7/25/2019 Sels de Bases Faibles Monovalentes Et d'Acide Fort

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de bases faibles monovalentes et d'acide fort

//C|/Users/Hp/Desktop/chimie%20des%20solutions/Sels%20de%20bases%20faibles%20monovalentes%20et%20d'acide%20fort.htm[21/02/2016 16:52:

Acides/Bases

Table des

matires

1. INTRODUCTION2. Dissociation de l'eau

et pH3. Concept d'acide et de

base4. Acides polyprotiques

et bases polyvalentes5. Hydrolyse des sels

1. Sels d'acidesfaiblesmonoprotiqueset de base forte

2. Sels de basesfaiblesmonovalenteset d'acide fort

3. Sels d'acidesfaiblesbiprotiques

4. Sels d'acidesfaiblestriprotiques

5. Exercices6. Rsum du calcul du

pH

Sels de bases faibles monovalentes et d'acide fort

Ces sels se dissocient totalement suivant l'expression gnrale :

BHXBH++ X- (22)

Pour les chlorures, X-sera l'ion Cl -prsent la concentration Cselde la solution. La

nature du cation BH+dpend de la base faible dont drive le sel : pour les sels drivs

l'ammoniac, BH+est l'ion ammonium NH4+.

Les ions B+captent une petite quantit d'ions OH-provenant de l'quilibre de dissocia

de l'eau ((1) p.B) ce qui dplace cet quilibre vers la droite. Il se forme donc des ions Hle milieu devient lgrement acide. On peut reprsenter cette raction d'hydrolyse du separ l'quation globale :

BHX + H2OB + X-+ H3O

+ (23)

La relation (13b) p.C5 caractrisant l'quilibre (12) p.C5 peut tre rsolue en considranque [BH+] Cselet [B] [H

+], ce qui fournit :

Ka= = ([B] [H+] / [BH+] [H+]2/ Csel

soit, en passant au logarithme

pH = 1/2 (pKa- log Csel)

(24)

Exemple : pour une solution 0,1 mol/L de chlorure d'ammonium (pKa= 9,25 pour

l'ammoniaque) pH = 1/2 (9,25 - log 0,1) = 5,12.

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