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Acid base equilibrium
2.Explain why citrus fruit, which contains citric acid, tastes less sour when their temperature decreased from 30℃ to 5℃.3.Explain why the pH of pure water is less than 7.0 at 323K4.A solution is formed by mixing equal volumes of 0.20M ethanoic acid and 0.20M sodium ethanoate solutions.(a) Identify... 顯示更多 2.Explain why citrus fruit, which contains citric acid, tastes less sour when their temperature decreased from 30℃ to 5℃. 3.Explain why the pH of pure water is less than 7.0 at 323K 4.A solution is formed by mixing equal volumes of 0.20M ethanoic acid and 0.20M sodium ethanoate solutions. (a) Identify all Bronsted acids and all Bronsted bases in the solution. (b)calculate the concentration of each chemical species, excluding H2O, present in the solution at 298K.(Ka of the ethanoic acid at 298K = 1.76×10^ -5)
最佳解答:
2. Citric acid is a tribasic (triprotic) acid. Each citric acid molecule has 3 ionizable H atoms. Denote citric acid as H3A. The pH of citric acid is mainly due to the first ionization of the acid, an endothermic reaction, i.e. H3A(aq) ≒ H+(aq) + H2A-(aq) .. ΔH = +ve When temperature decreases, according to Le Chatelier's principle, the equilibrium position will shifts to the left because the backward reaction exothermic. Due to the decrease in the concentration of H+(aq) ions, the fruit tastes less sour. 3. The self-ionization of pure water is endothermic, i.e. H2O(l) ≒ H+(aq) + OH-(aq) .. ΔH = +ve At 298 K, [H+] = 1 x 10-7 M, and thus pH = -log(1 x 10-7) = 7 When temperature is increased to 323 K, according to Le Chatelier's principle, the equilibrium position will shifts to the right because the forward reaction is endothermic. Due to the increases in H+(aq) ions, [H+] > 1 x 10-7 M. Hence, pH = -log[H+] < -log(1 x 10-7) = 7. pH < 7 4. (a) Bronsted acids: CH3COOH, H+, H2O Bronsted bases: CH3COO-, OH-, H2O (b) CH3COOH(aq) ≒ CH3COO-(aq) + H+(aq) At equilibrium: [CH3COOH] ≈ [CH3COOH]o = 0.2 M [CH3COO-] ≈ [CH3COO-]o = 0.2 M Ka- = (0.2)[H+]/(0.2) = 1.76 x 10-5 [H+] = 1.76 x 10-5 M [OH-] = Kw/[H+] = (1 x 10-14)/(1.76 x 10-5) = 5.68 x 10-10 M Ans: [CH3COOH] = 0.2 M Ans: [CH3COO-] = 0.2 M Ans: [H+] = 1.76 x 10-5 M Ans: [OH-] = 5.68 x 10-10 M =
其他解答:
H+ and OH- itself is not Bronsted acids or Bronsted bases the concentration is also wrong they should be [CH3COOH] = 0.1 M [CH3COO-] = 0.1 M as equal amount of them is mixed ,that 's mean the concentration has halfed|||||Less hydrogen ions are ionized as the temperature decreases from 30 degree celcius to 5 degree celcius At 323 K, more hydrogen ions are releasesd, decreasing pH value No reaction between 2009-03-20 17:20:50 補充： A-level~~~63D0B758E2D502CC
Acid base equilibrium
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發問:2.Explain why citrus fruit, which contains citric acid, tastes less sour when their temperature decreased from 30℃ to 5℃.3.Explain why the pH of pure water is less than 7.0 at 323K4.A solution is formed by mixing equal volumes of 0.20M ethanoic acid and 0.20M sodium ethanoate solutions.(a) Identify... 顯示更多 2.Explain why citrus fruit, which contains citric acid, tastes less sour when their temperature decreased from 30℃ to 5℃. 3.Explain why the pH of pure water is less than 7.0 at 323K 4.A solution is formed by mixing equal volumes of 0.20M ethanoic acid and 0.20M sodium ethanoate solutions. (a) Identify all Bronsted acids and all Bronsted bases in the solution. (b)calculate the concentration of each chemical species, excluding H2O, present in the solution at 298K.(Ka of the ethanoic acid at 298K = 1.76×10^ -5)
最佳解答:
2. Citric acid is a tribasic (triprotic) acid. Each citric acid molecule has 3 ionizable H atoms. Denote citric acid as H3A. The pH of citric acid is mainly due to the first ionization of the acid, an endothermic reaction, i.e. H3A(aq) ≒ H+(aq) + H2A-(aq) .. ΔH = +ve When temperature decreases, according to Le Chatelier's principle, the equilibrium position will shifts to the left because the backward reaction exothermic. Due to the decrease in the concentration of H+(aq) ions, the fruit tastes less sour. 3. The self-ionization of pure water is endothermic, i.e. H2O(l) ≒ H+(aq) + OH-(aq) .. ΔH = +ve At 298 K, [H+] = 1 x 10-7 M, and thus pH = -log(1 x 10-7) = 7 When temperature is increased to 323 K, according to Le Chatelier's principle, the equilibrium position will shifts to the right because the forward reaction is endothermic. Due to the increases in H+(aq) ions, [H+] > 1 x 10-7 M. Hence, pH = -log[H+] < -log(1 x 10-7) = 7. pH < 7 4. (a) Bronsted acids: CH3COOH, H+, H2O Bronsted bases: CH3COO-, OH-, H2O (b) CH3COOH(aq) ≒ CH3COO-(aq) + H+(aq) At equilibrium: [CH3COOH] ≈ [CH3COOH]o = 0.2 M [CH3COO-] ≈ [CH3COO-]o = 0.2 M Ka- = (0.2)[H+]/(0.2) = 1.76 x 10-5 [H+] = 1.76 x 10-5 M [OH-] = Kw/[H+] = (1 x 10-14)/(1.76 x 10-5) = 5.68 x 10-10 M Ans: [CH3COOH] = 0.2 M Ans: [CH3COO-] = 0.2 M Ans: [H+] = 1.76 x 10-5 M Ans: [OH-] = 5.68 x 10-10 M =
其他解答:
H+ and OH- itself is not Bronsted acids or Bronsted bases the concentration is also wrong they should be [CH3COOH] = 0.1 M [CH3COO-] = 0.1 M as equal amount of them is mixed ,that 's mean the concentration has halfed|||||Less hydrogen ions are ionized as the temperature decreases from 30 degree celcius to 5 degree celcius At 323 K, more hydrogen ions are releasesd, decreasing pH value No reaction between 2009-03-20 17:20:50 補充： A-level~~~63D0B758E2D502CC
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