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標題:
acid base equilicrium(focus on assumption problem)
發問:
(a)Calculate the pH of a 0.1M solution of propanoic,C2H5COOH (Ka of C2H5COOH=1.3^10-5 mol dm^-3)(b)Find the pH of 0.05 mole of sodium propanoate, C2H5COONa, is completely dissolved in 1 dm^3 in (a) above.For part (b) , I am not focus on the answer, just focus on how to assume, likeKa is very small... 顯示更多 (a)Calculate the pH of a 0.1M solution of propanoic,C2H5COOH (Ka of C2H5COOH=1.3^10-5 mol dm^-3) (b)Find the pH of 0.05 mole of sodium propanoate, C2H5COONa, is completely dissolved in 1 dm^3 in (a) above. For part (b) , I am not focus on the answer, just focus on how to assume, like Ka is very small then......??(please give the full steps of assumptions) 更新: sorry, typo: (Ka of C2H5COOH=1.3*10^-5 mol dm^-3)
最佳解答:
In both (a) and (b), it is assumed that the concentration of H3O+(aq) ions and OH-(aq) ions formed from the self-ionization of water are negligible when compared with the dissociation of C2H5COOH. ======== (a) start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq) start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0 MOO-(aq) + H0 M change aaa -y MOH(aq) + H2O(l) ≒ 2H+y M O-(aq) +H+y M eqm aaa(0.1 - y) M(aq) + H2O(l) ≒ aC2y M O-(aq) + H3y M Ka = y2 / (0.1 - y) = 1.3 x 10-5 mol dm-3 y2 + (1.3 x 10-5)y - (1.3 x 10-6) = 0 [H3O+] = y = 1.13 x 10-3 M pH = -log(1.13 x 10-3) = 2.95 ======== (b) start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq) start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0.05 MOO-( + H0 M change aaa -w MOH(aq) + H2O(l) ≒ 2H+w M O-(aq) +H+w M eqm aaa(0.1 - w) M(aq) + H2O(l) aC2(0.05 + w) M O-(aHw M Assumption: Since Ka is small and the presence of CH3COO-(aq) would shift the equilibrium position to the left, it is assumed that w is very small. Therefore: [C2H5COOH] = (0.1 - w) M ≈ 0.1 M Therefore: [C2H5COO-] = (0.05 + w) M ≈ 0.05 M Ka = (0.05 + w)w/(0.1 - w) ≈ 0.05w/0.1 = 1.3 x 10-5 mol dm-3 [H3O+] = w = 2.6 x 10-5 M pH = -log(2.6 x 10-5) = 4.59
其他解答:
acid base equilicrium(focus on assumption problem)
發問:
(a)Calculate the pH of a 0.1M solution of propanoic,C2H5COOH (Ka of C2H5COOH=1.3^10-5 mol dm^-3)(b)Find the pH of 0.05 mole of sodium propanoate, C2H5COONa, is completely dissolved in 1 dm^3 in (a) above.For part (b) , I am not focus on the answer, just focus on how to assume, likeKa is very small... 顯示更多 (a)Calculate the pH of a 0.1M solution of propanoic,C2H5COOH (Ka of C2H5COOH=1.3^10-5 mol dm^-3) (b)Find the pH of 0.05 mole of sodium propanoate, C2H5COONa, is completely dissolved in 1 dm^3 in (a) above. For part (b) , I am not focus on the answer, just focus on how to assume, like Ka is very small then......??(please give the full steps of assumptions) 更新: sorry, typo: (Ka of C2H5COOH=1.3*10^-5 mol dm^-3)
最佳解答:
In both (a) and (b), it is assumed that the concentration of H3O+(aq) ions and OH-(aq) ions formed from the self-ionization of water are negligible when compared with the dissociation of C2H5COOH. ======== (a) start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq) start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0 MOO-(aq) + H0 M change aaa -y MOH(aq) + H2O(l) ≒ 2H+y M O-(aq) +H+y M eqm aaa(0.1 - y) M(aq) + H2O(l) ≒ aC2y M O-(aq) + H3y M Ka = y2 / (0.1 - y) = 1.3 x 10-5 mol dm-3 y2 + (1.3 x 10-5)y - (1.3 x 10-6) = 0 [H3O+] = y = 1.13 x 10-3 M pH = -log(1.13 x 10-3) = 2.95 ======== (b) start aaaC2H5COOH(aq) + H2O(l) ≒ C2H5COO-(aq) + H3O+(aq) start aaaaa 0.1 MH(aq) + H2O(l) ≒ C2H0.05 MOO-( + H0 M change aaa -w MOH(aq) + H2O(l) ≒ 2H+w M O-(aq) +H+w M eqm aaa(0.1 - w) M(aq) + H2O(l) aC2(0.05 + w) M O-(aHw M Assumption: Since Ka is small and the presence of CH3COO-(aq) would shift the equilibrium position to the left, it is assumed that w is very small. Therefore: [C2H5COOH] = (0.1 - w) M ≈ 0.1 M Therefore: [C2H5COO-] = (0.05 + w) M ≈ 0.05 M Ka = (0.05 + w)w/(0.1 - w) ≈ 0.05w/0.1 = 1.3 x 10-5 mol dm-3 [H3O+] = w = 2.6 x 10-5 M pH = -log(2.6 x 10-5) = 4.59
其他解答:
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