II - Acids and Bases, pH pKa, Buffers

Brönstead-Lowry Acids and Bases

Acid = proton donor. If it readily gives up a proton, the acid HA is strong

Base = proton acceptor

H₂O₂ can be both → autoionization

Ionization of water

equilibrium K_eq = [products]/[reactants] = [OH⁻][H⁺]÷[H₂O]

To find the Ionic product of water Kw = K(eq) × [H₂O] = [OH⁻][H⁺] = 1.0₁₀⁻¹⁴

Relationship between [H3O⁺], [OH⁻] & Kw

Taking –log₁₀ on both sides of the equation (since -log (A x B) = -log A + -log B)

-log₁₀ K_w = -log₁₀ [H⁺] + -log₁₀ [OH⁻]. Let p = a shorthand for -log₁₀

log₁₀(1₁₀⁻¹⁴) = pH + pOH → 14.0 = pH + pOH = pK_w

Equilibrium constants for the dissociation of acids

Kₐ = [H₃O⁺][A⁻]÷[HA], where pKₐ= -log₁₀Kₐ

✧ The larger the Kₐ value, the smaller the pKₐ value, the stronger the acid

Ionization constants for acids

✧ Strong acids are ~100% ionized in water → Kₐ >> 1, product favored

✧ Weak acids are <100% ionized in water, Kₐ < 1, reactant favored. They don’t ionize appreciably

✦ pKₐ is the pH at which an acid is exactly half-ionized

Henderson-Hasselbalch Equation

Used to find the pH when we mix a weak acid and its conjugate base together. It’s rearranged from the Kₐ expression for an acid.

Using log(ab)= log(a)+ log(b) and letting p = -log:

pH = pKₐ + log([A⁻]/[HA] (pKₐ for acid HA, A⁻ is conj. base)

✧[HA] & [A⁻] must be orders of magnitude greater than ∆[H₃O⁺]

How can the Henderson-Hasselbalch Equation be used?

✦ calculate the ionization state of an ionizable species at a given pH

✦ determine the appropriate amounts of an acid and its conjugate base that are required to make a buffer at a given pH

✦ calculate the pKa of an acid if you know how much it is ionized at a given pH

Buffer

A buffer solution is a mixture of a weak acid and its conjugate base that resists a pH change when an acid or a base is added.

pH = pKa, [A⁻] = [HA] (highest buffering capacity)

pH > pKa, A⁻ predominates

pH < pKa, HA predominates

What makes a good buffer?

✦ It is able to absorb small additions of acid or base when [HA] and [A⁻] are large and within a factor of 10 of each other.

✦ For every power of 10 change in the [A⁻]/[HA] ratio changes by 1.

✦ Buffering capacity is lost when the pH differs from pKa by more than 1 pH unit

pK_a

The pK_a is the pH at which the two charge states of an ionizable group are present in equal amounts.

As the pH is decreased below the pK_a of an ionizable group, the protonated form predominates. Conversely, if the pH is above the pK_a,then the deprotonated form predominates.

Ionization of amino acids

This varies with pH.

✦ @ acidic pH, C and N groups are protonated → AA is a cation

✦ @ ∼ physiological pH, Carboxyl group is deprotonated but Amino group isn’t → dipolar form called zwitterion (an act as acid or base)

✦ @ alkaline pH, both C and N groups are deprotonated → AA is an anion

Isoelectric Point

The pH at which the net charge on a molecule (amino acid or protein) is 0.

Differences in pI depending on the number of ionizable groups

For a simple amino acid with only two protonatable groups, it’s the average of the α‑amino and α‑carboxyl pK_a values.

For an amino acid having a positively charged side chain in the fully protonated form, the pI is the average of pK₂ and pK₃ .

For an amino acid with a neutral dissociable group in the side chain, the pI is the average of pK₁ and pK₂ .

Acidic and basic properties of amino acids

Each ionizable group in the amino acid has its own pKa and thus, its own buffering range

e.g. Glycine can buffer around pH 2.3 (pKa of carboxylic acid) and at pH 9.6 (pKa of a- amino group)eg. Arginine can buffer at pH 1.8, 9.0 and 12.5 (guanidinium moiety)