Introtoacidbase
Transcript of Introtoacidbase
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Biochemistry
Study of chemistry in biological organisms
Understand how the chemical structure of a molecule is determining its function
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Focus on important biochemical macromolecules
– amino acids ----->proteins – fatty acids----->lipids – nucleotides---> nucleic acids– monosaccharides---> carbohydrates
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Focus on important processes
– Protein Function– Compartmentalization/regulation– Metabolism-– DNA synthesis/replication
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Protein Function
– What is a protein’s structure and what role does it play in the body?
– What are some important proteins in the body?
– What are some key principles behind protein’s functions?
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Enzymes
• What are enzymes?
• What is the role of enzymes in an organism?
• How do they work?
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Lipids
• What are lipids and their structures
• What are roles of lipids
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Membranes and Transport
• What is the structure of a membrane?
• What is compartmentalization and why is it important?
• How can molecules and information get across a membrane?
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Carbohydrates
• What the structures of carbohydrates and what is their role?
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Metabolism
• Glycolysis, Krebs cycle, Oxidative Phosphorylation, beta oxidation
• How does a cell convert glucose to energy?
• How does a cell convert fat to energy?
• Roles of ATP, NAD and FAD
• vitamins
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Nucleic Acids
• What are their structures?
• What their functions?
• How do they replicate?
• What is the relationship between nucleic acids and proteins?
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Connecting structure and function requires chemistry
• Chemistry knowledge needed:– Intermolecular forces– Properties of water– Equilibrium– Acid/Base Theory
• Definitions
• Buffers
• Relation of structure to pH
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Connecting structure and function requires chemistry
– Oxidation-Reductions– Thermodynamics: study of energy flow – Organic functional groups– Important organic reactions
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Intermolecular forces
• Hydrogen bonds
• Dipole/dipole interactions
• Nonpolar forces
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Dipole/Dipole interactions
• Polarity in molecules– Polar bonds– Asymmetry
• Positive side of one polar molecule sticks to negative side of another
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Dipole-Dipole interactions
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Hydrogen Bonding
• Special case of dipole dipole interaction– Hydrogen covalently attached to O, N, F, or Cl
sticks to an unshared pair of electrons on another molecule
• H-bond donors– Have the hydrogen
• H-bond acceptors– Have the unshared pair
• Strongest of intermolecular forces
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Hydrogen bonding
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Hydrogen bonding
• Affect the properties of water
• Water has a higher boiling point than expected
• Water will dissolve only substances that can interact with its partially negative and partially positive ends
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Nonpolar forces
• Nonpolar molecules stick together weakly
• Use London dispersion forces
• Examples are carbon based molecules like hydrocarbons
• Velcro effect– Many weak interactions can work together to
be strong
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Dissolving process
• Solute—solute + solvent—solvent - 2 solute---solvent
• Have to break solute—solute interactions as well as solvent—solvent interactions
• Replace with solute-solvent interactions
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Like dissolves like
• Hydrophobic = nonpolar
• Hydrophilic = polar
• Overall, like dissolves like means that polar molecules dissolve in polar solvents and nonpolar solutes dissolve in nonpolar solvents
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Like dissolves like
• Salt dissolving in water
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Amphipathicity
• Some molecules have both a hydrophilic and hydrophobic part
• soap is an example
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Amphipathicity
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Equilibrium
Two opposing processes occurring at the same rate:
walking up the down escalator
treadmill
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Equilibrium
For chemical equilibrium, It is when two opposing reactions occur at the same rate.
mA + nB <= pC + q D– Two reactions:
• Forward: mA + nB - pC + qD• Reverse: pC + qD - mA + nB
– Equilibrium when rates are equal
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Reaction Rates
Rate of reaction depends on concentration of reactants
For the reaction: mA + nB => pC + qD
Forward rate (Rf) = kf[A]m[B]n
Reverse rate (Rr) = kr[C]p[D]q
(rate constants kf and kr as well as superscripts have to be determined experimentally)
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Equilibrium
• When rates are equal:– Rf = Rr so (from previous slide)
• kf[A]m[B]n = kr[C]p[D]q
– Putting constants together: (Law of Mass Action)• kf = [C]p[D]q = Keq
kr [A]m[B]n
• Keq is the equilibrium constant
• Solids and liquids don’t appear…they have constant concentration
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Equilibrium in quantitative terms
• The equilibrium state is quantified in terms of a constant called the Equilibrium Constant Keq. It is the ratio of products/reactants
• It is determined by Law of Mass Action
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Possible Situations at Equilibrium
• 1. There are equal amounts of products and reactants. K=1 or close to it
• 2. There are more products than reactants due to strong forward reaction– equilibrium lies right)– K >>1
• 3. There are more reactants than products due to strong reverse reaction – equlibrium lies left– K <<1
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Keq Constant Expression
• Given the following reactions, write out the equilibrium expression for the reaction
• CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O(l) + CO2(g)
• 2SO2(g) + O2(g) --->2SO3(g)
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Answers
[CaCl2][CO2]
[HCl]2
[SO3]2
SO2]2 [O2]
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Le Chatelier’s Principle
• When a system at equilibrium is stressed out of equilibrium, it shifts away from the stress to reestablish equilibrium.– Shifts away from what is added– Shifts towards what is removed
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Le Chatelier’s Examples
• N2 + 3 H2 => 2 NH3
– If we add nitrogen or hydrogen, it shifts to the right, making more ammonia
– Removal of ammonia accomplishes the same thing
– Shifts to the left if add ammonia
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Le Chatelier’ and Regulation of Metabolism
• What the diet industry doesn’t want you to know!
– Food - ABCDenergy
• A fat
– What happens if energy is used up?
– What happens if eat a big meal and don’t use energy
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Acid/Base Theory
• Definitions– Acid is a proton (H+) donor
• Produces H3O+ in water
• HCl + H2O - H3O+ + Cl-
– Base is a proton (H+) acceptor• Produces OH- in water
• NH3 + H2O > NH4+ + OH-
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Strong acids v weak acids
– Strong 100 percent ionized• No Equilibrium or equilibrium lies to the right
• K eq >>> 1 and is too large to measure
– Weak acids not completely ionized• Equilibrium reactions
• Have Keq
– For acids, Keq called a Ka
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Acetic Acid as Example of a Weak Acid
• HC2H3O2(aq) <---> H+(aq) + C2H3O2- (aq)
• K = [H+] [C2H3O2-]
[HC2H3O2]
• value is 1.8 x 10-5
• 1.8 x 10-5 <<< 1 www.freelivedoctor.com
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Weak acids, Ka and pKa
– pKa = - log Ka
– For weak acids, weaker will be less dissociated• Make less H3O+
• Eq lies further to left
• Lower Ka
– Since pKa and Ka inversely related: the lower the Ka, the higher the pKa, the weaker the acid
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pH
• pH= -log [H+]
• increasing the amount of H+ (in an acidic solution), decreases the pH
• increasing the amount of OH-decreases the amount of H+ (in a basic solution), therefore, the pH increases
• pH< 7 acidic
• pH>7 basicwww.freelivedoctor.com
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Conjugate Base Pairs
• Whatever is produced when the acid (HA) donates a proton (H+) is called its conjugate base (A-).
• Whatever is produced when the base (B) accepts a proton is called a conjugate acid (HB+).
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Conjugate Base Pairs
• HA(aq)+ H2O(l) H3O+(aq)+ A–(aq)
• Acid Base conjugate acid conjugate base
• differ by one H+ for acids/bases
• Example: HC2H3O2 and C2H3O2-
• acid conj. basewww.freelivedoctor.com
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Buffers
• A buffer is a solution that resists a change in pH upon addition of small amounts of acid or base.
• It is a mixture of a weak acid/weak base conjugate pair– Ex: HA/ A-
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Buffer with added acid• Weak base component of the buffer
neutralizes added acid
• A- + H+ -- HA
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Buffers with added base
• Weak acid component of the buffer neutralizes added base
• Equation: OH- + HA --> H2O + A-
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Relationship of pH to structure
• We can think of a weak acid, HA, as existing in two forms.– Protonated = HA– Deprotonated = A-
• Protonated is the acid
• Deprotonated is the conjugate base– Titrated form
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Henderson-Hasselbach Equation
• pH = pKa + log ([A-] / [HA])
• Can be used quantitatively to make buffers
• Ka is the equilibrium constant for the acid
– HA(aq) + H2O(l) < H3O+(aq) + A-
(aq)
– Ka = [H3O+][A-][HA]
– Higher Ka = more acidic acid
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Henderson Hasselbach continued
• pH = pKa + log ([A-] / [HA])
• pKa = -logKa
Since negative, lower pKa = more acidic
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Henderson Hasselbach and structure
In a titration if we add base to the acid:
HA + OH- - H2O + A-
For every mole of HA titrated, we form a mole of A-
So, if we add enough OH- to use up half the HA (it is half-titrated) we end up with equimolar HA and A-
Looking at the equation:
pH = pKa + log ([A-] / [HA])
If [A-] = [HA] then [A-] / [HA] = 1 and log ([A-] / [HA]) = log (1) – 0
So pH = pKa
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So what?We can now relate the pH of the solution to the
structure of weak acid using Henderson-HasselbachpH = pKa + log ([A-] / [HA])
If pH = pKa, we have equal amounts of protonated and
deprotonated forms
If, pH < pKa, means log term is negative so [HA]>[A-] and protonated form dominates
If pH > pKa, means log term is postive so [HA] < [A- and deprotonated form dominates.
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