1422-hw06-acids-bases2-2009-ANSWERS

1422-hw06-acids-base - CHEM 1422 Homework 6 Acids Bases 2 Due Tuesday ANSWER KEY 1(5 pts 250 mL of 0.2 M acetic acid(HOAc reacts with 250 mL of 0.2

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CHEM 1422 - Homework # 6 ANSWER KEY Acids & Bases # 2 Due Tuesday, April 14, 2009 1. (5 pts) 250 mL of 0.2 M acetic acid (HOAc) reacts with 250 mL of 0.2 M NaOH. What is the pH of the resulting solution? pK a (HOAc) = 5 The first thing to realize is that one is reacting a weak acid (acetic acid) with a strong base (NaOH). This will generate the salt of a weak acid (sodium acetate, NaOAc), which is a weak base . There are equivalent amounts of acid and base present (same volume and same concentration) and since one is doubling the solution volume (250 to 500 mL) the concentration is being halved to 0.1 M. So what I am asking you to calculate is the pH of a 0.1 M solution of sodium acetate (the salt of a weak acid). This will be a basic equilibrium for which you need to use a K b value. Initial: 0.1 M -- 0 0 OAc ( aq ) + H 2 O( l ) HOAc( aq ) + OH ( aq ) @eq: 0.1 – x -- x x Plug the @equilibrium values into the equilbrium expression and solve for x: 9 10 1 ) 1 . 0 ( ) )( ( x x x b K assume that x << 0.1 since K b is so small, 9 10 1 ) 1 . 0 ( ) )( ( x x or x 2 = 1 × 10 10 , or x = [OH ] = 1 × 10 5 so the pOH = 5. BUT THIS IS NOT YOUR ANSWER , since I asked for the pH!! pH = 14 pOH = 9 so the pH = 9 2. (5 pts) Identify whether the following 1:1 solutions will be acidic , basic , or neutral : a) NaOAc/HOAc (pK a = 4.7) ACIDIC (buffer solution, pH = 4.7) b) NH 3 /NH 4 NO 3 (pK a = 9.3) BASIC (buffer solution, pH = 9.3) c) NaCl/KNO 3 Neutral ( not a buffer!! neutral salts!!)
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This note was uploaded on 11/12/2011 for the course CHEM 1422 taught by Professor Staff during the Fall '08 term at LSU.

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1422-hw06-acids-base - CHEM 1422 Homework 6 Acids Bases 2 Due Tuesday ANSWER KEY 1(5 pts 250 mL of 0.2 M acetic acid(HOAc reacts with 250 mL of 0.2

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