concentration at equilibrium can be determined from the pH which is always

Concentration at equilibrium can be determined from

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concentration at equilibrium can be determined from the pH, which is always given at equilibrium. The initial concentration of H3O+and A-is zero. HA H2O H3O+A-Initial 0.100 0 0 Change Equilibrium 6.31 x 10-4In order to reach equilibrium the HA dissociates and, correspondingly, H3O+and A-form. The concentration of HA that dissociates, and correspondingly, the concentrations of H3O+ and A-that form are equal to X. HA H2O H3O+A-Initial 0.100 0 0 Change -X -X +X +X Equilibrium 6.31 x 10-40 + X = 6.31 x 10-4X = 6.31 x 10-4HA H2O H3O+A-Initial 0.100 0 0 Change -6.31 x 10-4+6.31 x 10-4+6.31 x 10-4Equilibrium 0.100-(6.31 x 10-4) 6.31 x 10-4+6.31 x 10-4The concentrations at equilibrium are substituted into the equilibrium expression to find K: ([H3O+][A-])/[HA] = K ([6.31 x 10-4][ 6.31 x 10-4])/[0.100 - (6.31 x 10-4)] = K 4.01 x 10-6= K
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14629. Chem 162-2004 Final Exam + Answers Chapter 14 - Acids and Bases Acid and base concepts and equilibria A 0.340-g sample of Ba(OH)2(molar mass = 171.3 g/mol) is dissolved in water. What volume of 0.150 M HCl is required to completely neutralize the Ba(OH)2solution? A. 30.2 ml. B. 26.5 mL C. 25.0 ml D. 45.2 mL E. 53.0 mL Ba(OH)2+ 2HCl → 2H2O + Ba2++ 2Cl-Plan: g Ba(OH)2→ mol Ba(OH)2→ mol HCl → V HCl0.340 g Ba(OH)2 x (1 mol Ba(OH)2/171.3 g Ba(OH)2) x (2 mol HCl/mol Ba(OH)2) = 0.003970 mol HCl 0.003970 mol HCl/(0.150 mol/L) = 0.02646 L = 26.46 mL HCl 14. Chem 162-2004 Final Exam + Answers Chapter 14 - Acids and Bases Acid and base concepts and equilibria Calculate the pH of a 0.10 M solution of pyridine (C5H5N, Kb= 1.7 x 10-9)A. 7.83 B. 8.69 C. 11.30 D. 9.12 E. 10.71 RNH2+ H2O RNH3++ OH-RNH2H2O RNH3+OH-Initial 0.10 0 0 Change -X +X +X Equilibrium 0.10-X +X +X ([RNH3+][OH-])/[RNH2] = 1.7 x 10-9([X][X])/[0.10-X] = 1.7 x 10-9X = 1.304 x 10-5= [OH-] [H+] = (1 x 10-14)/(1.304 x 10-5) = 7.67 x 10-10pH = -log(7.67 x 10-10) = 9.11
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146314. Chem 162-2005 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka, Kb, pH, pOH, % dissoc calculations Sulfuric acid is a diprotic acid for which Ka1is very large and Ka2= 0.012. Which of the following species has the largest concentration in 0.10 M H2SO4? A. OH-B. SO42-C. HSO4-D. H+E. H2SO4Step 1: H2SO4is a strong acid with a very large K, so bring the reaction to completion. H2SO4+ H2O H3O++ HSO4-Initial 0.10 0 0 Change -0.10 +0.10 +0.10 Equilibrium 0 +0.10 +0.10 Step 2: HSO4-is a weak acid, so bring the reaction to equilibrium. HSO4-+ H2O H3O++ SO42-Initial 0.10 0.10 0 Change -X +X +X Equilibrium 0.10 - X 0.10 + X +X ([H3O+] x [SO42-])/([HSO4-]) = 0.012 ([0.10 + X] x [X])/([0.10 - X]) = 0.012 Simplify. Although 0.012 is not small enough to warrant dropping the ―X‖, since we are looking for a qualitative answer let‘s drop the X anyway.([0.10] x [X])/([0.10]) = 0.012 X ≈ 0.012A. Since [H3O+] >0.10M, then [OH-] < (1 x 10-14)/(0.10) < 10-13M, which is very small. B. [SO42-] ≈ 0.012C. [HSO4-] < 0.10 M D. [H+] = [H3O+] = (0.10 + X) >0.10 M E. [H2SO4] = 0
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146418. Chem 162-2005 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka, Kb, pH, pOH, % dissoc calculations A monoprotic weak acid has Ka= 2.0 x 10-4. What is the percent ionization of a 0.50 M solution of this acid? A. 2.0% B. 0.50% C. 5.0% D. 10% E. 1.0% HA + H2O H3O++ A-Initial 0.50 0 0 Change -X +X +X Equilibrium 0.50 - X +X +X ([H3O+] x [A-])/[HA] = 2.0 x 10-4([X] x [X])/[0.50 - X] = 2.0 x 10-4Simplify: ([X] x [X])/[0.50] = 2.0 x 10-4X = 1.0 x 10-2Molar percent dissociation = ((1.0 x 10-2)/0.50) x 100 = 2.0 molar percent dissociation 24. Chem 162-2005 Exam II Acids and Bases - Chapter 14 Solutions of acids and bases, Ka, Kb, pH, pOH, % dissoc calculations Acid HX has Ka = 2.0 x 10-5
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