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# Buffers - Enkele uitwerkingen

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• 1. ' & \$ % Buﬀers Systemen CH3COOH / CH3COONa pKa = 4.74 Ka = 1.8 10−5 Mr = 60 Mr = 82 afwegen NaH2PO4.2H2O / Na2HPO4.12H2O pKa = 6.7 Ka = 2.0 10−7 Mr = 156 Mr = 358 pKa = 7.21 NH4Cl / NH3 (0, 2M) pKa = 9.26 Ka = 5.55 10−10 Mr = 53.5 pKb = 4.74 Kb = 1.8 10−5 pH Berekenen pH = pKa + log Cb Cz → CH3COO−/HPO2− 4 /NH3 → CH3COOH/H2PO− 4 /NH+ 4 of zuur + H2O Ka −−−−−− base + H3O+ Ka = ˆ H3O+ ˜ [base] [zuur] ⇒ ˆ H3O+ ˜ ev = Ka · [zuur] [base] Hoogste buﬀerend vermogen 1 10 ≤ Cb Cz ≤ 10 ⇒ pH = pKa ± 1
• 2. ' & \$ % Buﬀers Opgaven 1. Eerste opgave pH → Cb Cz = ? [1] Cb + Cz = 0, 1 [2] [1] + [2] → Cb en Cz ⇒ 100 mL → mol zuur ×Mr mol base 2. Tweede opgave Aantal mL HCl (0.1 M) tot buﬀeroplossing met pH = ? base + H3O+ −− geconjugeerd zuur + H2O x mol #mL × 0.1M −#mL × 0.1M −#mL × 0.1M +#mL × 0.1M x − (#mL × 0.1M) #mL × 0.1M Buﬀerformule gebruiken: pH = pKHA + log CA− CHA = pKHA + log x − (#mL × 0.1M) #mL × 0.1M ⇒ x = # mol base(×Mr = #g) 3. Derde opgave Aantal mL NaOH (0.1 M) tot buﬀeroplossing met pH = ? zuur + OH− −− geconjugeerde base + H2O x mol #mL × 0.1M −#mL × 0.1M −#mL × 0.1M +#mL × 0.1M x − (#mL × 0.1M) #mL × 0.1M Buﬀerformule gebruiken: pH = pKHA + log #mL × 0.1M x − (#mL × 0.1M) ⇒ x = # mol zuur(×Mr = #g)
• 3. ' & \$ % Buﬀerformule HA + H2O −− A− + H3O+ v´o´or reactie CHA mol/L CA− mol/L reactie −x mol/L + x mol/L + x mol/L bij evenwicht (CHA − x) mol/L (CA− + x) mol/L x mol/L KHA = h A− i h H3O+ i [HA] = “ CA− + x ” · x CHA − x = “ CA− ” · x CHA h H3O + i ev = x = KHA · CHA CA− =⇒ pH = pKHA + log CA− CHA