6. Zuren en basen
6. Zuren en basen 6. Zuren en basen
pH van oplossingen van zwakke basen • NH 3 + H 2O NH 4 + + OH - K B = [OH - ] e.[NH 4 + ]e /[NH 3] e = 1,78.10 -5 [NH 3] e = [NH 3] 0 en [NH 4 + ]e = [OH - ] e ⇒ [OH - ] = √ K B.[NH 3] 0 [NH 3] = 0,1 mol/l ⇒ pOH = 2,88 ⇒ pH = 11,12 [NH 3] = 0,01 mol/l ⇒ pOH = 3,38 ⇒ pH = 10,62 • 1 t.e.m. 3 Oefeningen p. 133
pH van oplossingen van zouten • MX M + + X - • Soms: M + = Brønstedzuur X - = Brønstedbase Zout (0,1 mol/l) Zuur K z Base K b pH NaCl Na(H 2O) 6 + 2.10 -15 Cl - 1.10 -21 7,0 Al(NO 3) 3 Al(H 2O) 6 3+ 1.10 -5 NO 3 - 1.10 -16 3,0 Na 2CO 3 Na(H 2O) 6 + 2.10 -15 CO 3 2- 1,78.10 -4 11,6 NH 4Ac NH 4 + 5,62.10 -10 Ac - 5,62.10 -10 7,0 NaHCO 3 HCO 3 - 5,62.10 -10 HCO 3 - 2,34.10 -8 8,3 • Als K
- Page 1 and 2: Hoofdstuk 6 Zuren en basen Chemie 5
- Page 3 and 4: Brønstedzuren en -basen • Brøns
- Page 5 and 6: Zuur-base-eigenschappen zouten •
- Page 7 and 8: Ionisatie van water • Water is ee
- Page 9 and 10: Zuurtegraad [H 3O + ] pH pOH [OH -
- Page 11 and 12: Sterke en zwakke zuren • Sterk zu
- Page 13 and 14: Evenwicht HAc-oplossing • HAc + H
- Page 15 and 16: Sterke en zwakke basen • Sterke b
- Page 17 and 18: Evenwicht NH 3-oplossing • NH 3 +
- Page 19 and 20: Zuur-basekoppels (HCl) • HCl + H
- Page 21 and 22: Rangschikking zuur-basekoppels •
- Page 23 and 24: pH van oplossingen van zwakke zuren
- Page 25: pH van oplossingen van sterke basen
- Page 29 and 30: Oefeningen p. 139 1. Bereken pH: a,
- Page 31 and 32: Buffers • Oplossing van een zwak
- Page 33 and 34: Buffer + sterk zuur of base • Pro
- Page 35 and 36: Oefening buffer + sterk zuur / base
- Page 37 and 38: Neutralisatie NaOH + HCl • [NaOH]
- Page 39 and 40: Neutralisatie NH 3 + HCl • [NH 3]
- Page 41 and 42: Neutralisatie NH 3 + HAc • [NH 3]
pH van oplossing<strong>en</strong> van zwakke bas<strong>en</strong><br />
• NH 3 + H 2O NH 4 + + OH -<br />
K B = [OH - ] e.[NH 4 + ]e /[NH 3] e = 1,78.10 -5<br />
[NH 3] e = [NH 3] 0 <strong>en</strong> [NH 4 + ]e = [OH - ] e<br />
⇒ [OH - ] = √ K B.[NH 3] 0<br />
[NH 3] = 0,1 mol/l ⇒ pOH = 2,88 ⇒ pH = 11,12<br />
[NH 3] = 0,01 mol/l ⇒ pOH = 3,38 ⇒ pH = 10,62<br />
• 1 t.e.m. 3<br />
Oef<strong>en</strong>ing<strong>en</strong> p. 133
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