Atequilibrium170 mo lofhiarepresent calculatekeq

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Unformatted text preview: Weak acids ionize part ially in so lutions to reach the fo llowing equilibrium: + – ˆˆ HA (aq) + H2O (l) ‡ˆ† H3O (aq) + A (aq) or simply, + – ˆ ˆˆ HA (aq) ‡ˆ† H (aq) + A (aq) · The equilibrium constant for these reactions is called acid ionization constant (Ka) and has the fo llowing form: K a = [H + ][A ­ ] [HA] Examples: + ­ 1. At 25 °C, a 0.100 M solut ion of HC2H3O2 is 1.34% ionized and has an [H ] of 1.34 x 10 3 M. Calculate Ka for acetic acid. + ˆˆ HC2H3O2 ‡ˆ† H (aq) + C2H3O2 – (aq) Initial D Equilibrium K a = 0.100 M – x 0.100 –x 0 + x x 2 [H + ][C2 H 3O ­2 ] x = [HC2 H 3 O 2 ] 0.100 ­x x = [H + ] = [C2 H 3O ­2 ] = 1.34 x 10­3 M 12 0 + x x Chemistry 52 Chapter 16 + 2. What is the [H ] in a 0.50 M HC2H3O2 solut ion? The ionizat ion constant, Ka, for ­ HC2H3O2 is 1.8 x 10 5 . ˆˆ HC2H3O2 ‡ˆ† Initial D Equilibrium Ka = 3. Calculate the percent ionization for the solution in Problem 2. [H + ] or [A ­ ] % ionization = x100 [HA] 13 Chemistry 5...
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This document was uploaded on 03/18/2014 for the course CHEM 52 at Los Angeles Mission College.

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