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Chemie samenvatting HO17 - Buffers en titraties

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Chemie samenvatting HO17 over buffers en titraties.

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Samenvatting Chemie HO17
Buffers en titraties
 17.1 Neutralisatie reactie
Neutralisatie-reactie: reactie tussen [H3O+] en [OH-] (zuur en base).
Deze reactie loopt niet altijd volledig, dat hangt er van af welke zuren en basen je gebruikt.
Mate van neutralisatie  Kn

Sterk zuur + sterke base  volledige dissociatie (in water)
Bijv.
HCl (aq) + NaOH(aq)  H2O(l) + NaCl(aq)
= H3O+(aq) + OH-(aq)  2H2O(l)
1 1 1 14
K n= = = −14 = 1.0 * 10 Kn is groot, dus evenwicht licht rechts
¿¿ Kw 1,0∗10

Zwak zuur + sterke base  zwak zuur is bijna niet uiteengevallen (in water)
Bijv.
CH3COOH(aq) + OH(aq)  CH3COO-(aq) + H2O(l)

De reacties die hierbij horen zijn:
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) Kn = Ka = 1,8*10-5
H3O+(aq) + OH(aq)  2H2O(l) Kn = 1/Kw = 1,0*1014
Ka-waarde is in het boek te vinden bij appendix C

De netto reactie is nu:
CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
H3O+(aq) + OH(aq)  2H2O(aq) +
CH3COOH(aq) + OH(aq)  CH3COO-(aq) + H2O(aq)

K netto =K 1∗K 2 enz . Dus: (1,8*10-5) * (1,0*1014) = 1,8*109  hoge waarde van Kn, dus
evenwicht zal rechts liggen.

Sterk zuur + zwakke base  zwakke base is bijna niet uiteengevallen (in water)
Bijv.
H3O+(aq) + NH3(aq)  H2O(l) + NH4+(aq)

De reacties die hierbij horen zijn:
H2O(l) + NH3(aq)  OH-(aq) + NH4+(aq) Kb = 1,8*10-5
H3O+(aq) + OH(aq)  2H2O(l) Kn = 1/Kw = 1,0*1014

De netto reactie is nu:
H2O(l) + NH3(aq)  OH-(aq) + NH4+(aq)
H3O+(aq) + OH(aq)  2H2O(l) +
H3O+(aq) + NH3(aq)  H2O(l) + NH4+(aq)
Dus, Knetto = (1,8*10-5) * (1,0*1014) = 1,8*109  hoge waarde van Kn, dus evenwicht zal rechts
liggen.

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Subjects

  • buffers
  • titraties
  • neutralisatie reactie
  • common ion effect
  • evenwicht
  • buffer oplossingen
  • henderson hasselbalch
  • titratie
  • equivalentiepunt

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