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CHEM數求救!!!(20分)
發問:
1. 40.0 cmcube of an aqueous solution of a weak acid,HX,was titrated with a strong base,MOH(aq),at 298K. The initial pH,before the addition of base,was 2.70. At the equivalence point of the titration,the pH was 8.90.Calculatei) the intial concentration of the acid.ii) the volume of the base added to... 顯示更多 1. 40.0 cmcube of an aqueous solution of a weak acid,HX,was titrated with a strong base,MOH(aq),at 298K. The initial pH,before the addition of base,was 2.70. At the equivalence point of the titration,the pH was 8.90. Calculate i) the intial concentration of the acid. ii) the volume of the base added to reach the equivalence point. iii) the concentration of the base. [the dissociation constant of the weak acid and the ionic product of water at 298 are respectively Ka= 1.8x10^-5 mol dm^-3 , Kw= 1.0x10^-14 mol^-2 dm^-6] 2. Solid silver nitrate was slowly dissolved in a solution Q containing ethanedioate (C2O4 2-) and chromate(VI) ions (CrO4 2-) of concentrations 0.025M and 1.44x10^-5 M respectively. i)When a permanent precipitate of silver ethanedioate first appeared, the concentration of silver ions in the solution was 2.10x10^-5M. Calculate the solubility product Ksp of silver ethanedioate. ii)The dissolving of solid silver nitrate in solution Q was continued until a permanent red ppt of silver chromate(VI) first appeared. Calculate the concentration of silver ions and ethanedioate ions at that instant. Also calculate the number of silver ethanedioate precipitated from 1.00 dm^3 of the solution. [Ksp of silver chromate(VI) is 1.2x10^-12 mol^3 dm^-9] May any CHEM experts here help me to work out the answers!!! I really need them...THANKS SO MUCH!!!
最佳解答:
1. i) Consider the aqueous HX before titration (pH 2.70). At eqm, [H+] = 10-pH = 10-2.7 M staartaaa HX aa+a H2O ≒ H3O+ aaa+aaa X- start aaa y M + aaaaaaaaa 0 M aaaaaa0 M changea -10-2.7 M aH2O a+10-2.7 M aaa+10-2.7 M eqm a((y - 10-2.7) M a H a 10-2.7 M aaa 10-2.7 M Ksp = (10-2.7)(10-2.7)/(y - 10-2.7) = 1.8 x 10-5 Hence, initial concentration of HX, y = 0.2232 M ii) pH 8.9 is due to the hydrolysis of X- at the equivalence point. At equivalence point, pOH = 14 - 8.9 = 5.1 [OH-] = 10-pOH = 10-5.1 staartaaa X- aa+a H2O ≒ aOH- aaa+aaa HX aa(eqm constant = Kh) start aaa w M + aaaaaaaa 0 M aaaaaaaa0 M changea -10-5.1 M aH2O a+10-5.1 M aaa+10-5.1 M eqm a((w - 10-5.1) M a H a 10-5.1 M aaa 10-5.1 M Kh = Kw/Ka = (1 x 10-14)/(1.8 x 10-5) Kh = (10-5.1)(10-5.1)/(w - 10-5.1) = (1 x 10-14)/(1.8 x 10-5) Hence, [X-] at equivalence point, w = 0.1136 M HX + MOH → MX + H2O No. of moles of HX used = MV = 0.2232 x (40/1000) = 0.008928 mol No. of moles of MX formed = 0.008928 mol Vol. of MX solution = mol/M = 0.008928/0.1136 = 0.07859 dm3 = 78.59 cm3 Vol. of MOH added = 78.59 - 40 = 38.59 cm3 iii) HX + MOH → MX + H2O No. of moles of HX used = 0.008928 mol No. of moles of MOH used = 0.008929 mol Vol. of MOH added = 38.59 cm3 = 0.03859 dm3 Molarity of MOH = mol/V = 0.008928/0.03859 = 0.2314 M ========== 2. i) Consider the moment when Ag2C2O4 starts precipitation: [Ag+] = 2.1 x 10-5 M [C2O42-] = 0.025 M Ksp(Ag2C2O4) = [Ag+]2 [C2O42-] = (2.1 x 10-5)2 x 0.025 = 1.103 x 10-11 M3 ii) Consider the moment when AgCrO4 starts precipitation: Consider the precipitation of Ag2CrO4: [CrO42-] = 1.44 x 10-5 M [Ag+]2 = Ksp(Ag2CrO4)/[CrO42-] = (1.2 x 10-12)/(1.44 x 10-5) Hence, [Ag+] = 2.887 x 10-4 M Consider the precipitation of Ag2C2O4: [Ag+] = 2.887 x 10-4 M [C2O42-] = Ksp(Ag2C2O4)/[Ag+]2 = (1.103 x 10-11)/(2.887 x 10-4)2 = 0.000132 M [C2O42-]o = 0.025 M Decrease in [C2O42-] = 0.025 - 0.000132 = 0.02487 M Decrease in no. of moles of C2O42- = MV = 0.02487 x 1 = 0.02487 mol No. of moles of AgC2O4 formed = 0.02487 mol 2008-06-14 23:58:20 補充： The last line should read: No. of moles of Ag2C2O4 formed = 0.02487 mol
i) HA <==> H+ + A- since pH=2.7 [H+] = 10^-2.7 = 0.001995 M (0.001995)(0.001995) / (C-0.001995) = 1.8 X 10^-5 ==> C = 0.22 M no of mole of HA = (0.22)(40X10^-3) = 8.8X10^-3 ii) A- + H2O <==> HA + OH- Since pH = 8.9 [H+] = 10^-8.9 = 1.259X10^-9 and [OH-] = 10^-14 / 1.259X10^-9 = 7.94X10^-6 (7.94X10^-6)(7.94X10^-6)/[A-] = Kh = 1X10^-14 / Ka = 5.56X10^-10 [A-] = 0.113 M Total vol= no.of mole of A- / 0.113 = 8.8X10^-3 / 0.113 = 77.9 cm3 Vol. of basse = 77.9 - 40 = 37.9 cm3 iii) Conc. of base = 8.8X10^-3 / 37.9X10^-3 = 0.232 M 2008-06-14 22:40:55 補充： i) Ksp = (2.1X10^-5)^2 X(0.025) = 1.1 X 10^-11 ii) [Ag+]^2 X 1.44X10^-5 = 1.2X10^-12 [Ag+] = 2.89X10^-4 [Ag+][C2O4 2-] = 1.1X10^-11 (2.89X10^-4)[C2O4 2-] = 1.1X10^-11 [C2O4 2-] = 3.81 X 10^-8
CHEM數求救!!!(20分)
發問:
1. 40.0 cmcube of an aqueous solution of a weak acid,HX,was titrated with a strong base,MOH(aq),at 298K. The initial pH,before the addition of base,was 2.70. At the equivalence point of the titration,the pH was 8.90.Calculatei) the intial concentration of the acid.ii) the volume of the base added to... 顯示更多 1. 40.0 cmcube of an aqueous solution of a weak acid,HX,was titrated with a strong base,MOH(aq),at 298K. The initial pH,before the addition of base,was 2.70. At the equivalence point of the titration,the pH was 8.90. Calculate i) the intial concentration of the acid. ii) the volume of the base added to reach the equivalence point. iii) the concentration of the base. [the dissociation constant of the weak acid and the ionic product of water at 298 are respectively Ka= 1.8x10^-5 mol dm^-3 , Kw= 1.0x10^-14 mol^-2 dm^-6] 2. Solid silver nitrate was slowly dissolved in a solution Q containing ethanedioate (C2O4 2-) and chromate(VI) ions (CrO4 2-) of concentrations 0.025M and 1.44x10^-5 M respectively. i)When a permanent precipitate of silver ethanedioate first appeared, the concentration of silver ions in the solution was 2.10x10^-5M. Calculate the solubility product Ksp of silver ethanedioate. ii)The dissolving of solid silver nitrate in solution Q was continued until a permanent red ppt of silver chromate(VI) first appeared. Calculate the concentration of silver ions and ethanedioate ions at that instant. Also calculate the number of silver ethanedioate precipitated from 1.00 dm^3 of the solution. [Ksp of silver chromate(VI) is 1.2x10^-12 mol^3 dm^-9] May any CHEM experts here help me to work out the answers!!! I really need them...THANKS SO MUCH!!!
最佳解答:
1. i) Consider the aqueous HX before titration (pH 2.70). At eqm, [H+] = 10-pH = 10-2.7 M staartaaa HX aa+a H2O ≒ H3O+ aaa+aaa X- start aaa y M + aaaaaaaaa 0 M aaaaaa0 M changea -10-2.7 M aH2O a+10-2.7 M aaa+10-2.7 M eqm a((y - 10-2.7) M a H a 10-2.7 M aaa 10-2.7 M Ksp = (10-2.7)(10-2.7)/(y - 10-2.7) = 1.8 x 10-5 Hence, initial concentration of HX, y = 0.2232 M ii) pH 8.9 is due to the hydrolysis of X- at the equivalence point. At equivalence point, pOH = 14 - 8.9 = 5.1 [OH-] = 10-pOH = 10-5.1 staartaaa X- aa+a H2O ≒ aOH- aaa+aaa HX aa(eqm constant = Kh) start aaa w M + aaaaaaaa 0 M aaaaaaaa0 M changea -10-5.1 M aH2O a+10-5.1 M aaa+10-5.1 M eqm a((w - 10-5.1) M a H a 10-5.1 M aaa 10-5.1 M Kh = Kw/Ka = (1 x 10-14)/(1.8 x 10-5) Kh = (10-5.1)(10-5.1)/(w - 10-5.1) = (1 x 10-14)/(1.8 x 10-5) Hence, [X-] at equivalence point, w = 0.1136 M HX + MOH → MX + H2O No. of moles of HX used = MV = 0.2232 x (40/1000) = 0.008928 mol No. of moles of MX formed = 0.008928 mol Vol. of MX solution = mol/M = 0.008928/0.1136 = 0.07859 dm3 = 78.59 cm3 Vol. of MOH added = 78.59 - 40 = 38.59 cm3 iii) HX + MOH → MX + H2O No. of moles of HX used = 0.008928 mol No. of moles of MOH used = 0.008929 mol Vol. of MOH added = 38.59 cm3 = 0.03859 dm3 Molarity of MOH = mol/V = 0.008928/0.03859 = 0.2314 M ========== 2. i) Consider the moment when Ag2C2O4 starts precipitation: [Ag+] = 2.1 x 10-5 M [C2O42-] = 0.025 M Ksp(Ag2C2O4) = [Ag+]2 [C2O42-] = (2.1 x 10-5)2 x 0.025 = 1.103 x 10-11 M3 ii) Consider the moment when AgCrO4 starts precipitation: Consider the precipitation of Ag2CrO4: [CrO42-] = 1.44 x 10-5 M [Ag+]2 = Ksp(Ag2CrO4)/[CrO42-] = (1.2 x 10-12)/(1.44 x 10-5) Hence, [Ag+] = 2.887 x 10-4 M Consider the precipitation of Ag2C2O4: [Ag+] = 2.887 x 10-4 M [C2O42-] = Ksp(Ag2C2O4)/[Ag+]2 = (1.103 x 10-11)/(2.887 x 10-4)2 = 0.000132 M [C2O42-]o = 0.025 M Decrease in [C2O42-] = 0.025 - 0.000132 = 0.02487 M Decrease in no. of moles of C2O42- = MV = 0.02487 x 1 = 0.02487 mol No. of moles of AgC2O4 formed = 0.02487 mol 2008-06-14 23:58:20 補充： The last line should read: No. of moles of Ag2C2O4 formed = 0.02487 mol
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其他解答:i) HA <==> H+ + A- since pH=2.7 [H+] = 10^-2.7 = 0.001995 M (0.001995)(0.001995) / (C-0.001995) = 1.8 X 10^-5 ==> C = 0.22 M no of mole of HA = (0.22)(40X10^-3) = 8.8X10^-3 ii) A- + H2O <==> HA + OH- Since pH = 8.9 [H+] = 10^-8.9 = 1.259X10^-9 and [OH-] = 10^-14 / 1.259X10^-9 = 7.94X10^-6 (7.94X10^-6)(7.94X10^-6)/[A-] = Kh = 1X10^-14 / Ka = 5.56X10^-10 [A-] = 0.113 M Total vol= no.of mole of A- / 0.113 = 8.8X10^-3 / 0.113 = 77.9 cm3 Vol. of basse = 77.9 - 40 = 37.9 cm3 iii) Conc. of base = 8.8X10^-3 / 37.9X10^-3 = 0.232 M 2008-06-14 22:40:55 補充： i) Ksp = (2.1X10^-5)^2 X(0.025) = 1.1 X 10^-11 ii) [Ag+]^2 X 1.44X10^-5 = 1.2X10^-12 [Ag+] = 2.89X10^-4 [Ag+][C2O4 2-] = 1.1X10^-11 (2.89X10^-4)[C2O4 2-] = 1.1X10^-11 [C2O4 2-] = 3.81 X 10^-8
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