| Lösungen |
|
| 1)
a) 36,5 g HCl = 1 mol => 0,365 g HCl = 0,01 mol => 10-2 mol HCl (starke Säure: Konzentration H3O+-Ionen = Konzentration Säure) cH3O+ = 10-2 mol/l => pH 2 b) 3,65 g HCl = 0,1 mol in 100 ml => 1 mol in 1 Liter => cH3O+ = 100 mol/l => pH 0 c) 63 g HNO3 = 1 mol => 0,063 g HNO3 = 10-3 mol => cH3O+ = 10-3 mol/l => pH 3 d) 0,315 g HNO3 = 0,005 mol in 500 ml => 0,01 mol in 1 Liter => cH3O+ = 10-2 mol/l => pH 2 e) 40 g NaOH = 1 mol => 0,4 g = 0,01 mol => 10-2 mol NaOH => cOH- = 10-2 mol/l => pOH = 2 => pH 12 f) 2 g NaOH = 0,05 mol in 500 ml => 0,1 mol in 1 Liter => cOH- = 10-1 mol/l => pOH 1 => pH 13 g) 56 g KOH = 1 mol => 0,56 g = 0,01 mol => 10-2 mol KOH => cOH- = 10-2 mol/l => pOH = 2 => pH 12 2) a) pH 3 => cH3O+ = 10-3 mol/l => 10-3 mol HCl = 0,0365 g HCl b) pH 12 = pOH 2 => cOH- = 10-2 mol/l => 10-2 mol KOH = 56 x 10-2 g/l = 0,56 g/l 250 ml = 1/4 von 1000 ml = 0,14 g KOH |
|
|
|
|
|