http://en.wikipedia.org/wiki/Acid_Queen

Bio 98 - Lecture 2

Acid-Base Equilibria, pH and Buffers

pH profiles of enzymatic reactions

UCI Bio199 Independent Research

Pepsin

Amylase

H2O H+ + OH-

Pure water is only slightly ionized

H+ ions (protons) do not persist free in solution, they are immediately hydrated to hydronium ions (H+ + H2O H3O+).

Grotthuss

proton wire

Achieving equilibrium

Kw = [H+] [OH-] = 10-14 M2

[H+] = [OH-] = 10-7 M = 0.1 M

[H+] [OH-] [H2O]

Concentration of “water in water” ([H2O]) is 55.6 M [next slide], thus

Pure water has equal quantities of H+ and OH- ions, or, put differently, pure water has equal [H+] and [OH-].

Constant ion product!

= 1.8x10-16 MKeq =

H2O H+ + OH-

[H2O]

Concentration is measured in moles per liter (mol/l) or simply M.

1 l = 1,000 ml of water has a mass of 1,000 g.

1 mole of water has a mass of 18 g (hydrogen 1 Da, oxygen 16 Da).

Thus 1 liter of water (1,000 g) contains 1,000 g / 18 g moles of water.

[H2O] = (1,000 g / 18 g) M = 55.6 M

pH = log[H+]

1= -log [H+]

pH = -log(10-7) = -(-7) = 7

The pH scale

In neutral/pure water[H+] = [OH-] = 10-7 M, so

Logarithm (base 10) refresher:if log10(x)=y then x=10y

Acid: proton donor

Base: proton acceptor

pH = -log [H+]

HCl H+ + Cl-

HCl is a strong acid that completely dissociates inwater. 1 M HCl will thus yield 1 M [H+] and the pH will be pH = -log [H+] = -log(1) = 0

NaOH is a strong base that completely dissociates inwater. 1 M NaOH will thus yield 1 M [OH-]. Since [H+] [OH-] = 10-14 M and must remain constant[H+] = 10-14 M and the pH will be pH = -log [H+] = -log(10-14) = 14

Life is compatible only in a narrow pH range around pH 7.

Strong acids and bases

Dissociation of a weak acid or weak base

[H+] [A-] Ka = ––––––––– = acid dissociation constant [HA]

R-C-OH R-C-O- + H+ (C-term/Asp/Glu)

R-NH3+ R-NH2 + H+ (N-term/Lys)

HA A- + H+ (general)

O O

Weak acids and weak bases

+ H+

HA A-

Keq = Ka = [H+] [A-][HA]

= 1.7x10-5 M

pKa = -log(Ka) = -log(1.7x10-5 M) = 4.8

Acetic acid is a weak acid as it does not completely dissociate in water.

-

[H+] [OH-][H2O] = 1.8x10-16 M with [H2O] = 55.6 M!Keq = Recall for water:

Ka = [A-] [H+]

[HA]

when [A-] and [HA] are equal then Ka = [H+].

And thus pKa = 4.8 = -log(Ka) = -log [H+] = pH

(since pH is defined as -log [H+])

pKa and pH

pH = pKa + log [A-][HA]

Ka = [H+] [A-]

[HA]

Start at low pH and begin to add HO-. The product of [H+] [HO-] must remain constant, so adding HO- means [H+] must decrease and thus pH increases. At the pKa, [A-] and [HA] are equal, so adding more HO- does not change the ratio of [A-] to [HA] very much and thus the pH does not change very much (shallow slope of titration curve from ~1 pH unit below pKa to ~1 pH unit above).

Titration curves

+/-1 pH unit

Measuring pKa values

NH4+ H+ + NH3

[NH4+ ]

[H+] [NH3]Ka =

pKa = pH when [NH4+] = [NH3]

-

Ka = [H+] [A-]

[HA]

take the -log on both sides

The Henderson-Hasselbalch Equation

-log Ka = -log [H+] -log [A-][HA]

pH = pKa + log [A-][HA]

= pKa + log [Proton acceptor][Proton donor]

HA H+ + A-

pKa = pH -log [A-][HA]

Apply definition p(x) = -log(x)

and finally solve for pH…

Acetic acid has a pKa of 4.8. How many ml of 0.1 M acetic acid and 0.1 M sodium acetate are required to

prepare 1 liter of 0.1 M buffer with a pH of 5.8?

Substitute the values for the pKa and pH into the Henderson-Hasselbalch equation:

5.8 = 4.8 + log [Acetate]

[Acetic acid]

1.0 = log [Acetate] 10x then *[Acetic acid][Acetic acid]

10 [Acetic acid] = [Acetate]

For each volume of acetic acid, 10 volumes of acetate must be added (total of 11 volumes).

Acetic acid needed: 1/11 x 1,000 ml = 91 mlAcetate needed: 10/11 x 1,000 ml = 909 ml

on both sides

At the pKa, [HA] = [A-] so the system is able to absorb the addition of HO- or H+. If we add HO- near the pH where [HA] = [A-] (ie pH ~= pKa) then HA can release H+ to offset the HO- added but the ratio of HA to A- does not change much. If we add H+ then A- can absorb H+ to form HA. Hence, the pH does not change much.

How does a buffer work?

-

Buffers are vitally important in biochemical systems since pH needs to be controlled. Living systems must be “buffered” to resist large variations in pH.

Phosphate buffering

H3PO4 H+ + H2PO4- pKa1 = 2.2

H2PO4- H+ + HPO4

2- pKa2 = 7.2

HPO42- H+ + PO4

3- pKa3 = 12.7

Carbonate buffering

CO2 + H2O H2CO3

H2CO3 H+ + HCO3- pKa1 = 6.4

HCO3- H+ + CO3

2- pKa2 = 10.2

Phosphate buffering

CO2 + H2O H2CO3

H2CO3 H+ + HCO3-

pKa = 6.4

HCO3- CO3

-2 + H+

pKa = 10.2 (not relevant, far from pH 7.4)

Carbon dioxide - carbonic acid - bicarbonate buffer

Carbon dioxide - carbonic acid - bicarbonate buffer

What happens to blood pH when you hyperventilate?What happens to blood pH when you hypoventilate?

If blood pH drops due to metabolic production of H+ then [H2CO3] increases by protonation of HCO3

-, H2CO3 rapidly loses water to form CO2(aq), which is expelled as CO2(g).

If the blood pH rises, [HCO3-] increases

by deprotonation of H2CO3, then breathing rate changes and CO2(g) is converted to CO2(aq) and then to H2CO3

in the capillaries in the lungs.

What is the pH of 0.15 M acetic acid? The pKa of acetate is 4.8, so the Ka = 10-4.8 M = 1.58x10-5 M.

[H+] [A-] Ka = _________

[HA] [H+]2 [H+]2

Ka = ––––– = ––––––––– = 1.58x10-5 M [HA] 0.15 M - [H+]

[H+]2 +1.58x10-5 M [H+] + (-2.37x10-6 M2) = 0 (ax2+bx+c = 0)

[H+] = 1.53x10-3 M and thus pH = 2.8

R-C-OH R-C-O- + H+

OO and[H+]=[A-]

[HA]=0.15-[H+]

ax2 + b x + c = 0 Quadratic Formula

What is the pH of 0.15 M acetic acid? The pKa of acetate is 4.8, so the Ka = 10-4.8 M = 1.58x10-5 M.

[H+] [A-] Ka = _________

[HA] [H+]2 [H+]2

Ka = ––––– = ––––––––– = 1.58x10-5 M [HA] 0.15 M - [H+] Assumption: [H+] << 0.15 M!

[H+]2 = 0.15 M * 1.58x10-5 M

[H+] = 1.54x10-3 M or 0.00154 M and thus pH = 2.8

R-C-OH R-C-O- + H+

OO and[H+]=[A-]

[HA]=0.15-[H+]

[H+]2 = 2.37x10-6 M2 Assumption: [H+] << 0.15 M!

Your 199 prof asks you to make a pH 7 phosphate buffer. You already have 0.1 M KH2PO4. What concentration of K2HPO4 do you need?

pH = 7 = pKa + log

7 = 7.2 + log(x / 0.1 M) -0.2 = log(x / 0.1 M)10-0.2 = x / 0.1 M

x = 0.063 M = [K2HPO4]

[HPO42-]

[H2PO4-]

H2PO4- HPO4

2- + H+ pKa = 7.2

KH2PO4 H2PO4

- + K+ and K2HPO4 HPO4

2- + 2 K+

Phosphate buffering