9.1 Chapter 17 Buffers (1 per page)

9.1 Chapter 17 Buffers (1 per page) - Chapter 17 Additional...

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hapter 17: Additional Aspects of Chapter 17: Additional Aspects of Acid-Base Equilibria
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Objectives alculate % ionization for weak acids Calculate % ionization for weak acids & bases as a function of pH. nderstand how buffers work and Understand how buffers work and perform buffer calculations. Examine the use and mechanism of pH indicators. Study titrations.
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17-1 Common Ion Effects in Acid-Base Equilibria eak acids and bases are protonated at Weak acids and bases are protonated at high [H 3 O + ] (low pH). eak acid: [R- OOH] >> [R- OO - weak acid: [R COOH] [R COO ] weak base: [R-NH 3 + ] >> [R-NH 2 ] Weak acids and bases are deprotonated at p 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”
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wimming Pools are Swimming Pools are often disinfected with leach: bleach: odium hypochlorite sodium hypochlorite aOCl NaOCl
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NaOCl Na + + OCl - conjugate ase of HClO OCl - + H O HOCl + OH - base of HClO 2 uncharged, asily crosses charged, does ot easily cross easily crosses cell membranes not easily cross cell membranes E. coli
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17-1 Common Ion Effects OCl - + H 2 O HOCl + OH - Suppose we add a strong acid: Cl + H l - H neutralize HCl + H 2 O Cl + H 3 O + Le Chatelier’s Principle HOCl
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17-1 Common Ion Effects OCl + H Cl - H + HOCl + H 2 O OCl + H 3 O HCl + H 2 O Cl - + H 3 O + Ionization of HOCl and HCl both produce H 3 O + : common ion Adding HCl affects the HOCl OCl - equilibrium: pushes reaction to HOCl.
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pH affects the balance of [HClO] versus [ClO - ], therefore bacterial growth in swimming pools.
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Question What percent of ClO - is converted to ClOina1.0*10 -3 solution of NaClO, if HClO in a 1.0 10 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 0 - 4 0 - 4 K b (ClO - ) = 10 14 2.9*10 -8 = 3.5*10 -7 10 14 K a =
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Step 2: Calculate extent of hydrolysis of water. Assume all OH - comes from ClO - . ClO - + H 2 O HClO + OH - itial 0 -3 initial 10 -x x x final 10 -3 -x x x K = [HClO][OH - ] lO - = x 2 0 - ) b [ClO ] (10 3 -x)
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x 2 ssume x<<10 (10 ) 10 K b = 3.5*10 -7 = (10 -3 -x) assume x<<10 -3 , (10 -3 -x) -3 5 7 -3 0 * 9 0 * 5 0 10 1.9 10 3.5 10 x =[HClO] uick pH calculation: M [HClO] =[OH - ] pOH= -log{x} Quick pH calculation: = 4.7 pH = 14 – pOH = 9.3
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Step 3: Percent of ClO - converted to HClO %conv = [HClO]*100% aClO] itial = x 0 -3 *100%= 1.9% [NaClO] initial 10 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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Question Now suppose HCl is added until the pH aches 6 5 What percent of ClO - now reaches 6.5 . What percent of ClO is now converted to HClO ? increasing [H 3 O + ] pushes weak acids towards their protonated form
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Step 1: find [OH - ] pH = 6.5 pOH= 14 – pH 14 5 = 14 – 6.5 = 7.5 [OH - ] = 10 -7.5 M 2*10 -8 = 3. 210 M
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Step 2: Use the known value of [OH - ] ClO - + H 2 O HClO + OH - initial 10 -3 -x x final 10 -3 -x x 3.2*10 -8 = [HClO][OH - ] = x(3.2*10 -8 ) K b [ClO - ] (10 -3 -x) rearranging: 10 -3 K b –x K b = x(3.2*10 -8 )
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10 -3 K b –x K b = x(3.2*10 -8 ) = 10 -3 K b 2*10 K = 3.5*10 -7 x 3.2*10 -8 + K b b =92*10 - ClO] x = 9.
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9.1 Chapter 17 Buffers (1 per page) - Chapter 17 Additional...

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