notes_Chapter 8 - Chapter 8 Applications of Aqueous...

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Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids 8.8 Solubility Equilibria and The Solubility Product 8.9 Precipitation and Qualitative Analysis 8.10 Complex Ion Equilibria Le Châtelier’s principle for the dissociation equilibrium for HF HF (aq) H + (aq) + F - (aq) Adding any salt with F - : “Common ion effect” K a = [H+][F-] [HF] Molecular model: F-, Na+, HF, H 2 O Like Example 8.1 (P 278-9) - I Nitrous acid, a weak acid (K a = 4.0 x 10 -4 ), is only 2.0% ionized in a 1.0 M solution. Calculate the [H + ], the pH, and the percent dissociation of HNO 2 in a 1.0 M solution that is also 1.0 M in NaNO 2 ! HNO 2(aq) H + (aq) + NO 2 - (aq) [H + ] [NO 2 - ] [HNO 2 ] K a = = 4.0 x 10 -4 Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO 2 ] 0 = 1.0 M [HNO 2 ] = 1.0 – x (from dissolved HNO 2 ) [NO 2 - ] 0 = 1.0 M [NO 2 - ] = 1.0 + x (from dissolved NaNO 2 ) [H + ] 0 = 0 [H + ] = x (neglect the contribution from water) Like Example 8.1 (P 274-5) - II K a = = = 4.0 x 10 -4 [H + ] [NO 2 - ] [HNO 2 ] ( x ) ( 1.0 + x ) (1.0 – x ) Clearly, x must be small as compared to 1.0 to get right so small: x (1.0) (1.0) = 4.0 x 10 -4 or x = 4.0 x 10 -4 = [H + ] Therefore pH = - log [H + ] = - log ( 4.0 x 10 -4 ) = 3.40 The percent dissociation is: 4.0 x 10 -4 1.0 x 100 = 0.040 % Nitrous acid Nitrous acid alone + NaNO 2 [H + ] 2.0 x 10 -2 4.0 x 10 -4 pH 1.70 3.40 % Diss 2.0 0.040 Example 8.2 (P279-82) - I A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2 , K a = 1.8 x 10 -5 ) and 0.50 M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of this solution, and the pH after 0.010 M of solid NaOH is added to 1.0 L of this buffer and to pure water. HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 (aq) K a = 1.8 x 10 -5 = [H + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HC 2 H 3 O 2 ] 0 = 0.50 [HC 2 H 3 O 2 ] = 0.50 – x [C 2 H 3 O 2 - ] 0 = 0.50 [C 2 H 3 O 2 - ] = 0.50 +x [H + ] 0 = ~ 0 [H + ] = x X mol/L of HC 2 H 3 O 2 dissociates to reach equilibrium
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[H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] (x) (0.50) 0.50 Example 8.2 (P279-82) - II K a = 1.8 x 10 -5 = = = ( x ) ( 0.50 + x) 0.50 - x x = 1.8 x 10 -5 The approximation by the 5% rule is fine: [H + ] = x = 1.8 x 10 -5 M and pH = 4.74 To calculate the pH and concentrations after adding the base: OH - + HC 2 H 3 O 2 H 2 O + C 2 H 3 O 2 - Before reaction: 0.010 mol 0.50 mol - 0.50 mol After rxn. (if 100%): 0.010 – 0.010 0.50 – 0.010 - 0.50 + 0.010 = 0 mol = 0.49 mol = 0.51 mol Note that 0.01 mol of acetic acid has been converted to acetate ion by the addition of the base. ~ K
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This note was uploaded on 02/06/2010 for the course CHEM 142 taught by Professor Iandearing during the Spring '10 term at University of Warsaw.

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notes_Chapter 8 - Chapter 8 Applications of Aqueous...

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