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Buffers and Buffer Problems

Last updated

Jun 3, 2019
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- BIS 105: Biomolecules and Metabolism (Murphy)
- Carbohydrates

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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)

Acid-Base Buffer Tutorial

http://www.mcb.ucdavis.edu/courses/bis102/acid-base/

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Henderson-Hasselbach equation

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