Lecture 5 - Acid Base Concepts

8/18/2019 Lecture 5 - Acid Base Concepts

1/27

The Ionization of Water

and Acid-BaseChemistry

Lemon tree in blossom, Jose Escofet 1989


2/27

Water ionizes to positively and negatively charged ions

• electron-rich molecules are called nucleophiles and they react with electron-poor

molecules called electrophiles

• the oxygen atom in a water molecule is a nucleophile because of its two pairs

of un-bonded (un-shared) electrons.

• the lone-pair electrons of the oxygen of one H2O molecule can attack the

hydrogen atom of another H2O molecule. This results in the breaking ofexisting bonds and forming of new bonds in the reacting molecules.

the movement of a pair of

electrons (arrow) results in

new O-H bond forming the

Hydronium molecule (thenew H atom comes from

the second H2O molecule).

the original electron pair in

the second H2O molecule

remains associated with theresulting hydroxide ion.


3/27

H2O

[H2O]

Keq

OH-H+ +

[OH-]

[H+

]=

[H2O] Keq [OH-]

[H+

]=

Kw [OH-] [H+]=

[OH-] [H+] === =1/2

(Kw) (1 x 10-14)

1/21 x 10-7 M

the dissociation of water can be simplified to:

and may be written as:

the Kw for pure H2O at 25oC and 1 atm pressure is 1 x 10-14

Since the concentrations of H+ and OH- are equal:

where Keq is the equilibrium constant for the reaction.The concentration of water is essentially unchanged.

Therefore the constants can be combined and

re-written as:

Keq [H2O] is referred to as Kw so:

Any solution that contains equal concentrations of H+ and OH- (like pure water) is said

to be neutral.

Solutions with [H+] > [OH-] are acidic.

Solutions with [H+] < [OH-] are basic.


4/27

since for any solution:

[OH-] [H+] = 1 x 10-14 M2 we can calculate the concentration of one ion

if the concentration of the other is known.

if a solution has a [H+] of 1 x 10-3 M then:

[OH-] [H+] = 1 x 10-14 M2

[OH-] (1 x 10-3 M) = 1 x 10-14 M2

[OH-] = 1 x 10-11 M

[OH-] = 1 x 10

-14 M2

(1 x 10-3 M)

note that the [H+] (1 x 10-3 M) is much higher than the [OH-] (1 x 10-11).

A hundred million times higher to be exact!

Because these number range so widely, we convert them to a logarithmic scale, called pH


5/27

Strong Acids and Bases completely dissociate in water

HCl Cl-

H+ +

NaOH OH-Na+ +

This means that in an aqueous solution of 0.1 M HCl:

H+

== =[Cl-] [H+] 0.1 M 1 x 10-1 M

The strong acid hydrochloric acid and the strong base sodium hydroxide completely

dissociate in water

and the [OH-] will be 1 x 10-13 M

What is the pH of this solution? The pH is the negative log of the hydrogen ion concentration.

pH = -log10[H+]

pH = -log (1 x 10-1M)

pH = -(-1) = 1

Conversely, one can calculate the [H+] from the pH:

pH = -log[H+]

[H+] = antilog (-pH)

[H+] = antilog (-1)

[H+] = 1 x 10-1 M


6/27

The pH scale

pH, which is defined as the negative logarithm of the

Hydrogen ion concentration, is a convenient scale

turning small and awkward numbers intowhole numbers

Moran p43


7/27

Practice Questions (Solutions will be posted on D2L).

1. What is the concentration of OH- in an aqueous solution if the H+ concentration

is 1 x 10-8M? Is this solution acidic or basic?

2. What is the [H+] in an aqueous solution of 0.2 M HCl. What is the pH?

3. You add 10 ml of 0.1 M HCl to 990 ml water. What are the [H+] and [OH-]

in this solution? What is the pH of the solution?

Note, for examinations, it is your responsibility to:

-bring a calculator

-ensure it is a functional calculator

-know how to use the calculator


8/27

Weak Acids and Bases partially dissociate in water

Organic acids (which are the kind that are most important to biological systems) do not

completely dissociate like HCl.

HA A-H+ +

weak acid conjugate base of HA

The dissociation of a weak acid can be represented by the following equation:

For example, acetic acid (CH3COOH) dissociates to the conjugate base CH3COO- and H+

H+ +

weak acid, acetic acid

(CH3COOH)

conjugate base, acetate ion

(CH3COO- )

-

http://www.google.ca/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=CQCwfRANC3yjDM&tbnid=uF3MF2H8GskOfM:&ved=0CAUQjRw&url=http://www.chemistryinnovation.co.uk/stroadmap/roadmap.asp-id=459.htm&ei=zo4xUqyTIJOu2gWPxYGQCw&bvm=bv.52109249,d.b2I&psig=AFQjCNGyh34x3qGfhPlTH8--VC0uTVuvug&ust=1379065896755734http://www.google.ca/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=CQCwfRANC3yjDM&tbnid=uF3MF2H8GskOfM:&ved=0CAUQjRw&url=http://www.chemistryinnovation.co.uk/stroadmap/roadmap.asp-id=459.htm&ei=zo4xUqyTIJOu2gWPxYGQCw&bvm=bv.52109249,d.b2I&psig=AFQjCNGyh34x3qGfhPlTH8--VC0uTVuvug&ust=1379065896755734http://www.google.ca/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&docid=CQCwfRANC3yjDM&tbnid=uF3MF2H8GskOfM:&ved=0CAUQjRw&url=http://www.chemistryinnovation.co.uk/stroadmap/roadmap.asp-id=459.htm&ei=zo4xUqyTIJOu2gWPxYGQCw&bvm=bv.52109249,d.b2I&psig=AFQjCNGyh34x3qGfhPlTH8--VC0uTVuvug&ust=1379065896755734


9/27

[HA] K

a

[A-] [H+]

=

where Ka is the acid dissociation constant. The larger

the Ka, the stronger the acid.

The strength of a weak acid (its ability to release H+ ions) can be determined using

following expression:

for acetic acid, where Ka = 1.76 x 10-5 M we can calculate the [H+] in a 0.1 M solution

of acetic acid and also its pH:

[CH3COOH] Ka

[CH3OO-] [H+]=1.76 x 10-5 =

[CH3COOH]

[CH3OO-] [H+]

1.76 x 10-5 = 0.1 M

x2

=

simplifying assumption:

concentration of acid remainsnearly constant.

x = 1.33 x 10-3 M = [H+]

pH = -log (1.33 x 10-3M) = 2.88


10/27

Because Ka values vary over a wide range, they are expressed using a log scale:

pKa = -log10KaThe lower the pKa the stronger the acid.

McKee p87


11/27


4. What is the pH of a 0.2 M acetic acid solution where the Ka for acetic acid

is 1.33 x 10-5 M.

5. What is the pH of a 0.1 M formic acid solution where the pKa for formic acid is 3.77.


12/27

Buffers

• Buffers are solutions that resist changes in pH

• Regulating pH is a universal and essential activity in living organisms

• Human blood has a pH of 7.4 and can vary only between pH 7.35 and 7.45.

pH below 7.35 results in acidosis, eventual CNS depression, coma, death.

pH above 7.45 (alkalosis) results in muscle spasms, convulsions.

• There are 3 main buffering system in the body and in cells

Bicarbonate buffer

Phosphate buffer

Protein buffer (see later).

• Buffers are also often used in the lab and are made by mixing appropriate

concentrations of a weak acid with its conjugate base.


13/27

A Solution that resists changes in pH is said to be buffered

A buffered solution can resist pH changes because an equilibrium exists between the

weak acid and its conjugate base.

Consider a buffer composed of the weak acid acetic acid (CH3COOH)

and its conjugate base, the acetate ion (CH3COO-), which is initially at equilibrium.

KaIf more H+ ions are added

to this solution, it drivesthe equilibrium in the

direction of acetic acid

formation and the pH

changes little.

If OH-

ions are added

to this solution, they combine

with the H+ ions forming

water and this drives

the equilibrium in the

opposite direction. The pHchanges little.

Ka

McKee p89


14/27

This is why, when we titrate (in the example below) a conjugate base by adding acid

this kind of curve is obtained:

acid

Notice the pH changesless rapidly where

pH = pKa (4.73) and

+/- one unit.


15/27

Water itself is weakly buffered

H2O OH-

H+

+

H+

if we add extra H+ ions to water

they will combine with OH- ionsto form water, driving the equilibrium

towards H2O formation.

• the problem is that the dissociation constant of water is so low, there are few OH-

ions available to react with H+ ions. Thus when acid is added to

pure water, the pH plummets immediately (we therefore view water as

unbuffered).

• Chemical reactions often produce or consume H+ ions. When we study these

reactions in the lab, we do not want the pH of the reaction liquid to change.

• Thus we add a buffer: a mixture of weak acid and its conjugate base to resist the

change in pH that would otherwise occur and perhaps ruin (or confound)

the experiment.

Kw

h d lb l h


16/27

The Henderson-Hasselbalch Equation

• When choosing or making a buffer in the laboratory, pH and pKa are useful concepts

•

The relationship between these quantities is expressed in theHenderson-Hasselbalch equation which is derived as below.

[HA] Ka

[A-] [H+]

=[HA]

Ka

[A

-

]

[H+] =

[HA] Ka

[A-]

[H+] =-log -log -log

solving for [H+] gives

taking the -log of each side gives

pH = pKa[HA]

[A-]

-log

pH = pKa

[HA]

[A-]

+ log

remembering the definitions ofpH and pKa gives

inverting the log term thus changing

its sign, the H-H equation is

obtained.

[A ]


17/27

pH = pKa[HA]

[A-] + log

• A titration curve reveals that the best buffering situation (where a solution best

resists changes in pH when acid or base is added) occurs when the

concentrations of the acid [HA] and its conjugate base [A-] are equal.

• This situation occurs when the pH of the solution equals the pKa of the acid:

pH = pKa[HA]

[A-]

+ log

pH = pKa + log (1) since log(1) = 0, then

pH = pKa

the most effective buffering occurs 1 pH unit above and below an acid’s pKa

S th ti d t ll i h i l b d t k b ff


18/27

Synthetic and naturally occurring chemicals can be used to make buffers

if the experimental conditions demand that the solution be buffered between

pH 3.5 and 5.5, then an acetate buffer (acetic acid pKa 4.73) would be good.

if the solution needs to be buffered between pH 6 and 8 ( a typical pH rangewhen dealing with biological systems), the acetate buffer would be unsuitable and

one would look for a buffer system with a pKa around 7

U i h H d H lb l h i f l b ff


19/27

Using the Henderson-Hasselbalch equation to formulate buffer

cetic acid has a pKa of 4.8. How many ml of 0.1 M acetic acid and 0.1 M Na acetate are

required to prepare 1 litre of 0.1 M buffer solution having a pH of 5.8?

pH = pKa[HA] [A

-] + log

5.8 = 4.8

[HA]

[A-]

+ log

antilog (1) =[HA]

[A-] 10 =

[HA]

[A-]

1

10 =

10 volumes of acetate ion (Na acetate) must be added for every 1 volume of acetic acidequaling a total of 11 volumes.

acetic acid needed

acetate needed

11

1 x 1000 ml = 91 ml

11

10

x 1000 ml = 909 ml

substitute desired pH and pKa into H-H

equation

solve for ratio of [A-]/[HA]


20/27


6. You are studying a enzyme in the laboratory that is only active between pH 5 and 8. You

wish to study the activity in a buffer of pH 7. Pick the best buffer system for making your

buffer:

Buffer system pKa

Tris/HCl 8.1

Acetic acid/acetate 4.8

Lactic acid/lactate 3.9

Carbonate 6.4

Bicarbonate 10.2

Phosphate 7.2

7. Formic acid has a pKa of 3.8. How many ml of 0.2 M formic acid and 0.2 M Na formate

would you need to mix to make 0.5 litres of 0.2 M buffer solution having a pH of 3.8?

8. Lactic acid has a pKa of 3.9. How many ml of 0.4 M lactic acid and 0.2 M Na lactatewould you need to mix to make 0.5 litres of 0.2 M buffer solution having a pH of 3.9?


21/27

Some acids have multiple ionizable H atoms

CH3COOH CH3COO-H+ +

Acetic acid has one ionizable H atom:

pKa = 4.8

Phosphoric acid has 3 ionizable H atoms:

H3PO4 H2PO4-H+ + pKa1 = 2.2

H2PO4- HPO4

2-H+ + pKa2 = 7.2

HPO42- PO4

3-H+ + pKa3 = 12.7


22/27

Many important biological acids

and bases have 2 or more

ionizable groups including

the amino acids, which can

also act as buffers.

Phosphoric acid (H3PO4) is

called a polyprotic acid.

Moran p48

N t ll i b ff t i bi l


23/27

Naturally occurring buffer systems in biology

• Bicarbonate buffer - used to regulate the pH of blood, which must be very

strictly controlled. Involves the bicarbonate ion HCO3- and CO2. The amount

of CO2 in the blood is strictly controlled by the amount exhaled whenbreathing.

• Phosphate buffer - used to regulate the pH inside of cells.

•

Protein buffers - since there is a high concentration of protein in the body andin cells, the amino acids that make up proteins have a considerable buffering

effect on pH. All amino acids have at least two ionizable groups, the amino

group and the carboxylate group (as shown in this Aa below).

Some amino acids also have ionizable side chains as well, as we will see later.

C C

CH3

H OH

O

H3N+ C C

CH3

H O-

O

H3N+ C C

CH3

H

O

H2N

O-

pKa1 = 2.3 pKa2 = 9.7


24/27

H2N – C – CO

OHH

R

–

–

All possess:

-an α Carbon to which is attached

-an amino group

-a carboxylic acid group

-a hydrogen atom

-a unique sidechain (R)

Hints to get started on becoming familiar with Amino acids


25/27

H3N – C – C

O

O-H

R

–

–+

The amino and carboxyl group can be ionized depending on the surrounding pH.

At the pH typical of cytoplasm (pH 6.8 – 7.4, physiological pH) the amino group will be

protonated and the carboxyl group will be deprotonated.

Thus we typically draw amino acids in this form:

The

zwitterionic form of

an amino acid

i i d f f i id fully ionized form of amino acid


26/27

H3N – C – C

O

O-H

R

–

–+

H2N – C – C

O

OHH

R

–

–

non-ionized form of amino acid

does not exist in appreciable amounts

fully ionized form of amino acid

-major form at physio pH (6.8-7.2)

note all amino acids

have ionized carboxyl

group above pH 2

and ionized aminogroup below

pH 9-10

Key Concepts


27/27

Key Concepts• Water itself ionizes to a very small degree into H+ (H3O

+) ions and OH- ions

• pH (a logarithmic scale) is used because the H+ concentrations can vary over such

a vast range.

• pKa (a logarithmic scale) is a measure of the ability of a weak acid to dissociate

into H+ ions and the conjugate base of the weak acid

• Buffers are mixtures of weak acids and their conjugate bases. They resist changes

in pH.

• The Henderson-Hasselbalch equation relates pH and pKa and is useful for creating

buffers in the laboratory.

•

Naturally occurring buffers regulate pH in the blood, in cells, and other body fluids.

• pH influences the ionization state of many biomolecules and thus influences their

chemical characteristics and the chemistry of the cell.

• H+ ions participate ( are produced and consumed) in many chemical reactions in

the cell

Lecture 5 - Acid Base Concepts

Documents

Transcript of Lecture 5 - Acid Base Concepts