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BCHS 3304: General Biochemistry I, Section 07553 Spring 2003 1:00-2:30 PM Mon./Wed. AH 101 http://www.uh.edu/sibs/facul PowerPoint Presentation
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BCHS 3304: General Biochemistry I, Section 07553 Spring 2003 1:00-2:30 PM Mon./Wed. AH 101 http://www.uh.edu/sibs/facul - PowerPoint PPT Presentation


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BCHS 3304: General Biochemistry I, Section 07553 Spring 2003 1:00-2:30 PM Mon./Wed. AH 101 http://www.uh.edu/sibs/faculty/glegge Instructor: Glen B. Legge, Ph.D., Cambridge UK Phone: 713-743-8380 Fax: 713-743-2636 E-mail: glegge@uh.edu Office hours:

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slide1

BCHS 3304: General Biochemistry I,Section 07553

Spring 2003 1:00-2:30 PM Mon./Wed. AH 101

http://www.uh.edu/sibs/faculty/glegge

Instructor:

Glen B. Legge, Ph.D., Cambridge UK

Phone: 713-743-8380

Fax: 713-743-2636

E-mail: glegge@uh.edu

Office hours:

Mon. and Wed. (2:30-4:00 PM) or by appointment

353 SR2  (Science and Research Building 2)

1

sibs program
SIBS program
  • Monday Chat room on Webct: 8:00-10:00 PM Tuesday Workshop: 5:00-7:00 PM in 101 AH Wednesday Office Hours: 3:00-4:45 PM in 114 S Wednesday Workshop: 5:00-7:00 PM in 116 SR1
  • Students must activate their webct accounts.
  • SIBS will not print out exam reviews
  • Jerry Johnson (BCHS 3304 workshops) contact email: MYSTIK1775@aol.com
ionization of water
Ionization of water
  • Although neutral water has a tendency to ionize

H2O <-> H+ + OH-

  • The free proton is associated with a water molecule to form the hydronium ion

H3O+

  • High ionic mobility due to proton jumping
proton jumping large proton and hydroxide mobility
Proton Jumping Large proton and hydroxide mobility
  • H3O+ : 362.4 x 10-5 cm2•V-1•s-1
  • Na+: 51.9 x 10-5 cm2•V-1•s-1
  • Hydronium ion migration; hops by switching partners at 1012 per second
equilibrium expression
Equilibrium expression
  • Described by:

K = [H+][OH-]

[H2O]

  • Where K is the dissociation constant
  • Considering [H2O] constant yields

Kw = [H+][OH-]

slide7
Kw

Kw = [H+][OH-]

  • Where Kw is the ionization constant of water
  • For pure water ionization constant is

10-14 M2 at 25º

  • For pure water

[H+] = [OH-] = (Kw)1/2 = 10-7 M

acids and bases
Acids and bases
  • For pure water (neutral)

[H+] = [OH-] = (Kw)1/2 = 10-7 M

  • Acidic if [H+] > 10-7 M
  • Basic if [H+] < 10-7 M
acids and bases1
Acids and Bases

Lowery definition:

  • Acid is a substance that can donate a proton.
  • Base is a substance that can accept a proton.

HA + H2O H3O+ + A- /OH-

Acid Base Conjugate Conjugate

Acid Base

or

HA A- + H+

Acid Conjugate Conjugate

Base Acid

slide10

If you establish equilibrium, changes in [H+] will shift the ratio of HA and A-.

By adding more H+ , A- will be consumed forming HA.

If there is sufficient [A-], the extra H+ will also be consumed and the [H+] will not change.

acid strength is specified by its dissociation constant
Acid strength is specified by its dissociation constant

Molar concentration

for: HA + H2O H3O+ + A-

  • reactants products

HAH3O+

A- H2O

a measure of relative proton affinities for each conjugate acid base pair.

These ratios are

what to do about the water
What to do about the water!

The concentration of H2O remains almost unchanged especially in dilute acid solutions.

What is the concentration of H2O?

Remember the definition: Moles per liter

1 mole of H2O = 18 g = 18 ml

1000 g/liter

slide13

From now on we will drop the a, in Ka

Weak acids (K<1)

Strong acids (K>1)

Strong acid completely dissociates: Transfers all its protons to

H2O to form H3O+

HA H+ + A-

weak acids
Weak Acids

Weak acids do not completely dissociate: They form an equilibrium:

If we ADD more H+, the equilibrium shifts to form more HA using up A- that is present.

dissociation of h 2 o
Dissociation of H2O

Water also dissociates [H2O] = 55.5

Ionization constant for water

slide16

Since there is equal amounts of [H+] and [OH-]

This is neutral

At [H+] above this concentration the solution isACIDIC

At [H+] below this concentration the solution is BASIC

slide17

[H+] pH

10-7 = 7

10-3 = 3

10-2 = 2

10-10 = 10

5x10-4 = 3.3

7x10-6 = 5.15

3.3x10-8 = 7.48

pH = -Log[H+]

It is easier to think in log of concentrations but it takes practice!!

slide19

Observation

If you add .01 ml i.e 1/100 ml of 1M HCl to 1000 ml of water, the pH of the water drops from 7 to 5!!

i.e 100 fold increase in H+ concentration: Log = 2 change.

Problem:

Biological properties change with small changes in pH, usually less than 1 pH unit.

How does a system prevent fluctuations in pH?

buffers
Buffers

A buffer can resist pH changes if the pH is at or near a weak acid pK value.

Buffer range: the pH range where maximum resistance to pH change occurs when adding acid or base. It is = +1 pH from the weak acid pK

If pK is 4.8 the buffering range is 3.8 5.8

Why?

henderson hasselbalch equation
Henderson - Hasselbalch equation

From

Rearrange

Take (-)Log of each

slide22

Above and below this range there is insufficient amount of conjugate acid or base to combine with the base or acid to prevent the change in pH.

slide23

For weak acids

HA A- + H+

This equilibrium depends on concentrations of each component

.

If [HA] = [A-] or 1/2 dissociated

: pH = pK

Then

By definition the pK is the pH where [HA] = [A-] :

50% dissociated

slide24

The buffer effect can be seen in a titration curve.

To a weak acid salt, CH3C00-, add HC1 while monitoring pH vs. the number of equivalents of acid added.

or

do the opposite with base.

Buffer capacity: the molar amount of acid which the buffer can handle without significant changes in pH.

i.e

1 liter of a .01 M buffer can not buffer 1 liter of a 1 M solution of HCl

but

1 liter of a 1 M buffer can buffer 1 liter of a .01 M solution of HCl

slide28

Table 2-3

Dissociation constants and pK’s of Acids & buffers

Acid K pK

Oxalic 5.37x10-2 1.27

H3PO4 7.08x10-3 2.15

Succinic Acid 6.17x10-5 4.21 (pK1)

Succinate 2.29x10-6 5.65 (pK2)

H2PO4- 1.51x10-7 6.82

NH4+ 5.62x10-10 9.25

Glycine1.66x10-10 9.78

slide30

Exams Problems

You will be asked to solve Henderson - Hasselbalch type problems:

You may be asked the pH, pK, the ratio of acid or base or solve for the final concentrations of each.

slide31

The 6 step approach

1. Write the Henderson + Hasselbalch equation.

2. Write the acid base equation

3. Make sure either an H+ or OH- is in the equation.

3. Find out what you are solving for

4. Write down all given values.

5. Set up equilibrium conditions.

6. Plug in H + H equation and solve.

slide32

What is the pH of a solution of that contains 0.1M CH3C00- and 0.9 M CH3C00H?

1) pH = pK + Log [A-]

[HA]

2) CH3C00H CH3C00- + H+

3) Find pH

4) pK = 4.76 A- = 0.1 M HA = 0.9 M

5) Already at equilibrium

6) X = 4.76 + Log 0.1

0.9

Log 0.111 = -.95 X = 4.76 + (-.95) X = 3.81

slide33

What would the concentration of CH3C00- be at pH 5.3 if 0.1M CH3C00H was adjusted to that pH.

  • 1) pH = pK + Log [A-]
  • [HA]
  • 2) CH3C00H CH3C00- + H+
  • 3) Find equilibrium value of [A-] i.e [CH3C00-]
  • 4) pH = 5.3; pK = 4.76
  • 5) Let X = amount of CH3C00H dissociated at equilibrium
  • [A-] = [X]
  • [HA] = [0.1 - X]
  • 6) 5.3 = 4.76 + Log [X]
      • [0.1 - X]
      • Now solve.
blood buffering system
Blood Buffering System
  • Bicarbonate most significant buffer
  • Formed from gaseous CO2

CO2 + H2O <-> H2CO3

H2CO3 <-> H+ + HCO3-

  • Normal value blood pH 7.4
  • Deviations from normal pH value lead to acidosis