9.2 Chapter 17 Buffers (6 per page)

9.2 Chapter 17 Buffers (6 per page) - Objectives Calculate...

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11/03/2011 1 Chapter 17: Additional Aspects of Acid Base Equilibria Acid-Base Equilibria Objectives Calculate % ionization for weak acids & bases as a function of pH. Understand how buffers work and perform buffer calculations perform buffer calculations. Examine the use and mechanism of pH indicators. Study titrations. 17-1 Common Ion Effects in Acid-Base Equilibria Weak acids and bases are protonated at high [H 3 O + ] (low pH). weak acid: [R-COOH] >> [R-COO - ] weak base: [R-NH 3 + ] >> [R-NH 2 ] Weak acids and bases are deprotonated at low [H 3 O + ] (high pH). weak acid: [R-COOH] << [R-COO - ] weak base: [R-NH 3 + ] << [R-NH 2 ] [H 3 O + ] can be altered by adding other acids and bases: “Common Ion Effect” Swimming Pools are often disinfected with bleach: sodium hypochlorite NaOCl NaOCl Na + + OCl - OCl - + H 2 O HOCl + OH - uncharged, easily crosses cell membranes charged, does not easily cross cell membranes conjugate base of HClO E. coli 17-1 Common Ion Effects OCl - + H 2 O HOCl + OH - Suppose we add a strong acid: HCl + H 2 O Cl - + H 3 O + neutralize Le Chatelier’s Principle HOCl
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11/03/2011 2 17-1 Common Ion Effects HOCl + H 2 O OCl - + H 3 O + HCl + H O C - +H O + Ionization of HOCl and HCl both produce H 3 O + : common ion Adding HCl affects the HOCl OCl - equilibrium: pushes reaction to HOCl. HCl + H 2 O Cl + H 3 O pH affects the balance of [HClO] versus [ClO - ], therefore bacterial growth in swimming pools. Question What percent of ClO - is converted to HClO in a 1.0*10 -3 M solution of NaClO, if K a for HClO is 2.9*10 -8 ? ClO - + H 2 O HClO + OH - Step 1: Find K b for ClO - K a (HClO)* K b (ClO - ) = K w = 10 -14 K b (ClO - ) = 10 -14 2.9*10 -8 = 3.5*10 -7 10 -14 K a = ClO - + H 2 O HClO + OH - Step 2: Calculate extent of hydrolysis of water. Assume all OH - comes from ClO - . initial 10 -3 -x x x final 10 -3 -x x x K b = [HClO][OH - ] [ClO - ] = x 2 (10 -3 -x) assume x<<10 -3 , (10 -3 -x) 10 -3 K b = 3.5*10 -7 = x 2 (10 -3 -x) 5 7 -3 10 * 1.9 10 * 3.5 10 x M =[HClO] =[OH - ] pOH= -log{x} = 4.7 pH = 14 – pOH = 9.3 Quick pH calculation: Step 3: Percent of ClO - converted to HClO %conv = [HClO]*100% [NaClO] initial = x 10 -3 *100%= 1.9% If we add 10 -3 M NaClO to pure water, the pH rises to 9.3 and only 1.9% of ClO - is converted to active HClO.
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11/03/2011 3 Question Now suppose HCl is added until the pH reaches 6.5 . What percent of ClO - is now converted to HClO ? increasing [H 3 O + ] pushes weak acids towards their protonated form Step 1: find [OH - ] pH = 6.5 pOH=14 – pH pOH= 14 – pH = 14 – 6.5 = 7.5 [OH - ] = 10 -7.5 M = 3.2*10 -8 M ClO - + H 2 O HClO + OH - initial 10 -3 -x x Step 2: Use the known value of [OH - ] final 10 -3 -x x 3.2*10 -8 K b = [HClO][OH - ] [ClO - ] = x(3.2*10 -8 ) (10 -3 -x) rearranging: 10 -3 K b –x K b = x(3.2*10 -8 ) 10 -3 K b K b = x(3.2*10 -8 ) x = 10 -3 K b 3.2*10 -8 + K b K b = 3.5*10 -7 x = 9.2*10 -4 M [HClO] Step 3: Percent of ClO - converted to HClO %conv = [HClO]*100% [NaClO] initial = x 10 -3 *100%= 92% Decreasing the pH shifts ClO - to HClO More toxic to bacteria 17-1 Summary Both strong and weak acids produce H 3 O + ions.
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9.2 Chapter 17 Buffers (6 per page) - Objectives Calculate...

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