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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 AcidBase 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 Châtelier’s 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 2889)  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 104 [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 2889)  II K a = = = 4.0 x 104 [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 104 or x = 4.0 x 104 = [H+] Therefore pH =  log [H+] =  log ( 4.0 x 104 ) = _________ The percent dissociation is: 4.0 x 104 1.0 x 100 = _______ % Nitrous acid Nitrous acid alone + NaNO 2 [H+] 2.0 x 102 4.0 x 104 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 (P28992)  I A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2 , K a = 1.8 x 105) 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 105 = [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 (P28992)  II K a = 1.8 x 105 = = = [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 105 The approximation by the 5% rule is fine: [H+] = x = 1.8 x 105 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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 Spring '07
 ZOLLER,WILLIAMH
 Bases, pH

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