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curva di titolazione acido debole-base forte

curva di titolazione acido debole-base forte. CH 3 COOH H + + CH 3 COO -. [H + ] [ CH 3 COO - ] [ CH 3 COO H]. [H+] 1.34x10 -3. K a =. V(ml) pH 0 2.87. x x 0.1-x. K a =. x = 1.8 . 10 -5. 0.1. 100 mL di CH 3 COOH 0.1 M con NaOH 0.1 M.

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curva di titolazione acido debole-base forte

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  1. curva di titolazione acido debole-base forte CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] [H+] 1.34x10-3 Ka = V(ml) pH 0 2.87 x x 0.1-x Ka = x = 1.8.10-5. 0.1 100 mL di CH3COOH 0.1 M con NaOH 0.1 M i 0. 1 M / / e 0.1 -x x x x= 1.34x10-3 pH = - log [H+] = - log (1.34.10-3) = 2.87

  2. 100 mL di CH3COOH 0.1 M con 1 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (9.9.10-4 + x) x 0.098-x i 9.9x10-3 / 10-4 0.101 0.101 Ka = 9.9.10-4x +x2 0.098 1.8.10-5 = -9.9.10-4± (9.9.10-4)2+ 4.1.76.10-6 x= = 9.2.10-4 2 nNaOH = 1 x 10-3 x 0.1 = 10-4 nCH3COOH = 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 + 9.9x 10-4 x2 + 9.9.10-4x - 1.76.10-6= 0 pH = - log (9.2.10-4) pH= 3.03

  3. [H+] 1.34x10-3 9.2 x 10-4 V(ml) pH 0 2.87 1 3.03

  4. 100 mL di CH3COOH 0.1 M con 5 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (4.8.10-3 + x) x 0.090-x i 9.5x10-3 / 5x10-4 0.105 0.105 Ka = 4.8.10-3x +x2 0.090 1.8.10-5 = -4.8.10-3± (4.8.10-3)2+ 4.1.63.10-6 x= = 3.2.10-4 2 nNaOH = 5 x 10-3 x 0.1 = 5x10-4 nCH3COOH = 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 + 4.8x 10-3 x2 + 4.8.10-3x - 1.63.10-6= 0 pH = - log (3.2.10-4) pH= 3.5

  5. [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 V(ml) pH 0 2.87 1 3.03 5 3.5

  6. 100 mL di CH3COOH 0.1 M con 10 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (9.1.10-3 + x) x 0.082-x i 9 x10-3 / 10-3 0.11 0.11 Ka = 9.1.10-3x +x2 0.082 1.8.10-5 = -9.1.10-3± (9.1.10-3)2+ 4.1.47.10-6 x= = 1.6.10-4 2 nNaOH = 10 x 10-3 x 0.1 = 10-3 nCH3COOH = 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 + 9.1x 10-3 x2 + 9.1.10-3x - 1.47.10-6= 0 pH = - log (1.6.10-4) pH= 3.8

  7. [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8

  8. 100 mL di CH3COOH 0.1 M con 25 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (2.10-2 + x) x 0.06-x i 7.5 x10-3 / 2.5x10-3 0.125 0.125 Ka = 2.10-2 x 0.06 1.8.10-5 = 1.8.10-5.0.06 2.10-2 x = = 5.4.10-5 nNaOH = 25 x 10-3 x 0.1 = 2.5x10-3 nCH3COOH = 0.100 x 0.1 = 10-2 i 10-2 2.5x10-3 / f 7.5x 10-3 / 2.5x10-3 2.5x10-3 e 0.06 -x x x + 2 x 10-2 pH = - log (5.4.10-5) pH= 4.27

  9. [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 5.4 x 10-5 V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8 25 4.27

  10. 100 mL di CH3COOH 0.1 M con 50 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (3.3.10-2+x)x 3.3.10-2-x i 5 x10-3 / 5x10-3 0.15 0.15 Ka = 3.3.10-2 x 3.3.10-2 1.8.10-5 = nNaOH = 50 x 10-3 x 0.1 = 5 x10-3 nCH3COOH = 0.100 x 0.1 = 10-2 i 10-2 5x10-3 / f 5x 10-3 / 5x10-3 5x10-3 e 3.3x10-2 -x x x + 3.3 x 10-2 pH = - log (1.8.10-5) x = 1.8.10-5 pH= 4.74

  11. [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 5.4 x 10-5 1.8 x 10-5 V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8 25 4.27 50 4.74

  12. 100 mL di CH3COOH 0.1 M con 75 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O CH3COOH H+ + CH3COO- [H+] [CH3COO-] [CH3COOH] Ka = (4.3.10-2+x)x 1.4.10-2-x i 2.5 x10-3 / 7.5x10-3 0.175 0.175 Ka = 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 = nNaOH = 75 x 10-3 x 0.1 = 7.5 x10-3 nCH3COOH = 0.100 x 0.1 = 10-2 i 10-2 7.5x10-3 / f 2.5x 10-3 / 7.5x10-3 7.5 x10-3 e 1.4x10-2 -x x x + 4.3 x 10-2 pH = - log (5.9.10-6) pH= 5.23 x = 5.9.10-6

  13. [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 5.4 x 10-5 1.8 x 10-5 5.9 x 10-6 V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8 25 4.27 50 4.74 75 5.23

  14. 100 mL di CH3COOH 0.1 M con 100 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O [HO-] [CH3COOH] [CH3COO-] Kb = CH3COO-+ H2O CH3COOH + OH- x x 5.10-2-x i 10-2 / / 0.2 Kb = x2 5.10-2 5.55.10-10 = x = 5.55.10-10.5.10-2 nNaOH = 0.100 x 0.1 = 10-2 nCH3COOH = 0.100 x 0.1 = 10-2 i 10-2 10-2 / f / / 10-2 10-2 e 5 x10-2 -x x x pOH = - log (5.27.10-6) pOH= 5.28 x = 5.27.10-6 pH = 14 -pOH= 8.72

  15. V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8 25 4.27 50 4.74 75 5.23 100 8.72 [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 5.4 x 10-5 1.8 x 10-5 5.9 x 10-6 1.9 x 10-9 punto equivalente moli di acido = moli di base

  16. 100 mL di CH3COOH 0.1 M con 110 mL NaOH 0.1 M CH3COOH+NaOH CH3COO- +Na+ + H2O NaOH Na+ + OH- i 10-3 / / 0.21 nNaOH = 0.110 x 0.1 = 1.1x10-2 nCH3COOH = 0.100 x 0.1 = 10-2 i 10-2 1.1x10-2 / f / 10-3 10-2 10-2 La soluzione contiene una base forte ed una base debole in concentrazioni simili. Trascuro il contributo all’OH- da parte dello ione acetato f / 4.76 x10-3 4.76 x10-3 pOH = - log (4.76.10-3) pOH= 2.32 pH = 14 -pOH= 11.68

  17. V(ml) pH 0 2.87 1 3.03 5 3.5 10 3.8 25 4.27 50 4.74 75 5.23 100 8.72 110 11.68 [H+] 1.34 x10-3 9.2 x 10-4 3.2 x 10-4 1.6 x 10-4 5.4 x 10-5 1.8 x 10-5 5.9 x 10-6 1.9 x 10-9 2.1 x 10-12 . . . . . .

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