Autoprotolyse

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2.6.2. D’après l’équation chimique [HCO3–(aq)]éq = [H3O+]éq Or [H3O+] = 10–pH

ainsi [HCO3–(aq)]éq = 10–5,7 = 2,0(10–6 mol.L–1

2.6.3. D’après 2.2. KA = [pic]avec [HCO3–(aq)]éq = [H3O+]éq = 10–pH,

il vient KA = [pic] donc [CO2,H2O]éq = [pic]. Et Enfin KA = 10–pKA,

soit [CO2,H2O]éq = [pic] et donc [CO2,H2O]éq = 10(6,4 –2(5,7) = 10–5,0 = 1,0(10–5 mol.L–1

2.6.4. [CO2,H2O]éq = [pic] = c – [H3O+]éq = c – 10–pH

c = [CO2,H2O]éq + 10–pH c = 1,0(10–5 + 10–5,7 = 1,2(10–5 mol.L–1

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pH

pKA

6,4

CO2,H2O

HCO3–