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pH = pKa + log

pH = pKa + log. [CH 3 COO - ] = =. [CH 3 COOH] = =. = × = = 2. Donc pKa = pH – log 2 = 4,7. pH = 5 ± 0,2. Donc Δ pH = 0,2 On en déduit d’après le texte que Δ pKa = 0,2 aussi. Donc pKa = 4,7 ± 0,2

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pH = pKa + log

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  1. pH = pKa + log [CH3COO-] = = [CH3COOH] = = = × = = 2 Donc pKa = pH – log 2 = 4,7

  2. pH = 5 ± 0,2 Donc ΔpH = 0,2 On en déduit d’après le texte que ΔpKa = 0,2 aussi. Donc pKa = 4,7 ± 0,2 On peut donner l’encadrement suivant : 4,5 < pKa < 4,9

  3. Une solution tampon est une solution dont le pH varie peu lors d’un ajout modéré d’acide ou de base ou lors d’une dilution modérée (ajout d’eau). Quelques gouttes de vinaigre font donc varier le pH de l’eau distillée mais pas celui d’une solution tampon. Le pH du vinaigre est compris entre 2,0 et 3,0 tandis que le pKa du couple CH3COOH / CH3COO- est de 4,8. pH < pKa donc c’est la forme acide CH3COOH qui est majoritaire dans le vinaigre.

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