notes_Chapter_8 - Applications of Aqueous Applications of...

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Unformatted text preview: Applications of Aqueous Applications of Aqueous Equilibria Equilibria Chapter 8 Chapter 8 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 A base swirling in a solution containing phenolphthalein Le Chteliers principle for the dissociation equilibrium for HF HF (aq) H+ (aq) + F- (aq) Molecular model: F-, Na+, HF, H 2 O Like Example 8.1 (P 288-9) - I Nitrous acid, a very weak acid, is only 2.0% ionized in a 0.12 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) K a = = 4.0 x 10-4 [H+] [NO 2-] [HNO 2 ] Initial Concentration (mol/L) Equilibrium Concentration (mol/L) [HNO 2 ] = 1.0 M [HNO 2 ] = 1.0 x (from dissolved HNO 2 ) [NO 2-] = 1.0 M [NO 2-] = 1.0 + x (from dissolved NaNO 2 ) [H+] = 0 [H+] = x (neglect the contribution from water) Like Example 8.1 (P 288-9) - II K a = = = 4.0 x 10-4 [H+] [NO 2-] [HNO 2 ] ( x ) ( 1.0 + x ) (1.0 x ) Assume x is small as compared to 1.0: 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 ) = _________ The percent dissociation is: 4.0 x 10-4 1.0 x 100 = _______ % 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 Human blood is a buffered solution Source: Visuals Unlimited CO 2 (g) , H 2 CO 3 (aq) and HCO 3- (aq) are the buffering components in Blood that hold the pH to a range that will allow Hemoglobin to transport oxygen from the lungs to the cells of the body for metabolism. Example 8.2 (P289-92) - I A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2 , K a = 1.8 x 10-5) and o.50 M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of this solution, and the pH when 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 ] = 0.50 [HC 2 H 3 O 2 ] = 0.50 x [C 2 H 3 O 2-] = 0.50 [C 2 H 3 O 2-] = 0.50 +x [H+] = 0 [H+] = x X mol/L of HC 2 H 3 O 2 dissociates to reach equilibrium ~ Example 8.2 (P289-92) - II K a = 1.8 x 10-5 = = = [H+][C 2 H 3 O 2-] [HC 2 H 3 O 2 ] ( x ) ( 0.50 + x) 0.50 - x (x) (0.50) 0.50 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 reaction: 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....
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notes_Chapter_8 - Applications of Aqueous Applications of...

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