QuestionDetails: Determine the pH of each of the following solutions A) 0.095 M
ID: 684488 • Letter: Q
Question
QuestionDetails: Determine the pH of each of the following solutions A) 0.095 M hypochlorous acid Ka= 3.0 x 10-8 B) 0.0085 Mphenol Ka = 1.3 x 10-10 C) 0.095 M hydroxylamine Ka= 1.1 x 10-8 QuestionDetails: Determine the pH of each of the following solutions A) 0.095 M hypochlorous acid Ka= 3.0 x 10-8 B) 0.0085 Mphenol Ka = 1.3 x 10-10 C) 0.095 M hydroxylamine Ka= 1.1 x 10-8 QuestionDetails: Determine the pH of each of the following solutions A) 0.095 M hypochlorous acid Ka= 3.0 x 10-8 B) 0.0085 Mphenol Ka = 1.3 x 10-10 C) 0.095 M hydroxylamine Ka= 1.1 x 10-8Explanation / Answer
A) HOCl H+ + Cl- I(M) 0.095M 0 0 C(M) -x x x E(M) 0.095-x x x Ka = [H+] [Cl-] / [HOCl] 3.0 x 10-8 = x2 / 0.095-x bysolving x = [H+] = 5.33 x 10-5 M pH = -log [H+] = 4.273. B) PhOH H+ + PhO- I(M) 0.0085M 0 0 C(M) -x x x E(M) 0.0085-x x x Ka = [H+] [PhO-] / [PhOH] 1.3x 10-10 = x2 / 0.0085-x bysolving x = [H+] = 1.051 x 10-6 M pH = -log [H+] = 5.978. c) NH2OH NH2+ +OH- I(M) 0.095M 0 0 C(M) -x x x E(M) 0.095-x x x Kb = x2 / 0.095-x 1.1x 10-8 = x2 / 0.095-x bysolving x = [OH-] = 3.23 x 10-5 M pOH = -log [H+] = 4.490 pH = 14 - 4.490 = 9.509 C(M) -x x x E(M) 0.095-x x x Kb = x2 / 0.095-x 1.1x 10-8 = x2 / 0.095-x bysolving x = [OH-] = 3.23 x 10-5 M pOH = -log [H+] = 4.490 pH = 14 - 4.490 = 9.509 C(M) -x x x E(M) 0.095-x x x Kb = x2 / 0.095-x 1.1x 10-8 = x2 / 0.095-x bysolving x = [OH-] = 3.23 x 10-5 M pOH = -log [H+] = 4.490 pH = 14 - 4.490 = 9.509Related Questions
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