1 / 17

curva di titolazione base debole-acido forte

curva di titolazione base debole-acido forte. H 2 O + NH 3 OH - + NH 4 +. [OH - ] [ NH 4 + ] [ NH 3 ]. [H+] 7.41x10 -12. K b =. V(ml) pH 0 11.13. x x 0.1-x. K b =. 100 mL di NH 3 0.1 M con HCL 0.1 M.

blake
Download Presentation

curva di titolazione base debole-acido forte

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. curva di titolazione base debole-acido forte H2O + NH3 OH- + NH4+ [OH-] [NH4+] [NH3] [H+] 7.41x10-12 Kb = V(ml) pH 0 11.13 x x 0.1-x Kb = 100 mL di NH3 0.1 M con HCL 0.1 M i 0. 1 M / / e 0.1 -x x x x= 1.34x10-3 pOH= - log (1.34.10-3) = 2.87 pH = 14- pOH =11.13

  2. 100 mL di NH3 0.1 M con 1 mL di HCL 0.1 M NH3+ HCl NH4+ +Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (9.9.10-4 + x) x 0.098-x i 9.9x10-3 / 10-4 0.101 0.101 Kb = 9.9.10-4x +x2 0.098 1.8.10-5 = nHCl = 1 x 10-3 x 0.1 = 10-4 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 10-4 / f 9.9x 10-3 / 10-4 10-4 e 0.098 -x x x + 9.9.10-4 x2 + 9.9.10-4x - 1.76.10-6= 0 pOH = - log (9.2.10-4) pOH= 3.03 pH= 14-pOH= 10.97

  3. [H+] 7.41x10-12 1.07x10-11 V(ml) pH 0 11.13 1 10.97

  4. 100 mL di NH3 0.1 M con 5 mL di HCL 0.1 M NH3+ HCl NH4+ +Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (4.8.10-3 + x) x 0.090-x i 9.5x10-3 / 5x10-4 0.105 0.105 Kb = 4.8.10-3x +x2 0.090 1.8.10-5 = nHCl = 5 x 10-3 x 0.1 = 5x10-4 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 5x10-4 / f 9.5x 10-3 / 5x10-4 5x10-4 e 0.090 -x x x + 4.8.10-3 x2 + 4.8.10-3x - 1.63.10-6= 0 pOH = - log (3.2.10-4) pOH= 3.5 pH=14-pOH= 10.5

  5. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2

  6. 100 mL di NH3 0.1 M con 10 mL di HCL 0.1 M NH3+ HCl NH4+ +Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (9.1.10-3 + x) x 0.082-x i 9 x10-3 / 10-3 0.110 0.110 Kb = 9.1.10-3x +x2 0.082 1.8.10-5 = nHCl = 10 x 10-3 x 0.1 = 10-3 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 10-3 / f 9x 10-3 / 10-3 10-3 e 0.082 -x x x + 9.1. 10-3 x2 + 9.1.10-3x - 1.47.10-6= 0 pOH = - log (1.6.10-4) pOH= 3.8 pH= 14-pOH= 10.2

  7. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2

  8. 100 mL di NH3 0.1 M con 25 mL di HCL 0.1 M NH3+ HCl NH4+ + Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (2.10-2 + x) x 0.06-x i 7.5x10-3 / 2.5x10-3 0.125 0.125 Kb = 2.10-2 x 0.06 1.8.10-5 = 1.8.10-5.0.06 2.10-2 x = = 5.4.10-5 nHCl = 25 x 10-3 x 0.1 = 2.5x10-3 nNH3= 0.100 x 0.1 = 10-2 i 10-2 2.5x10-3 / f 7.5x 10-3 / 2.510-3 2.5x10-3 e 0.06 -x x x + 2.10-2 pOH = - log (5.4.10-5) pOH= 4.27 pH=14-pOH= 9.73

  9. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 1.86x10-10 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2 25 9.73

  10. 100 mL di NH3 0.1 M con 50 mL di HCL 0.1 M NH3+ HCl NH4+ + Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (3.3.10-2+x)x 3.3.10-2-x i 5x10-3 / 5x10-3 0.15 0.15 Kb = 3.3.10-2 x 3.3.10-2 1.8.10-5 = nHCl= 50 x 10-3 x 0.1 = 5 x10-3 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 5x10-3 / f 5x 10-3 / 5x10-3 5x10-3 e 3.3.10-2 -x x x + 3.3 . 10-2 pOH = - log (1.8.10-5) x = 1.8.10-5 pOH= 4.74 pH=14-pOH= 9.26

  11. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 1.86x10-10 5.49x10-10 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2 25 9.73 50 9.26

  12. 100 mL di NH3 0.1 M con 75 mL di HCL 0.1 M NH3+ HCl NH4+ + Cl- [OH-] [NH4+] [NH3] Kb = H2O + NH3 OH- + NH4+ (4.3.10-2+x)x 1.4.10-2-x i 2.5x10-3 / 7.5x10-3 0.175 0.175 Kb = 4.3.10-2 x 1.4.10-2 1.8.10-5 = 1.4.10-2 .1.8.10-5 4.3.10-2 x = nHCl= 75 x 10-3 x 0.1 = 7.5 x10-3 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 7.5x10-3 / f 2.5x 10-3 / 7.5x10-3 7.5x10-3 e 1.4.10-2 -x x x + 4.3 . 10-2 pOH = - log (5.9.10-6) x = 5.9.10-6 pOH= 5.23 pH =14-pOH= 8.77

  13. [H+] 1.34x10-3 9.2 x 10-4 3.1610-11 6.3x10-11 1.86x10-10 5.49x10-10 1.7x 10-9 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2 25 9.73 50 9.26 75 8.77

  14. 100 mL di NH3 0.1 M con 100 mL di HCL 0.1 M NH3+ HCl NH4+ +Cl- [H+] [NH3] [NH4+] Ka = NH4+ H+ + NH3 i 10-2 / / 0.2 x x 5.10-2-x Ka = x2 5.10-2 5.55.10-10 = nHCl = 0.100 x 0.1 = 10-2 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 10-2 / f / / 10-2 10-2 e 5.10-2 -x x x pH = - log (5.27.10-6) pH= 5.28 x = 5.27.10-6

  15. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 1.86x10-10 5.49x10-10 1.7x 10-9 5.27x10-6 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2 25 9.73 50 9.26 75 8.77 100 5.28 punto equivalente moli di acido = moli di base

  16. 100 mL di NH3 0.1 M con 110 mL di HCL 0.1 M NH3+ HCl NH4+ + Cl- HCl H+ + Cl- i 10-3 / / 0.21 nHCl = 0.110 x 0.1 = 1.1x10-2 nNH3 = 0.100 x 0.1 = 10-2 i 10-2 1.1x10-2 / / f / 10-3 10-2 10-2 f / 4.76 x10-3 4.76 x10-3 pH = - log (4.76.10-3) pH= 2.32

  17. [H+] 7. 41x10-12 1.07 x10-11 3.1610-11 6.3x10-11 1.86x10-10 5.49x10-10 1.7x 10-9 5.27x10-6 4.76x10-3 V(ml) pH 0 11.13 1 10.97 5 10.5 10 10.2 25 9.73 50 9.26 75 8.77 100 5.28 110 2.32 0 50 100 150 volume di acido aggiunto/mL

More Related