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Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf

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Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3 Solution Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3.

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Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3 Solution Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3

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Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf

  • 1. Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3 Solution Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M NH3 + H2O <=> NH4+ + OH- I 0.25 0 0 C -a +a +a E 0.25-a a a Kb = [NH4+][OH-]/[NH3] = a2/(0.25 - a) = 1.8 x 10-5 a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0 a = 0.002112 [OH-] = a = 0.002112 M [H+] = Kw/[OH-] where Kw is the ionic product of water = 10-14/0.002112 = 4.735 x 10-12 M pH = -log[H+] = 11.3