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

if you know either the [H+] or the [OH-] you can plug in and find the

other.

[H+][10-5]=10-14

[H+]=10-14/10-5

[H+]=10-9M

is the measure of hydrogen ions in a solution

pH= -log[H+]

e.g. if the [H+] in a solution is 10-9 find pH.

pH=-log(10-9)

pH=9

e.g. if a solution has 10-5M of [OH-], what is the pH?

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

[H+]=10-14/10-5=10-9

they are substances that completely dissociate when

placed in an aqueous solution. e.g. NaCl, HCl.

NaCl <--> Na+ + Cl-

HCl + H+ ---> H3O+ + Cl-

HCl ---> H+ + Cl- (conjugate base)

H2O + H+ ---> H3O+ (conjugate acid)

K=[H3O+][Cl-]/[HCl][H2O]

because water's M is very high

Ka=[H3O+][Cl-]/[HCl]

since HCl dissociates completely, we remove it from the above equation.

Ka=[H3O+][Cl-]

[H3O+]=[HCl] initial

they are substances which partially dissociate in an

aqueous solution. e.g. acetic acid

since they only partially dissociate, it's much harder for us to determine the

pH.

pH.

CH3COOH + H2O <--> CH3COO- + H3O+

K=[CH3COO-][H3O+]/[CH3COOH][H2O]

we remove [H2O] since it's molarity is very high

Ka=[CH3COO-][H3O+]/[CH3COOH]=1.74x10-5M

(very small number and that it's does not dissociate well)

HA <--> H+ + A- dissociation of a weak acid.

pH = pKa + log [A-]/[HA] (should be MEMORIZED)

using this equation pH of a solution can be calculated, provided pKa, [A-],

[HA] are known.

e.g. calculate pH when 0.01M A- (congugate base) and 0.02M HA (acid)

are in a solution, pKa of the HA (acid) is 4.78

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

pH= 4.78 + log ( 0.01)/( 0.02)

pH= 4.78 + (-0.3)

pH= 4.48

another scenario, [A-]=0.01, [HA]=0.01

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

pH= 4.78 + log (0.01)/(0.01)

pH= 4.78

the pH is equal to the pKa when the concentration of the conjugate

base (A-) is equal to the acid (HA) in a solution.

an analytical method used to determine the amount of acid in a

solution

they are solutions that tend to resist change. They are made from

weak acid and its conjugate base.

are chosen based on how close to the pKa value is to the desired pH

e.g. pKa of acetic acid is 4.74.

it's good buffer from pH range 3.74-5.74 (+-1)

if the rxn will proceed or not.

For the rxn:

^G = ^H -T^S

is related to equilibrium constant for a rxn

A +B <--> C + D

^G = ^G' + RT ln [C][D]/[A][B] - Keq

^G= free energy change

^G'= standard free energy change --> is free energy change under

standard conditions

R= gas constant

T= Temperature in Kelvin

^G''- for biochemical reactions, the standard free energy change at pH 7

is denoted by ^G''

if ^G is negative (-) the rxn is spontaneous (don't need to provide energy

for the rxn to proceed).

if ^G is positive (+) the rxn is non spontaneous (needs energy) if ^G = 0, it rxn is at equilibrium

e.g. the equilibrium constant at 54.5'C is 0.27, calculate ^G' and answer if this

rxn is spontaneous or not.

^G'= -RT ln keq

^G'= (-8.314)(54.5+273) ln(0.27)

^G'= 3.5KJ/mol

Phosphocreatine + H2O --> creatine + phospate ^G' = 42.8 KJ/mol

what happens if the concentration of reactants and products are 0.001M temp

31 degrees C ?

^G = ^G' + RT ln keq

^G = -42.8 KJ/mol + 8.314 J/mol*K * 301K ln (0.001)

^G= -42.8KJ/mol + (-17.8 KJ/mol

^G= -60.5 KJ/mol

- so it is spontaneous

keq= [creatine][phosphate]/[phosphocreatine]

keq= (0.001)(0.001)/0.001

keq= 0.001

there is a difference of -17 so concentration does play a factor

are very versatile macromolecules in living systems and they have

crucial function in essentially all biological processes

three dimensional structures, and the

function of a protein is dependent on its three dimensional structure.

consist of a central carbon atom (alpha carbon) linked to amino

group , a carboxylic group, a R group, and a hydrogen atom

this is the backbone of amino acids, and this is the same for all amino acids,

only the R group changes

amino

acids are chiral and they may exist in L or D isomer

‣ amino acids found in proteins are L-isomes, except few exceptions

the pka tells you that pka=pH when half of the H molecules has

dissociated (chart/graph)

everything will remain the same • except for the COO- because we are

below pH 2.3 the H remains and does not dissociate

pka (2.3) > pH (1)-- H does not dissociate

‣ consider various charge forms of the molecule

‣ find the isoelectric form

‣ take average of 2 pkas immediately on either side of the isoelectric form

9 amino acids

• glycine-Gly-G

• alanine-Ala-A

• valine-Val-V

• leucine-Leu-L

• isoleucine-Ile-I

• methoionine-Met-M

• tryptophan-Trp-W

• phenlyalanine-Phe-F

• proline -Pro-P

• serine - Ser-S

• threonine - Thr-T

• tyrosine - Tyr-T

• cystine- Cys-C

• asparagine - Asn- N

• glutamine - Gln - Q

• aspartic acid - Asp- D

glutamic acid - Glu- E

• lysine - Lys- K

• arginine - Arg - R

• histidine- His- H

Hydrophic group

Gly

G

Hydrophobic group

Ala

A

Hydrophobic Group

Val

V

Hydrophobic Grouo

Leu

L

Hydrophobic Group

Ile 

I

Methoionine

Hydrophobic Group

Met

M

Hydrophobic Group

Trp

W

Hydrophobic Group

Phe

F

Hydrophobic Group

Pro

P

Polar Uncharged Group

Ser

S

Threonine

Polar Uncharged Group

Thr

T

Polar Uncharged Group

Tyr

T

Polar Uncharged Group

Cys 

C

Polar Uncharged Group

Asn

N

Polar Uncharged Group

Gln

Q

Polar Charged (acidic)

Asp

D

Polar charged (acid)

Glu

E

Polar Charged (basic)

Lys

K

polar charged (basic)

Arg

R

polar charged (basic)

His

H