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Buffers and Buffer Problems

Henderson-Hasselbach equation

General solution when you know the concentrations of conjugate acid and base
pKa = pH – log ([ A-]/[HA])
 
Rearrange:
pH = pKa + log([A-]/[HA]) = pKa + log ([conjugate base]/[conjugate acid])
 

Example

What pH do you get when to 0.1 M HA, you add 0.02 M NaOH?
HA ↔ H+ + A-
0.1-0.02 0.02
pH = pKa + log [(0.02)/(0.1-0.02)]
Apply to acetic acid, pKa = 4.76:
pH = 4.76 - 0.60 = 4.16

Example

How much HCl must be added to 0.1 M Tris to give pH 7.8?
pKa (Tris) = 8.1
(Tris is tris-hydroxy-amino-methane and generally is the base form, e.g. A-)
A- + H+ ↔ HA
0.1-X X
pH = pKa + log ([ A-]/[HA])
7.8 = 8.1 + log ([ A-]/[HA])
log ([ A-]/[HA]) = -0.3
[ A-]/[HA] = 10-0.3 = 0.5
[ A-] = 0.5[HA]
0.1-X = 0.5X
X = [HA] = 0.1/1.5 = 0.067 M

Titrations

Example

Add NaOH to 0.1 M CH3COOH
pH = pKa + log([ A-]/[HA])
when [ A-] ~ [HA], then [ A-]/[HA] ~ 1, and log([ A-]/[HA]) ~ 0 and pH ~ pKa
Also, log([ A-]/[HA]) is most resistant to changes in HA
i.e., when pH ~ pKa: buffers!
 
So expect most resistance, lowest d(pH)/d(NaOH) at 0.05 M
[CH3COO-] / [CH3COOH] = 0.05 / (0.1 – 0.05) = 1
pKa for ammonium = 9.25, imidazole = 6.99, acetate =4.76 (note the shapes are all the same)
 
 TITRATION CURVE DIAGRAM GOES HERE
 
pKa for ammonium = 9.25, imidazole = 6.99, acetate =4.76 (note the shapes are all the same)
 
Phosphate dissociation and disproportionation:
H3PO4 ↔ H2PO4- ↔ HPO4-2 ↔ PO4-3
pK1 = 2.15 pK2 = 7.2  pK3 = 12.4
Put H2PO4- into water:
H2PO4- ↔ Η+ + HPO4-2
H2PO4- + Η+ ↔ H3PO4
so [HPO4-2] = [H3PO4
[pH = pK1 + log[H2PO4-1]/[H3PO4] = pK1 + log[H2PO4-] - log[H3PO4
[pH = pK2 + log[HPO4-2]/[H2PO4-1] = pK2 + log[HPO4-2] - log[H2PO4-]
 
Add equations
2pH = pK1 + pK2
pH = ½(pK1 + pK2)

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