University College - Chemistry Chem 112A Lecture Notes - Lecture 9: Ph
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8 Feb 2017
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0. 1 x e x 0. 1 + x. Ka = x(0. 1 + x)/(0. 1-x) = 1. 76 x 10-5 = [h3o+][oac-]/[hoac] ph = -log(x) = 4. 75, so solution is acidic. In this case ka > kb, if the opposite had been true, then the solution would be basic. Disturb a system at equilibrium by adding an acid or base, then determine the new ph. [h3o+] = # mol/v = 0. 1 m *0. 1 l /1. 1l = 9. 1 x 10-3 m. [oac-] = [hoac] = 0. 1m * 1l/1. 1l = 9. 1 x 10-2 m. I: 9. 1 x 10-2 9. 1 x 10-3 9. 1 x 10-2. Ka = 1. 76 x 10-5 = x(0. 0819 + x)/(0. 1001 x) X = [h3o+] = 2. 15 x 10-5 ph = 4. 67. Note that solution has not changed that much, despite the addition of hcl. Same system, but with the addition of a base: [oh-] = 100ml . 1 m naoh/1. 1l = 9. 1 x 10-3 m.
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