weak base Ammonia (NH₃). The k_b = 1.78*10^-5 NH₃ +H₂O --> NH₄⁺ + OH^-

Quizplus · Accepted Answer

The solution is as follows: The equilibrium equation is k_b=[NH₄][OH^-]/[NH₃] Since NH₃ will not fully dissociate, the concentrations of OH^- and HN ₃are unknown (x) and the concentration of NH₃ at equilibrium will be 0.5 - x. k_b = [x][x]/[0.500-x] Assuming x is so small it's negligible, that yields; k_b = x²/[0.500] (1.78 * 10^-5) = x²/[0.500] (1.78 * 10^-5) * 0.500 = x² x= .003M which is the concentration of OH^- Using the equilibrium equation for water, we have H₂O? H⁺ + OH^- The equilbrium value, Kw is 10^-14 Kw = [H⁺][OH^-] * 10^-14 = [H⁺][0.003] [H⁺] = 3.33*10^-12 pH = -log [H⁺] = -log (3.33 * 10^-12) = 11.48

Find the PH of the a 0