1 Biomolecules in Water Water, the Biological Solvent Hydrogen Bonding and Solubility Cellular Reactions of Water Ionization, pH and pK The Henderson-Hasselbalch

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Biomolecules in WaterBiomolecules in Water

Water, the Biological Solvent

Hydrogen Bonding and Solubility

Cellular Reactions of Water

Ionization, pH and pK

The Henderson-Hasselbalch Equation

Buffer Systems

Water, the Biological Solvent

Hydrogen Bonding and Solubility

Cellular Reactions of Water

Ionization, pH and pK

The Henderson-Hasselbalch Equation

Buffer Systems

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WaterWater

The solvent of choice for biological systems.

• Medium for metabolism.

• Able to absorb large amounts of heat.

• Solvent for many materials.

• Used for transport - blood, cerebrospinal fluid, lymph, urine.

• Serves as a reactant or product of many biochemical reactions.

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Structure of waterStructure of water

Water is a polar moleculeWater is a polar moleculeElectronegativities of hydrogen (2.1) and oxygen (3.5) result in a polar bond.

O

HH

104.5o

Water’s ‘bent’ shaperesults in the moleculehaving a + and - end.+

-

+

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Structure of waterStructure of water

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Hydrogen bondingHydrogen bonding

The small size of hydrogen along with the shape and polarity of the water molecule all add up to a relatively strong attraction between water molecules.

Hydrogen bonding is the strongest of the intermolecular forces.

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Hydrogen bondingHydrogen bonding

This interaction can occur between water and many biomolecules.

It can also occur between two biomolecules.

O

CH2CH3H

O

HH

Example Example DNA relies on H bonding to hold thedouble helix together.

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Hydrogen bonding and solubility Physical properties of water

Hydrogen bonding and solubility Physical properties of water

Due to hydrogen bonding, water is unique for its molecular weight and size.

Property H2O NH3 CH4 H2S

Molecular weight 18 17 16 32Boiling point (oC) 100 -33 -161 -60.7Melting point(oC) 0 -78 -183 -85.5Viscositya 1.01 0.25 0.10 0.15

aUnits are centipoise.

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Water as a solventWater as a solvent

Water will dissolve biomolecules that are polar or ionic - hydrophilichydrophilic.

It has a high ion solvating ability.

Na+ 0.14 MK+ 0.004MCl- 0.10 M

Concentrationsin blood.

Concentrationsin blood.

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Dissolution of NaClDissolution of NaCl

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Water as a solventWater as a solvent

Nonpolar compounds like fats are not very soluble in water - hydrophobichydrophobic.

Some materials have both polar and nonpolar ends - amphiphilicamphiphilic.

One end tends to dissolve in polar solvents and the other in nonpolar ones.

CC

CC

CC

CC

CC

CC

CC

CC

O

OH

Example - saturated fatty acid

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How detergents workHow detergents work

COO--OOC COO-

COO-

COO-

-OOC

-OOC

-OOC

Nonpolar taildissolves in oil.

Polar ‘heads’are attractedto the water.

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MicellesMicelles

Amphiphilic molecules tend to organize into micelle structures.

The polar heads will point towards the aqueous environment and the nonpolar tails will be on the inside.

OH

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Cellular reactions of waterCellular reactions of water

Since water is the working solvent for biological systems, it is appropriate to review the acid/base chemistry of this solvent.

AutoionizationAutoionization

pK, pKw, pKapK, pKw, pKa

TitrationsTitrations

Buffers Buffers

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Autoionization of waterAutoionization of water

Water is an amphiproticamphiprotic substance that can act either as an acid or a base.

HC2H3O2(aq) + H2O(l) H3O+ + C2H3O2-(aq)

acid base acid base

H2O(l) + NH3(aq) NH4+

(aq) + OH-(aq)

acid base acid base

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Autoionization of waterAutoionization of water

AutoionizationAutoionizationWhen water molecules react with one

another to form ions.

H2O(l) + H2O(l) H3O+(aq) + OH-

(aq)

(10-7M) (10-7M)

Kw = [ H3O+ ] [ OH- ]

= 1.0 x 10-14 at 25oC

ion productof water

ion productof water

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Acid dissociation constant, KaAcid dissociation constant, Ka

The strength of a weak acid can be expressed as an equilibrium.

HA (aq) + H2O(l) H3O+(aq) + A-

(aq)

The strength of a weak acid is related to its equilibrium constant, Ka.

Ka = [A-][H3O+]

[HA]

We omit water. It’salready includedin the constant.

We omit water. It’salready includedin the constant.

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pH and pOHpH and pOH

We need to measure and use acids and bases over a very large concentration range.

pH and pOH are systems to keep track of these very large ranges.

pH = -log[H3O+]

pOH = -log[OH-]

pH + pOH = 14

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pH scalepH scale

A log based scale used to keep track of the large change important to acids and bases.

When you add an acid, the pH gets smaller.

When you add a base, the pH gets larger.

14 7 0

10-14 M 10-7 M 1 M

Very basic Neutral Very acidic

H+ H+ H+

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Analytical methods based on measurement of volume.

• If the concentration of an acid is known, the concentration of the base can be found.

• If we know the concentration of the base, then we can determine the amount of acid.

• All that is needed is some calibrated glassware and either an indicator or pH meter.

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BuretBuret - volumetric glassware used for titrations.

It allows you to add a known amount of your titrant to the solution you are testing.

If a pH meter is used, the equivalence point can be measured.

An indicator will give you the endpoint.

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Note the color change which indicates that the ‘endpoint’ has been reached.

Start End

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Indicator examplesIndicator examples

Acid-base indicators are weak acids that undergo a color change at a known pH.

phenolphthalein

pH

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Indicator examplesIndicator examples

methyl red

bromthymol blue

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Titration of weak acids or basesTitration of weak acids or bases

Titration of a weak acid or base with a strong titrant is a bit more complex than the strong acid/strong base example.

We must be concerned with conjugate acid/base pairs and their equilibrium.

ExampleExampleHA + H2O H3O+ + A-

acidacid basebase

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First, we’ll only be concerned about the titration of a weak acid with a strong base or a weak base with a strong acid.

• We still have the same four general regions for our titration curve.

• The calculation will require that you use the appropriate KA or KB relationship.

• We’ll start by reviewing the type of calculations involved and then work through an example.

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0% titration0% titrationIf your sample is an acid then use

KA =

At this point [H3O+] = [A-].

You can solve for [H3O+] by using either the quadratic or approximation approach.

Finally, calculate the pH.

[H3O+][A-][HA]

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0% titration0% titrationIf your sample is a base then use

KB =

At this point [OH-] = [HA].

You can solve for [OH-] by using either the quadratic or approximation approach.

Then determine pH as 14 - pOH.

[OH-][HA][A-]

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5 - 95% titration5 - 95% titrationIn this region, the pH is a function of the K value and the ‘ratio’ of the acid and base forms of our analyte.

A common format for the equilibrium expression used in the region is the Henderson-Hasselbalch equationHenderson-Hasselbalch equation.

We can present it in two forms depending on the type of material we started with.

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5 - 95% titration5 - 95% titration

Starting with an acidStarting with an acid

pH = pKA + log

Starting with a baseStarting with a base

pH = 14 - ( pKB + log )

( We’re just determining the pOH and then converting it to pH. )

[HA][A-]

[A-][HA]

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Another approach that can be taken is to simply use the % titration values.

For an acidic sample, you would use:

pH = pKA + log % titration

100 - % titration

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These equations have their limits and may break down if:

KA or KB < 10-3

or you are working with very dilute solutions.

In those cases, you need to consider the equilibrium for water.

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100% titration - the equivalence point.100% titration - the equivalence point.At the point, we have converted all of our sample to the conjugate form.

If your sample was an acid, now solve for the pH using the KB relationship -- do the opposite if you started with a base.

Remember that KRemember that KBB + K + KAA = 14 = 14

You must also account for dilution of the sample as a result of adding titrant.

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Overtitration (> 100%)Overtitration (> 100%)

These calculations are identical to those covered in our strong acid/strong base example.

You simply need to account for the amount of excess titrant added and how much it has been diluted.

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ExampleExampleA 100 ml solution of 0.10 M benzoic acid is titrated with 0.10 M NaOH. Construct a titration curve.

For benzoic acidFor benzoic acidKA = 6.31 x 10-5

pKA = 4.20

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0% titration0% titration

KA =

[H3O+] = [A-]

[HA] + [A-] = 0.10 M

We’ll assume that [A-] is negligible compared to [HA].

[H3O+][A-][HA]

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KA = 6.31 x 10-5 =

x = (6.31 x 10-5 )(0.10)

= 0.025 M

pH = 2.60

x 2

0.10

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10% titration10% titrationHere we can use the Henderson-Henderson-HasselbalchHasselbalch equation in % titration format:

pH = pKA + log

pH = 4.20 + log (10 / 90)

= 3.25

We can calculate other points by repeating this process.

% titration100 - % titration

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% titration pH

0 2.6010 3.2420 3.6030 3.8340 4.0250 4.2060 4.3870 4.5780 4.8090 5.15

Note: At 50% titration, pH = pKA

Also, there was only a change of 1.91 pH units as we went from 10 to 90 % titration.

Note: At 50% titration, pH = pKA

Also, there was only a change of 1.91 pH units as we went from 10 to 90 % titration.

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% titration

pH

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At this point, virtually all of our acid has been converted to the conjugate base - benzoate.

We need to use the KB relationship to readily solve for this point.

KB =

KB = KW / KA = 1.58 x 10-10

[OH-][HA][A-]

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At the equivalence point:

[HA] = [OH-]We’ve diluted the

[HA] + [A-] = 0.05 M sample and the total volume

at this point is 200 ml

Finally, we can assume that [benzoic acid] is negligible compared to [benzoate].

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KA = 1.58 x 10-10 =

x = (1.58 x 10-10 )(0.050)

= 2.81 x 10-6 M

pOH = 5.55 pH = 14 - pOH = 8.45

x 2

0.050

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% titration

pH

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OvertitrationOvertitrationAll we need to do here is to account for the dilution of our titrant.

10 % overtitration (10 ml excess)10 % overtitration (10 ml excess)

[OH-] = 0.10 M

= 0.0048 M (when diluted)pOH = 2.32pH = 14-2.32 = 11.68

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ml titrant total volume [OH-] pH

110 210 0.0048 11.68120 220 0.0091 11.96130 230 0.013 12.11140 240 0.017 12.23150 250 0.020 12.30

This is identical to what we obtained for our strong acid/strong base example.

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% titration

pH

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BuffersBuffers

Buffer solutionBuffer solution

• A mixture of a conjugate acid - base pair.

• It corresponds to approximately 10-90% titration - a relative flat region of the curve.

• This mixture tends to resist changes in pH when an acid or base is added.

• Buffers are commonly used when pH must be maintained at a relatively constant value and in many biological systems.

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BuffersBuffers

Effect of adding an acid or base.Effect of adding an acid or base.We can use the Henderson-Hasselbalch equation to evaluate pH changes of buffered systems.

pH = pKA + log

The type and degree of buffering is a function of the pKA of our buffering material and the concentration of both forms.

[A-][HA]

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BuffersBuffers

Effect of adding an acid or base.Effect of adding an acid or base.The best way to appreciate the effect of a buffer is to work an example.

Example.Example. A total of 100 ml of 1.0 M HCl is added in 10 ml increments to 100 ml of the following solutions:1) Pure water where pH = 72) A solution containing 1.0 M

HA, 1.0 M A- where pKA = 7.0

Determine the pH after each addition.

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BuffersBuffers

Initially, each solution is at pH 7.00.

After adding 10 ml of 1.0 M HCl we have:

Pure water[H3O+] = =

0.091

pH = 1.04

This is a pretty big jump!

(10 ml)(1.0 M)(110 ml)

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BuffersBuffers

Addition of 10 ml 1.0 M HCl to our buffered system.

• We started with 0.10 moles of both the acid and conjugate base forms.

• The addition of our first 10 ml can be expected to react with the conjugate base, converting it to the acid form.

• After addition, we would have 0.09 moles of the base form and 0.11 moles of the acid form.

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BuffersBuffers

New concentrationsNew concentrationsSince all we need to be concerned about is the ratio of A- to HA in the 10-90% region then:pH = pKA + log

= 7.00 + log

= 6.91

A change of only 0.09 compared to 5.96 with no buffer.

[A-][HA]

0.09 mmol1.1 mmol

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BuffersBuffers

ml HCl pHadded unbuffered buffered 0 7.00 7.00 10 1.04 6.91 20 0.78 6.82 30 0.64 6.73 40 0.54 6.63 50 0.48 6.52 60 0.43 6.40 70 0.39 6.25 80 0.35 6.05 90 0.32 5.72

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BuffersBuffers

ml HCl added

pH

buffered

unbuffered

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BuffersBuffers

For the addition of 100 ml of HCl, we have converted virtually all of A- to HA so the calculation is different.

KA = 1.00 x 10-7 =

When we account for dilution,1.0 M = [HA] + [A-]

where [A-] is negligible.

This is a standard weak acid calculation.

[H+][A-][HA]

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BuffersBuffers

pH after addition of 100 ml HCl.pH after addition of 100 ml HCl.

[H+] = ( KA x [HA] )1/2

= (1.00 x 10-7 x 1.0 M )1/2

= 3.16 x 10-4

pH = 3.5

At this point, we have exceeded the buffer region for our system.

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BuffersBuffers

ml HCl added

pH

buffered

unbuffered

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BuffersBuffers

Obviously, a buffer only has a limited ability to reduce pH changes.

The pKA determines the range where a buffer is useful.

The concentration of our buffer system determines how much acid or base it can deal with.

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BuffersBuffers

Buffer capacityBuffer capacityThe number of moles of a strong acid or base that causes 1 liter of a buffer solution to undergo a pH change of 1.00.

Since pH = pKA + log

then +1 = log

So it is the amount of acid or base required to convert ~80% of one form to the other if we start with a 1:1 ratio.

[A-][HA]

[A-][HA]

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BuffersBuffers

ExampleExampleDetermine the buffer capacity of a 1:1 mixture of a weak acid at the following concentrations:

1.0 M0.5 M0.1M

0.05 M0.01M

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BuffersBuffers

Assume that [ HA ] = [ A- ] = C, where C is the initial concentration of either our acid or base.

Buffer capacity in general is then:

1 = log

[ A- ] = 10 [ HA ]

[A-] = 2 C - [ HA ]

[ HA ] = 0.22 C

[A-][HA]

We’ll only worryabout addition

of a base.

We’ll only worryabout addition

of a base.

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BuffersBuffers

Concentration, MBuffer capacity (mol)1.0 0.220.5 0.110.1 0.0220.05 0.0110.01 0.0022

The capacity may be smaller if you don’t start with a 1:1 mixture.

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Buffers and bloodBuffers and blood

Control of blood pHControl of blood pH

– Oxygen is transported primarily by hemoglobin in the red blood cells.

– CO2 transported both in plasma and the red blood cells.

CO2 (aq) + 2 H2O

H2CO3 (aq)

H3O+(aq) + HCO3

-(aq)

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Buffers and bloodBuffers and blood

The amount of CO2 helps control blood pH.

Too much COToo much CO22 - Respiratory arrest

pH goes down, acid level goes up.

acidosisacidosis

SolutionSolution - ventilate and give bicarbonate via IV.

Too little COToo little CO22 - Hyperventilation, anxiety

pH goes up, acid level goes down.

alkalosisalkalosis

SolutionSolution - rebreath CO2 in paper bag to raise level.

Documents

1 Biomolecules in Water Water, the Biological Solvent Hydrogen Bonding and Solubility Cellular Reactions of Water Ionization, pH and pK The Henderson-Hasselbalch