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

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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)

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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