The equation representing the neutralization of acetic acid, CH 3 COOH, by a base B is CH
Question:
The equation representing the neutralization of acetic acid, CH3COOH, by a base B is CH3COOH(aq) + B(aq) ⇌ CH3COO-(aq) + BH+(aq). Of the bases listed in Table 16.4, which would be effective for neutralizing essentially all of the CH3COOH in a sample, assuming that CH3COOH and B are initially present in equal amounts?
Table 16.4
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TABLE 16.4 lonization Constants of Some Weak Acids and Weak Bases in Water at 25 °C Acid Iodic acid Chlorous acid Chloroacetic acid Nitrous acid Hydrofluoric acid Formic acid Benzoic acid Hydrazoic acid Acetic acid Hypochlorous acid Hydrocyanic acid Phenol Hydrogen peroxide Base Diethylamine Ethylamine Ammonia Hydroxylamine Pyridine Aniline lonization Equilibrium HIO3 + H₂O HClO, + H,O CICH₂COOH + H₂O HNO₂ + H₂O HF + H₂O HCOOH + H,O C6H₂COOH + H₂O HN3 + H₂O CH3COOH + H₂O HOCI + H₂O HCN + H₂O C6H5OH + H₂O H₂O₂ + H₂O (CH,CH,),NH+H,O CH,CH,NH, + H,O NH,+H,O HONH, + H,O C₂H₂N + H₂O CoH;NH, + H,O H3O+ + 103- H3O+ + CIO₂ H3O+ + CICH₂COO- H3O+ + NO₂ H₂O+ + F- H_O* + HCOO- H3O+ + C6H₂COO™ H3O+ + N3 H3O+ + CH3COO- H₂O+ + OCI- H₂O+ + CN- H3O+ + C6H₂O- H3O+ + HO₂ lonization Constant K K₁ = 1.6 X 10-1 1.1 X 10-2 1.4 x 10-3 7.2 x 10-4 6.6 x 10-4 1.8 x 10-4 6.3 x 10-5 1.9 × 10-5 1.8 x 10-5 2.9 × 10-8 6.2 X 10-10 1.0 X 10-10 1.8 × 10-12 Kb = 4.3 x 10-4 (CH3CH₂)₂NH₂+ + OH-6.9 × 10-4 CH3CH₂NH3+ + OH- NH₂+ + OH- HONH3 + + OH- C-H₂NH+ + OH- C6H5NH₂+ + OH 1.8 x 10-5 9.1 x 10-9 1.5 × 10-9 7.4 X 10-10 pk pKa 0.80 1.96 2.85 3.14 3.18 3.74 4.20 4.72 4.74 7.54 9.21 10.00 11.74 = pKb = 3.16 3.37 4.74 8.04 8.82 9.13 Acid strength Base strength
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Related Book For
General Chemistry Principles And Modern Applications
ISBN: 9780132931281
11th Edition
Authors: Ralph Petrucci, Jeffry Madura, F. Herring, Carey Bissonnette
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