6. Zuren en basen

6. Zuren en basen 6. Zuren en basen

10.09.2013 Views

Evenwicht NaOH-oplossing • NaOH + H 2O Na(H 2O) 6 + + OH - • Volledige ionisatie, evenwicht naar rechts • Vb1: [NaOH] 0 = 0,1 mol/l ⇒ [OH - ] e = 0,1 mol/l ⇒ pOH = 1 ⇒ pH = 13 • Vb2: [NaOH] = 0,01 mol/l ⇒ pH = 12

Evenwicht NH 3-oplossing • NH 3 + H 2O NH 4 + + OH - • Beperkte ionisatie, evenwicht naar links • Vb1: [NH 3] 0 = 0,1 mol/l [NH 3] e = 0,09867 mol/l [NH 4 + ] e = [OH - ] e = 0,00133 mol/l ⇒ pOH = 2,88 ⇒ pH = 11,12 • K B = [OH - ] e.[NH 4 + ]e /[NH 3] e (base-cte) = 0,00133 ² / 0,09867 = 1,79 . 10 -5 • Kleine baseconstante: evenwicht naar links • Vb2: [NH 3] = 0,01 mol/l ⇒ pH = 10,62

Ev<strong>en</strong>wicht NH 3-oplossing<br />

• NH 3 + H 2O NH 4 + + OH -<br />

• Beperkte ionisatie, ev<strong>en</strong>wicht naar links<br />

• Vb1: [NH 3] 0 = 0,1 mol/l<br />

[NH 3] e = 0,09867 mol/l<br />

[NH 4 + ] e = [OH - ] e = 0,00133 mol/l<br />

⇒ pOH = 2,88 ⇒ pH = 11,12<br />

• K B = [OH - ] e.[NH 4 + ]e /[NH 3] e (base-cte)<br />

= 0,00133 ² / 0,09867 = 1,79 . 10 -5<br />

• Kleine baseconstante: ev<strong>en</strong>wicht naar links<br />

• Vb2: [NH 3] = 0,01 mol/l ⇒ pH = 10,62

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