Horvath Practice Problems 63

Horvath Practice Problems 63 - of an acetate bu f er that...

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With the K a value we can get the equilibrium concentrations, K a =1 . 4 × 10 5 = (0 . 028 + x )( x ) (0 . 071 x ) . (15.1) Rearranging and using the quadratic formula we get x =3 . 2 × 10 4 . Hence [ HC 6 H 4 NO 2 ]=0 . 071 3 . 2 × 10 4 =0 . 071 £ H + ¤ =0 . 0028 3 . 2 × 10 4 =0 . 0025 £ C 6 H 4 NO 2 ¤ =3 . 2 × 10 4 Bu f er Solutions An important application of the common ion e f ect occurs when the common ion is the conjugate pair to the original acid or base. These solutions are called pH bu f er solutions (or bu f ers for short) and they have the valuable property that one can add a strong acid or base to a bu f er and not change the pH signi f - cantly. The human body makes use of many bu f ers to control the pH of the extra- and intra- cellular environments. Example, the acetate bu f er (part 1) : Calculate the pH
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Unformatted text preview: of an acetate bu f er that is prepared by adding 25 ml of a 0.10 M acetic acid (HC 2 H 3 O 2 ) solution to 15 mL of 0.10 M of a sodium acetate (NaC 2 H 3 O 2 ) solution ( K a = 1 . 7 10 5 ). We f rst determine the initial concentrations of acetic acid and acetate ion, . 025 L . 10 mol HC 2 H 3 O 2 1 L 1 . 040 mL = 0 . 063 M HC 2 H 3 O 2 . 015 L . 10 mol C 2 H 3 O 2 1 L 1 . 040 mL = 0 . 038 M C 2 H 3 O 2 , then setup an ICE chart [ HC 2 H 3 O 2 ] H + C 2 H 3 O 2 I . 063 . 038 C x x x E . 063 x x . 038 + x . 62 THE COMMON ION EFFECT AND BUFFERS...
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.

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