Buffers - Enkele uitwerkingen
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This document discusses buffers and buffer calculations. It provides the definitions and acid dissociation constants (Ka or Kb values) for several common buffer systems, including acetic acid-sodium acetate, monosodium phosphate-disodium phosphate, and ammonium chloride-ammonia. It also gives the general equations for calculating the pH of a buffer solution and the concentrations needed to achieve maximum buffering capacity. Several sample buffer problems are presented, involving calculating concentrations or amounts needed to achieve a specific pH. The key equations for a generic acid-base buffer reaction and the Henderson-Hasselbalch equation are also summarized.
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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](https://image.slidesharecdn.com/zuurbase1-140427063750-phpapp02/85/Buffers-Enkele-uitwerkingen-1-320.jpg)
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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)](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Buffers - Enkele uitwerkingen
- 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