Download BioN01 Introduction, pH and buffer Summer 2014

Introduction into
biochemistry
Dr. Mamoun Ahram
Nursing
Summer semester, 2015
What is biochemistry?
It is the chemistry of living organisms.
It describes the structure, organization, and functions
of living matter in molecular terms.
Important terms
Electronegativity
Covalent bonds
Polar vs. non-polar covalent bonds
Non-covalent interactions
electrostatic interactions
hydrogen bonds (donor and acceptor)
van der Waals interactions
Hydrophobic interactions
Hydrophobic versus hydrophilic molecules
Backbone of a molecule
Nucleophile versus electrophile
Chemical elements in living creatures
Living organisms on Earth are composed mainly of 31
elements
Why can carbon form complex molecules?
Properties of carbon (1)
It can form four bonds, which
can be single, double, or triple
bonds.
Each bond is very stable.
strength of bonds: triple >
double > Single)
They link C atoms together in
chains and rings.
These serve as a
backbones.
Properties of carbon (2)
Carbon bonds have angles
giving molecules threedimensional structure.
In a carbon backbone, some
carbon atoms rotate around a
single covalent bond
producing molecules of
different shapes.
Properties of carbon (3)
The electronegativity of carbon is between other atoms.
It can form polar and non-polar molecules
Pure carbon is not water soluble, but when carbon forms
covalent bonds with other elements like O or N, the molecule
that makes carbon compounds to be soluble.
Nonpolar
Properties of water (1)
Water is a polar molecule
as a whole because of
the different
electronegativitiy between
Hydrogen and oxygen
It is angular
Water is highly cohesive.
Water molecules produce
a network.
Properties of water (2)
Water is an excellent solvent because It is small and it weakens
electrostatic forces and hydrogen bonding between polar
molecules.
Properties of water (3)
It is reactive because it is a nucleophile.
A nucleophile is an electron-rich molecule that is
attracted to positively-charged or electron-deficient
species (electrophiles)
Properties of water (4)
Water molecules are ionized to become a positivelycharged hydronium ion (or proton), and a hydroxide
ion:
Acids versus bases
Acid: a substance that produces H+ when dissolved in
water (HCl, HNO3, CH3COOH, H2SO4, H3PO3)
H+ Reacts with water producing hydronium ion (H3O+)
Base: a substance that produces OH- when dissolved in
water (NaOH, KOH)
Brønsted-Lowry acids and bases
The Brønsted-Lowry acid: any substance (proton
donor) able to give a hydrogen ion (H+-a proton) to
another molecule
Monoprotic acid: HCl, HNO3, CH3COOH
Diprotic acid: H2SO4
Triprotic acid: H3PO3
Brønsted-Lowry base: any substance that accepts a
proton (H+) from an acid
NH3
According to this…
The reaction does not need to occur in water
the Brønsted Lowry base does not need give OH- ions
in water.
The products of an acid base reaction can also behave
as acids and bases.
Water: acid or base?
Both
Products: hydronium ion (H3O+) and hydroxide
Amphoteric substances
Substances that can act as an acid in one reaction and
as a base in another are called amphoteric substances
Example: water
With ammonia (NH3), water acts as an acid because it
donates a proton (hydrogen ion) to ammonia
NH3 + H2O ↔ NH4+ + OH–
With hydrochloric acid, water acts as a base water is
acting as a bas
HCl+ H2O → H3O+ + Cl-
Acid/base strength
Acids differ in their ability
to release protons
Strong acids dissociate
100%
Bases differ in their ability
to accept protons
Strong bases have strong
affinity for protons
Note
For multi-protic acids
(H2SO3, H3PO4), each
proton is donated at
different strengths
Rule
The stronger the acid, the weaker the conjugate base
Reactions favors formation of weak acid
Strong acids and bases are one-way reactions
HCl → H+ + ClNaOH → Na+ + OHWeak acids and bases do not ionize completely
HC2H3O2 ↔ H+ + C2H3O2NH3 + H2O ↔ NH4+ + OH-
Equilibrium constant
Acid/base solutions are at constant equilibrium
We can write equilibrium constant (Keq) for such
reactions
The value of the Keq indicates direction of reaction
When Keq is greater than 1 the product side is favored
When Keq is less than 1 the reactants are favored
Equilibrium constant for acids is Ka and for bases Kb
Note: H3O+ = H +
Expression
Solutions can be expressed in terms of its
concentration or molarity
Acids and bases can also be expressed in terms of their
normality (N) or equivalence (Eq)
Molarity of solutions
We know that moles of a solution are the amount in grams
in relation to its molecular weight (MW or a.m.u.).
moles = grams / MW
A molar solution is one in which 1 liter of solution contains
the number of grams equal to its molecular weight.
M = moles / volume
Since (mol = grams / MW), you can calculate the grams of
a chemical you need to dissolve in a known volume of
water to obtain a certain concentration (M) using the
following formula:
grams = M x vol x MW
Exercise
How many grams do you need to make 5M NaCl
solution in 100 ml (MW 58.4)?
grams = 58.4 x 5 M x 0.1 liter = 29.29 g
Equivalents
When it comes to acids, bases and ions, it is useful to
think of them as equivalents.
An equivalent is the amount of moles or molar mass
(g) of hydrogen ions that an acid will donate
or a base will accept
A 1 g-Eq of any ion is defined as the molar mass of the
ion divided by the ionic charge.
Examples
1 mol HCl = 1 mol [H+] = 1 equivalent
1 mol H2SO4 = 2 mol [H+] = 2 equivalents
One equivalent of Na+ = 23.1 g
One equivalent of Cl- - 35.5 g
One equivalent of Mg+2 = (24.3)/2 = 12.15 g
Remember: One equivalent of any acid neutralizes one
equivalent of any base.
Exercise
Calculate milligrams of Ca+2 in blood if total
concentration of Ca+2 is 5 mEq/L.
1 Eq of Ca2+ = 40.1 g/2 =20.1 g
Grams of Ca2+ in blood =
= (5 mEq/L) x (1 Eq/1000 mEq) x (20.1 g/ 1 Eq)
= 0.1 mg/L
=100 mg/L
Normal solutions
Normality (N) considers both the molarity of the
solution and the equivalent content of the acid or base
N= n x M (where n is an integer)
For an acid solution, n is the number of H+ provided by
a formula unit of acid. Similarly, for a base solution is
the number of OH- a base can donate
Example
3 M H2SO4 solution is the same as a 6 N H2SO4 solution
1 M Ca(OH)2 solution is the same as a 2N Ca(OH)2
solution
Remember!
The normality of a solution is NEVER less than the
molarity
Exercise
What is the normality of H2SO3 solution made by
dissolving 6.5 g into 200 mL? (MW = 98)?
M = mol / MW or M = grams / (MW x vol)
= 6.5 g / (98 x 0.2 L)
= 6.5 / 19.6
= 0.33
N = M x n = 0.33 x 2 = 0.66 N
Normality and equivalents
Based on the equation above, since n eq of an acid is
neutralized by the same n eq of a base, then (N x liters) of
an acid is neutralized by (N x liters) of a base
Problems
If number of H+ (or OH-) is different, convert M to N, find the
answer in N, then convert N to M.
Note that each one produces 1 mole of H+ or OH-, so
1M of HCl is equal to 1M of NaOH, so 1M HCl produces
1M of H+, which is neutralized by 1M NaOH, which
produces 1M OHM1 x Vol1 = M2 x Vol2
0.12 x 22.4 = M2 x 12
M2 = (0.12 x 22.4) / 12
M2 = 0.224 M
Note that they 1 mole of HNO3 produces 1 mole of H+, but 1 mole
of Ba(OH)2 produces 2 moles of OH-. IN this case, it is better to
compare normalities (N), not molarities (M).
convert M to N first
N of HNO3 = 1 x 0.085 = 0.085 N
N of Ba(OH)2 = 2 x 0.12 = 0.24
N1 x Vol1 = N2 x Vol2
0.24 x 15 = 0.085 x Vol2
Vol2 = (0.24 x 15) / 0.085
Vol2 = 42.35 mL
Apply same method for third question
Ionization of water
Water dissociates into hydronium (H3O+) and hydroxyl
(OH-) ions
For simplicity, we refer to the hydronium ion as a
hydrogen ion (H+) and write the reaction equilibrium
as
Equilibrium constant
The equilibrium constant Keq of the dissociation of
water is
The equilibrium constant for water ionization under
standard conditions is 1.8 x 10-16 M
Kw
Since there are 55.6 moles of water in 1 liter, the
product of the hydrogen and hydroxide ion
concentrations results in a value of 1 x 10-14 for:
This constant, Kw, is called the ion product for water
[H+] and [OH-]
For pure water, there are equal concentrations of [H+]
and [OH-], each with a value of 1 x 10-7 M
Since Kw is a fixed value, the concentrations of [H+]
and [OH-] are inversely changing
If the concentration of H+ is high, then the
concentration of OH- must be low, and vice versa. For
example, if [H+] = 10-2 M, then [OH-] = 10-12 M
pH and buffer
What is pH?
Changes in [H+] have significant effects on many
biochemical processes
A logarithmic quantity, pH, was developed as a
convenient scale for working with levels of [H+] The pH
of a solution is a measure of its concentration of H+
(acidity)
The pH is defined as
Acidic and basic pH
On the pH scale, pH of 7.0 is considered neutral
pH values below 7.0 indicate acidic solutions
pH values above 7.0 indicate basic (alkaline) solutions
Acids and bases
An acid is a proton donor. A base is a proton acceptor
The species formed by the ionization of an acid is
proton(s) and conjugate base of the acid. Conversely,
protonation of a base yields its conjugate acid
The conjugate base of acetic acid is acetate ion
Strong acids
A strong acid dissociates completely, while a strong
base completely binds to a proton.
HA  H+ + AIn a 0.1 M solution of hydrogen chloride (HCl) in water,
[H+] = 0.1 M because all the HCl has dissociated into
H+ and Cl ions
For this solution pH = log 0.1 = 1.0
Strong bases
Similarly, a 0.01 M solution of NaOH will have an OHconcentration of 1 x 10-2
The pH is calculated as shown below:
[H+] = 1 x 10-14 / 1 x 10-2
= 1 x 10-12
The pH will be 12
Otherwise, you say if [OH-] = 10-2, then [H+] = 10-12 and
the pH = 12
Weak acids and bases
Many acids, such as the amino acids and acetic acid, do
not dissociate completely in water
Such acids are called weak acids, and similarly, some
bases do not completely dissociate and are called weak
bases
The ionization equilibrium of a weak acid is given as
Acid dissociation constant
Ka, the acid dissociation constant, is the
equilibrium constant for the reaction of a weak
acid (HA) converting into a proton and the
conjugate base (A-).
What is pKa?
As with pH, a logarithmic scale is useful for working
with Ka values. pKa is defined as:
pKa is the tendency of an acid to dissociate into H+ and
the conjugate base AEach weak acid and base have their own fixed pKa
values
Note: if Ka is very large, pKa is very small
The Henderson-Hasselbalch equation
What is the relation between pH, pKa, and the ratio of
acid to base?
A useful expression can be derived
When pKa = pH
At the point of the dissociation where the concentration
of the conjugate base [A-] is equal to that of the acid
[HA]:
pH = pKa + log[1]
The log of 1 = 0
pKa = pH
pKa is the pH where 50% of acid is dissociated into
conjugate base
Maintenance of equilibrium
Because a weak acid (or base) can exist in equilibrium
with its conjugate base, the equilibrium can be
maintained even when small amounts of [H+] or [OH-]
are added to it.
Example
If a small amount of [H+] is added to a solution of weak
acid, then the [H+] can combine with it to produce [HA],
maintaining [H+] level
If a small amount of [OH-] is added, then [OH-] will bind
to [H+] producing H2O, [HA] will dissociate to [H+] and
[A-] to compensate for the loss of [H+], maintaining [H+]
level
What is buffer?
The shift in equilibrium will then maintain the pH of
the solution
A solution with the ability to resist changes in pH is
called a buffer
Titration
The pKa values of weak
acids are determined by
titration.
This involves adding small
amounts of a strong acid
or base to the solution
and measuring the
resulting changes in pH
generating a plot of
titration curve.
Midpoint
For example, titration of
acetic acid gives the curve
shown to the right. Note
that the pKa value is the
midpoint of the curve.
At this point, pH is equal
to the pKa since there are
equal concentrations of
HA and A-.
Buffering capacity
The ability of a buffer to
minimize changes in pH is known
as its buffering capacity.
The buffering capacity of weak
acids and bases is one pH unit
from their pKa values.
In other words, a buffer is
effective at resisting changes in
pH at pH +/- 1 of the pKa.
For example, since the pKa of
acetic acid is 4.7, then the
buffering capacity ranges 3.7-5.7.
Conjugate bases
Acid
Conjugate base
CH3COOH
CH3COONa (NaCH3COO)
H3PO4
NaH2PO4
H2PO4- (or NaH2PO4)
Na2HPO4
H2CO3
NaHCO3
Properties of buffers
They all have the same
function.
Their pKa’s are different.
Buffering capacities are
same.
We choose a buffer
based on its pKa, which
should be close to the
desired pH.
Problems and solutions
the Henderson-Hasselbalch (HH) equation is useful for
determining the pH of a solution if the molar
proportion of A- to HA and the pKa of HA are known.
Example:
Consider a solution of 0.1 M acetic acid and 0.2 M
acetate ion. The pKa of acetic acid is 4.8. Hence, the pH
of the solution is given by
Similarly, the pKa of an acid can be calculated if the
molar proportion of A- to HA and the pH of the
solution are known.
Exercise
What is the pH of a buffer containing 0.1M HF and
0.1M NaF? (Ka = 3.5 x 10-4)
[H3O+]
= Ka x [HF]/[F-]
= 3.5 x 10-4 x (0.1/ 0.1)
= 3.5 x 10-4
pH
= -log 3.5 x 10-4
= 3.46
Can you solve it using the Henderson-Hasselbalch (HH)
equation?
Polyprotic weak acids
Some weak acids (buffers) can donate more than one
proton.
An example is phosphoric acid (H3PO4).
Phosphoric acid can donate up to three protons.
Different pKa
These protons do not dissociate at the same time.
Each proton has a certain pKa.
Hence
Each proton dissociate at certain pH
Excercise
What is the pH of a lactate buffer that contain 75%
lactic acid and 25% lactate? (pKa = 3.86)
What is the pKa of a dihydrogen phosphae buffer when
pH of 7.2 is obtained when 100 ml of 0.1 M NaH2PO3 is
mixed with 100 ml of 0.1 M Na2HPO3?
Excellent buffer
This makes phosphoric acid an excellent buffer at
different pH ranges.
It also the buffer intracellularly.
Buffers in human body
Carbonic acid-bicarbonate system (blood)
Dihydrogen phosphate-monohydrogen phosphate
system (intracellular)
Proteins (via the amino acids)
Blood buffering
Blood pH must be maintained at around 7.4.
Any dramatic change (up or down) can be dangerous.
By far the most important buffer for maintaining acidbase balance in the blood is the carbonic-acidbicarbonate buffer.
How?
During metabolism, cells produce CO2. This CO2 is
dissolved in blood and is converted to carbonic acid,
which dissociates into bicarbonate ion (conjugate base)
and H+.
Dissolved CO2 is in equilibrium with pressure of CO2
(pCO2) in lungs.
If blood pH changes, then the body changes the rate of
breathing in order to balance blood pH.
The Henderson-Hasselbalch equation for
blood buffering
The normal ratio of [HCO3-/CO2] in the blood is about 20 to 1
at pH 7.4. What is this ratio at pH 7.3 (acidosis) and at 7.5
(alkalosis)?
The equation
Relationships of the bicarbonate buffer system to the
lungs and the kidneys.
Roles of lungs and kidneys
Maintaining blood is balanced by the kidneys and the
lungs
Kidneys control blood HCO3 concentration ([HCO3])
Lungs control the blood CO2 concentration (PCO2)