6. Zuren en basen
6. Zuren en basen
6. Zuren en basen
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pH van oplossing<strong>en</strong> van sterke zur<strong>en</strong><br />
• HCl + H 2O H 3O + + Cl - K Z = 10 7<br />
[HCl] = 0,1 mol/l ⇒ pH = 1<br />
[HCl] = 0,01 mol/l ⇒ pH = 2<br />
• H 2SO 4 + H 2O H 3O + + HSO 4 - K Z = 10 3<br />
HSO 4 - + H2O H 3O + + SO 4 2- K Z = 1,2.10 -2<br />
2 de ionisatiegraad α = 0,5<br />
[H 2SO 4] 0 = 0,1 mol/l ⇒ [H 3O + ] e = 0,15 mol/l<br />
⇒ pH = 0,82
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