acids, bases, and salts - warszawski uniwersytet medyczny · pdf fileaccording to arrhenius,...

Acids, Bases, and Salts

History of Acids and Bases In the early days of chemistry chemists were organizing

physical and chemical properties of substances. They

discovered that many substances could be placed in two

different property categories:

Substance A

1. Sour taste

2. Reacts with carbonates to make CO2

3. Reacts with metals to produce H2

4. Turns litmus - red

5. Reacts with B substances to make

salt and water

Substance B

1. Bitter taste

2. Reacts with fats to make soaps

3. Do not react with metals

4. Turns litmus - blue

5. Reacts with A substances make

salt and water

Arrhenius was the first person to suggest a reason why

substances are in A or B due to their ionization in water.

The Swedish chemist Svante Arrhenius proposed the first

definition of acids and bases.

(Substances A and B became

known as acids and bases)

According to the Arrhenius model:

“acids are substances that dissociate in water to

produce H+ ions and bases are substances that

dissociate in water to produce OH- ions”

NaOH (aq) Na+ (aq) + OH- (aq) Base

HCl (aq) H+ (aq) + Cl- (aq) Acid

Arrhenius Theory

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What is H+?

+

e- +

Hydrogen (H) Proton (H+)

Unknown to Arrhenius free H+ ions do not exist in water. They

covalently react with water to produce hydronium ions, H3O+.

or:

H+ (aq) + H2O (l) H3O+ (aq)

This new bond is called a coordinate covalent bond since

both new bonding electrons come from the same atom

Hydronium Ion

Hydronium ion is the name for H3O+ and is often times

abbreviated as H+ (aq) they both mean the same thing.

What is the difference between a strong acid and a weak

acid?

Hydronium Ion




acid? Strong acids ionize 100% and weak ones do not!

Hydronium Ion





A single arrow is used to represent the ionization of a strong

acid. Double arrows (Equilibrium) are used to represent

weak acids.

For example: HCl (g) H+ (aq) + Cl - (aq)

HF (g) H+ (aq) + F -

Hydronium Ion







weak acids.


HF (g) H+ (aq) + F - (aq)

According to Arrhenius, is water an acid or base?

HOH (l) H+ (aq) + OH – (aq)

Hydronium Ion







weak acids.


HF (g) H+ (aq) + F -

According to Arrhenius, is water an acid or base?

HOH (l) H+ (aq) + OH – (aq)

Neither, he called it Neutral (same amount of OH- and H+

Hydronium Ion

Strong Acids and Bases

How can we identify strong acids or bases?

Copyright McGraw-Hill 2009 12

Method to Distinguish Types of Electrolytes

nonelectrolyte weak electrolyte strong electrolyte

How can we identify strong acids or bases?

Easy, memorize them!

Memorized Strong Acids

1. HClO4

2. H2SO4

3. HI

4. HBr

5. HCl

6. HNO3

Memorized Strong Bases

Hydroxides of group 1 and 2

metals, excluding Be and Mg

Strong Acids and Bases

Johannes Brønsted and Thomas Lowry revised

Arrhenius’s acid-base theory to include this behavior.

They defined acids and bases as follows:

“An acid is a hydrogen containing species that

donates a proton. A base is any substance that

accepts a proton”

HCl (aq) + H2O (l) Cl- (aq) + H3O+ (aq)

In the above example what is the Brønsted acid? What is

the Brønsted base?

Bronsted Lowry

Bronsted Lowry Theory

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http://www.chemcool.com/biography/Bronsted.gif

HCl (aq) + H2O (l) Cl - ( aq) + H3O+ (aq)

In reality, the reaction of HCl with H2O is an equilibrium

and occurs in both directions, although in this case the

equilibrium lies far to the right.

For the reverse reaction Cl - behaves as a Brønsted

base and H3O+ behaves as a Brønsted acid.

The Cl- is called the conjugate base of HCl. Brønsted

acids and bases always exist as conjugate acid-base

pairs.

Bronsted Lowry Theory

In pure water (no solute) water molecules behave as both an

acid and base!!

e.g.

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

This is called the self-ionization (autoionizaion) of water.

Although the equilibrium lies far to the left it is very important to

take into consideration, especially for living systems.

Does anyone know how we write the equilibrium constant for

this reaction?

Autoionization of Water

The auto-ionization of water is described by the

equation:

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

The equilibrium constant for this reaction is given by:

]OH][O3H[2]O2H[K

2]O2H[

]OH][O3H[

]O2H][O2H[

]OH][O3H[K

Kw = K[H2O]2 = 10-14

For pure water [OH-] = [H+] = 1 x 10-7 M


We define an aqueous solution as being neutral when the

[H+] = [OH-].

We define an aqueous solution as being acidic when

[H+] > [OH-].

We define an aqueous solution as being basic when

[H+] < [OH-].

However, in each case Kw = 1 x 10-14 M2

[H+] = 0.0000001 = 10-7


The mathematical definition of pH using [H+] for [H3O+] is

listed below:

pH = -log [H+], or [H+]= 1x10-pH (both are mathematically

equivalent)

How about the power for the OH -, what should this be called?

The mathematical definition of pH using [H+] for [H3O+] is listed

below:

pH = -log [H+], or [H+] = 1x10-pH (both are mathematically

equivalent)

pOH = -log [OH-], or [OH-] = 1x10-pOH (both are mathematically equivalent)


The mathematical definition of pH using [H+] for [H3O+] is

listed below:

pH = -log [H+], or [H+]= 1x10-pH (both are mathematically

equivalent)

How about the power for the OH -, what should this be called? Would you believe pOH?

Have you heard of pOH before?

pH + pOH = 14 for water solutions.


Now for some examples

1. Find the pH and pOH, when [H+] = 10-4

Now for some examples

1. Find the pH and pOH, when [H+] = 10-4

pH = 4 and pOH = 10, since they must add to 14

using the calculator pH = -log [H+], type in 10-4, push

the log button and pH = -(-4) = 4. Same for pOH

A pH Number line Number lines have been used in history and math classes,

so to keep up we use them in chemistry classes.

pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] > [OH -]

acidic



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] > [OH -]

acidic



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] = [OH -]

neutral

[H+] > [OH -] acidic



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2

[H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic



pH = 16

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2

[H+] =10-16

[OH -] =

[H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic



pH = 14

pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2

[H+] =10-14

[OH -] = 102

[H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic



pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2 [H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic



pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2 [H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic

acidic



pH = 12

pH = 7

pH = 2 [H+] = 10-2

[OH -] = 10-12

[H+] = 10-7

[OH -] = 10-7

[H+] =10-12

[OH -] = 10-2 [H+] < [OH -]

basic

[H+] = [OH -]

neutral

[H+] > [OH -]

acidic

acidic

basic

More Examples of pH from Daily Life

An acid is any ionic compound that

releases hydrogen ions (H+) in solution.

Weak acids have a sour taste.

Strong acids are highly corrosive (So don’t go around taste-testing acids.)

Examples:

• Ascorbic acid (C6H8O6, Vitamin C)

• Citric acid (C6H8O7, a weak organic acid in citrus fruits)

• Phosphoric acid (H3PO4, in pop…this stuff is also used to remove rust…hmmm)

Acids undergo characteristic double replacement reactions

with oxides, hydroxides, carbonates and bicarbonates.

e.g.

2HCl (aq) + CuO (s) CuCl2 (aq) + H2O (l)

2HCl (aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2H2O (l)

2HCl (aq) + CaCO3 (aq) CaCl2 (aq) + H2O (l) + CO2 (g)

2HC l (aq) + Sr(HCO3)2 (aq) SrCl2 (aq) + 2H2O (l) + 2CO2 (g)

Equations With Acuids

Bases undergo a double replacement reaction with acids

called neutralization:

NaOH (aq) + HCl (aq) H2O (l) + NaC l (aq)

In words this well known reaction is often described as:

“acid plus base = salt plus water”

Equations With Acuids

We have discussed the double replacement reactions and ionic

equations before. Since the acids and bases undergo double

replacement reactions called neutralization reactions, then they

can have ionic equations too.

e.g.

Formula equation:

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

Ionic equation:

H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H2O (l)

Net ionic equation:

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

Ionic Equations (a review)

Another property of acids is their reaction with certain metals to

produce hydrogen gas, H2 (g).

Zn (s) + 2HC l (aq) H2 (g) + ZnCl2 (aq)

This is an example of a single replacement reaction and is a

redox reaction.

Total ionic equation:

Zn (s) + 2H+ (aq) + 2Cl- (aq) H2 (g) + Zn2+ (aq) + 2Cl- (aq)

Net ionic equation:

Zn (s) + 2H+ (aq) H2 (g) + Zn2+ (aq)

Acidic Single Replacement Reactions

Salts Salts are the ionic product of an acid base neutralization

reaction.

Acidic Salts are formed from a strong acid and a weak base.

Neutral salts are formed from a strong acid and strong base.

Basic salts are formed from a strong base and a weak acid.

Give the acid and base the following salts were formed from

and label the salts as acidic, basic, or neutral.

1. NaCl

2. NaC2H3O2

3. NH4Cl


reaction.






1. NaCl

1. NaC2H3O2

1. NH4Cl

NaCl + HOH Reactants are?


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl

NaCl + HOH HCl + NaOH

NaC2H3O2 + HOH

S.A. s.b.


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl


NaC2H3O2 + HOH

S.A. s.b. Neutral salt


reaction.






1. NaCl

2. Na2CO3

3. NH4Cl


Na2CO3 + HOH H2CO3 + 2NaOH

Neutral salt s.a. s.b.


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl


NaC2H3O2 + HOH HC2H3O2 + NaOH w.a. s.b.



reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl


NaC2H3O2 + HOH HC2H3O2 + NaOH w.a. s.b. basic salt



reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl




NH4Cl + HOH


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl




NH4Cl + HOH NH4OH HCl +


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl




NH4Cl + HOH NH4OH HCl + s.a. w.b.


reaction.






1. NaCl

2. NaC2H3O2

3. NH4Cl



neutral salt s.a. s.b.

NH4Cl + HOH NH4OH HCl + s.a. w.b. acidic salt

Acid, Base, and Salt Hydrolysis

HBr (aq)


HBr (aq) H+ (aq) + Br - (aq)


HBr (aq) H+ (aq) + Br - (aq) Acidic, because H+ (aq)



0.1 Initial concentration



0.1 Initial concentration 0.0



0.1 Initial concentration 0.0 ?



0.1 Initial concentration 0.0 0.0




Final concentration ?




Final concentration 0.0




Final concentration 0.0 ?




Final concentration 0.0 0.1




Final concentration 0.0 0.1 ?




Final concentration 0.0 0.1 0.1





[H+] = ?





[H+] = 0.1





pH = ?

[H+] = 0.1





pH = ?

[H+] = 0.1 = 10-1





pH = 1

[H+] = 0.1 = 10-1





pH = 1

Ca(OH)2 (aq)





pH = 1

Ca(OH)2 (aq) Ca2+ (aq) + 2 OH- (aq)





pH = 1

Ca(OH)2 (aq) Ca2+ (aq) + 2 OH- (aq) acidic?





pH = 1

Ca(OH)2 (aq) Ca2+ (aq) + 2 OH- (aq) No, basic OH-





pH = 1


Initial concentration 0.1





pH = 1


Initial concentration 0.1 0.0





pH = 1


Initial concentration 0.1 0.0 ?





pH = 1


Initial concentration 0.1 0.0 0.0





pH = 1



Final concentration ?





pH = 1



Final concentration 0.0 ?





pH = 1



Final concentration 0.0 0.1





pH = 1



Final concentration 0.0 0.1 ?





pH = 1








pH = 1




[OH - ] = ?





pH = 1




[OH - ] = 0.2





pH = 1




pOH = ?

[OH - ] = 0.2





pH = 1




pOH = - log[OH-]

[OH - ] = 0.2





pH = 1




pOH = - log[OH-] = - log[0.2]

[OH - ] = 0.2





pH = 1




pOH = - log[OH-] = - log[0.2] = -(-0.698970004)

pOH = 0.7

[OH - ] = 0.2





pH = 1




pOH = - log[OH-] = - log[0.2] = -(-0.698970004)

pOH = 0.7

pH = ?

[OH - ] = 0.2





pH = 1




pOH = - log[OH-] = - log[0.2] = -(-0.698970004)

pOH = 0.7

pH = 14.0 - 0.07 = 13.3

[OH - ] = 0.2





pH = 1



final concentration 0.0 0.1 0.2

pH = 13.3





pH = 1




pH = 13.3

NaF (aq)





pH = 1




pH = 13.3

NaF (aq) Na+ (aq) + F – (aq)





pH = 1




pH = 13.3

NaF (aq) Na+ (aq) + F – (aq) Acidic, basic, or neutral?





pH = 1




pH = 13.3

NaF (aq) Na+ (aq) + F – (aq) Basic, since HF is w.a. and

NaOH is s.b.

Will sodium and fluorine ions react with water?





pH = 1




pH = 13.3


NaOH is s.b.


Na+ + HOH NaOH (sb) + H+





pH = 1




pH = 13.3


NaOH is s.b.


Na+ + HOH NaOH (sb) + H+

Na+ + HOH Na+ + OH- + H+

HOH OH- + H+ No Reaction, water

cannot make water





pH = 1




pH = 13.3


NaOH is s.b.


Na+ + HOH NaOH + H+ Cannot make strong

acids or bases from weak

ones s.b.





pH = 1




pH = 13.3


NaOH is s.b.




ones

F - + HOH HF + OH- w.a.





pH = 1




pH = 13.3


NaOH is s.b.




ones

F - + HOH HF + OH- Yes, HF weak acid and

OH- is formed, thus basic

salt! w.a.


NH4Cl (aq) NH4+

(aq) + Cl- (aq)


NH4Cl (aq) NH4+

(aq) + Cl- (aq) acidic, basic, or neutral?


NH4Cl (aq) NH4+


HCl + NH4OH NH4Cl + HOH


NH4Cl (aq) NH4+


HCl + NH4OH NH4Cl + HOH s.a. w.b.


NH4Cl (aq) NH4+

(aq) + Cl- (aq) Acidic!


Will the ions from the salt combine with water?

NH4+ + HOH NH4OH + H+


NH4Cl (aq) NH4+





w.b.


NH4Cl (aq) NH4+





w.b.

This reaction is OK,

since a w.b. is formed


NH4Cl (aq) NH4+





w.b.



Cl- + HOH HCl + OH-


NH4Cl (aq) NH4+





w.b.



Cl- + HOH HCl (sa) + OH-


NH4Cl (aq) NH4+





w.b.



Cl- + HOH H+ + Cl- + OH-


NH4Cl (aq) NH4+





w.b.



HOH H+ + OH- Again water

cannot make

water! NR


NH4Cl (aq) NH4+





w.b.



Cl- + HOH HCl + OH- s.a.


NH4Cl (aq) NH4+





w.b.




Cannot form s.a. from

weaker reactants, thus

N.R.


NH4Cl (aq) NH4+





w.b.




Cannot form s.a. from

weaker reactants, thus

N.R.

Since H+ was formed in the first reaction, then [H+] is now

greater than [OH-] making the solution acidic


NaCl (aq)


NaCl (aq) Na+ (aq) + Cl- (aq)


NaCl (aq) Na+ (aq) + Cl- (aq) Acidic, basic, or neutral?



HCl + NaOH NaCl + HOH



HCl + NaOH NaCl + HOH s.a. s.b.


NaCl (aq) Na+ (aq) + Cl- (aq) Neutral!





Now react each of the ions with water.

Na+ + HOH NaOH + H+





Na+ + HOH NaOH + H+ s.b.






Cannot form strong

bases from weaker

ones, thus N.R.






Cannot form strong

bases from weaker

ones, thus N.R.

Cl- + HOH HCl + OH-






Cannot form strong

bases from weaker

ones, thus N.R.

Cl- + HOH HCl + OH-

s.a.






Cannot form strong

bases from weaker

ones, thus N.R.

Cl- + HOH HCl + OH-

s.a.

Cannot form strong

acids from weaker

ones, thus N.R.

Buffers

Buffers are extremely important in chemistry and biology. They maintain a nearly consistent pH in various solutions.

Buffers Buffers are extremely important in chemistry and biology. They maintain a nearly consistent pH in various solutions.

Our blood must maintain a pH around 7.35-7.45. If the pH is above 7.45 you would have a condition called alkalosis. If the pH is below 7.35, then one would suffer from acidosis.



Acidosis leads to depression of the nervous system. Mild acidosis can result in dizziness, disorientation, or fainting; a more severe case can cause coma, or death.



What would happen to the pH of our blood if we were to eat acidic foods, such as apples, oranges, or limes? What might happen to the pH of our blood if some of the hydrochloric acid from our stomach were to seep into our blood?




What would happen to the pH of our blood if we were to eat acidic foods, such as apples, oranges, or limes? What might happen to the pH of our blood if some of the hydrochloric acid from our stomach were to seep into our blood? The pH would

be lower in both


Despite the possibility of pH increases or decreases, the body maintains a nearly constant pH of 7.4. The body uses buffers to maintain this remarkable feat.

What is a buffer and how does it work?



A buffer consists of a weak acid and the salt of its conjugate base, or a weak base and the salt of its conjugate acid.

Examples:

HF + NaOH NaF + HOH w.a. c.b.



A buffer consists of a weak acid and the salt of its conjugate base, or a weak base and the salt of its conjugate acid.

Examples:

HF + NaOH NaF + HOH w.a. c.b.

NH3 + HCl NH4Cl w.b. c.a.

1.0 L

HF (g) NaF (s)

Buffer preparation: Add 0.10 mole HF (g) and NaF (s) to 1.0 L of water.

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1.0 L

HF (g) NaF (s)


HF (g) H+ + F-

NaF (s) Na+ + F-

H+ Na+ F- HF

large small


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

HF H+

Na+ F-

Now add the strong acid HCl

HCl

large small


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

HF H+

Na+ F-


HCl

HCl H+ + Cl-

H+ Cl-

What will the pH be if just water and no buffer?

Large small


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

H+

HF

Na+ F-


HCl

HCl H+ + Cl-

H+ Cl-

What will the pH be if just water and no buffer? pH = 1, dead if this is your blood.

Large small


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

H+

HF

Na+ F-


HCl

HCl H+ + Cl-

H+ Cl-


Large small

What removes the H+ to keep the pH near 7?


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

H+

HF

Na+ F-


HCl

HCl H+ + Cl-

H+ Cl-


Large small

What removes the H+ to keep the pH near 7? The conjugate base, F-

H+ + F- HF (a weak acid, low H+ )


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

H+

HF

Na+ F-

Now add the strong base NaOH

NaOH

Na+ OH- Large small

What will the pH be if just water and no buffer?

NaOH Na+ + OH-


1.0 L


HF (g) H+ + F-

NaF (s) Na+ + F-

H+

HF

Na+ F-

Now add the strong base NaOH

NaOH

Na+ OH- Large small

What will the pH be if just water and no buffer? PH = 13, dead again

NaOH Na+ + OH-

What removes the OH- to keep the pH near 7? The acid HF

HF + OH- F- + HOH


Buffer Capacity

• The buffer capacity is the amount of

acid or base that can be added

before a significant change in pH

• This depends on the amounts of HA

and A- present in the buffer

• Most efficient buffer is when

[A-]

[HA] = 1

Henderson-Hasselbach Equation

• Derived from the equilibrium expression of

a weak acid and the pH equation.

• This equation allows you to determine the

pH of a buffer solution

• pKa = -log Ka

pH = pKa + log

[A-]

[HA]

The Common Buffer Problems • 1. Compute pH of a buffer given the actual

concentrations of the conjugate acid and conjugate

base.

• The pH is easily calculated with:

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

• Example: Determine the pH in which 1.00 mole of

H2CO3 (Ka = 4.2 x 10-7) and 1.00 mole NaHCO3

dissolved in enough water to form 1.00 Liters of solution.

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

pH = -log (4.2 x 10-7) + log (1.00M)/(1.00M)

pH = pKa = 6.4

** pKa of a weak acid can help determine pH of buffer

you will make if it is mixed in a 1:1 mole ratio!

Make a buffer problem

• How many moles of NaHCO3 should be added to 1 liter of

0.100M H2CO3 (Ka = 4.2 x 10-7 ) to prepare a buffer with a

pH of 7.00?

pH = pKa + log [HCO3-]

[H2CO3]

7.00 = -log(4.2 x 10-7) + log [HCO3-]

(0.100)

0.60 = log[HCO3-]

0.001

100.6 = [HCO3-]

0.100

[HCO3-] = 0.40 moles

should be added!

• The pH of body fluids is maintained by three major buffer

systems. The carbonic acid–bicarbonate system, the

dihydrogen phosphate-hydrogen phosphate system, and

a third system that depends on the ability of proteins to

act as either proton acceptors or proton donors at

different pH values.

• The carbonic acid–bicarbonate system is the principal

buffer in blood serum and other extracellular fluids. The

hydrogen phosphate system is the major buffer within

cells.

Buffers in the body

Normal functions of proteins (especially enzymes)

heavily depend on an optimal pH.

pH7.35-pH7.45

Regulation of acid-base balance

1) Chemical Buffers

2) Respiratory Control of pH

3) Renal Control of pH

H2O

pH 7.0 Buffer

pH 7.0

acid

pH 3.0 pH 6.8

acid

H2O

pH 7.0 Buffer

pH 7.0

base

pH 11.0 pH 7.2

base

3) The Protein Buffer

There are three major buffers in body fluid.

1) The Bicarbonate (HCO3-) Buffer

2) The Phosphate Buffer

Chemical Buffers

1) The Bicarbonate (HCO3-) Buffer System

H + HCO3- H2CO3 H2O + CO2

- reversible depending on the equilibrium between

the substrates and products.

- The lungs constantly remove CO2.

Each day, acid produced in the body is excreted in the urine. The kidney returns HCO3

- to the extracellular fluids, where it becomes part of the bicarbonate reserve.

2) The Phosphate Buffer System

H + HPO42– H2PO4

– + H H3PO4

3) The Protein Buffer System

- more concentrated than either bicarbonate or

phosphate buffers

- accounts for about three-quarters of all chemical

buffering ability of the body fluids.

- The carboxyl groups release H+ when pH rises and

amino groups bind H+ when pH falls.

NH2-CH2-CH2 CH2-CH2-COOH

H+ H+

Properties of Chemical Buffers

- respond to pH changes within a fraction

of a second.

- Bind to H but can not remove H out of

the body

- Limited ability to correct pH changes

pH

stimulate peripheral/central chemoreceptors

pulmonary ventilation

removal of CO2 and pH

H2CO3

H

+ HCO3- H2O + CO2

Titration Titration is an experimental procedure to determine the concentration of an unknown acid or base.

The figure on the left shows the glassware for a titration experiment. A buret clamp holds the buret to a ring stand and below the buret is a flask containing the solution to be titrated, which includes an indicator. The purpose of the indicator is to indicate the point of neutralization by a color change.

The picture on the left shows the tip of a buret, with air bubble, which is not good, and also shows the stop-cock. Note the position of the stop-cock is in the “off” position. This picture shows the color of the phenolphthalein indicator at the end-point. In this experiment a 23.00 mL aliquot of 0.1000 M NaOH titrant is added to 5.00 mL of an unknown HCL solution. The acid solution in the beaker starts out clear and becomes pink when all of the HCL has been consumed.

NaOH + HCl NaCl + HOH

© 2015 Pearson Education, Inc.

Titration of a Strong Acid with a Strong Base

• To analyze titration curve, break them down into four sections:

1. Initial pH: the pH of the solution before any titrant is added. (-log [HA]).

2. B/n initial pH and equivalence point: pH increases slowly at first and then rapidly increases as the equivalence point is approached. The pH before the equivalence point is determined by unreacted acid.

3. Equivalence point: Equal moles of strong acid and strong

base have been reacted. Only salt water remains (pH 7)

4. After the equivalence point: pH is determined by the

excess titrant (base) being added.

Titration of a Strong Base with a Strong Acid

• It looks like you “flipped over” the strong acid being titrated by a strong base.

• Start with a high pH (basic solution); the pH = 7 at the equivalence point; low pH to end.

A student measured exactly 15.0 mL of an unknown

monoprotic acidic solution and placed in an Erlenmeyer

flask. An indicator was added to the flask. At the end of

the titration the student had used 35.0 mL of 0.12 M

NaOH to neutralize the acid. Calculate the molarity of the

acid.

0.035LNaOH0.12molNaOH

1L1molacid

1molbase=0.0042molacid

M=0.0042mol

0.015L=0.28Macid

Titration of a Strong Base with a Strong Acid

Vb • Cb = Va • Ca

Vb – capacity of used base

Va – capacity of used acid

Cb – concentration of used base

Ca – concentration of used acid

Titration of a Weak Acid with a Strong Base

Must assume that

the neutralization

reaction goes to

completion!

Indicators Indicators are weak organic (carbon containing) acids of

various colors depending on the formula of the acid.

Below is a generic acid.

HA H+ + A- colorless pink

173

Indicators: Phenolphthalein

The pH range of indictors

indctors pKin

d

pH

litmus 6.5 5-8

methylorange 3.7 3.1-4.4

phenophthaline 9.3 8.3-10.0

Indictors dose not change colour

sharply at one particular pH, they

change over a narrow range of pH

Titration curves for strong base with strong acid

indctors pKind pH



Both of ph.ph and M.O. are

useful

Titration curves for weak base with strong

acid

• This time we are going to use hydrochloric acid

as the strong acid and ammonia solution as the

weak base.

indctors pKind pH



MO is useful and ph.ph

is useless

Titration curves for strong base with week acid

Ph.ph is useful M.O is

useless

indctors pKind pH



Titration curves for weak base with week acid

Both ph.ph and MO are

useless

indctors pKind pH



Color versus pH of Many Different indicators

How can we make an indicator?

Step One

Red Cabbage

Step Two

Cook the Cabbage

Step Three

Filter the Juice

http://images.google.com/imgres?imgurl=http://www.coffeemakersetc.com/images/Paper_Filters.jpg&imgrefurl=http://www.coffeemakersetc.com/bunn-coffee-filters-20111-series-gallon-urns-c-109_29_35.html&usg=__XztN17uTxz0ZX46CLKI9DDpKRBQ=&h=500&w=500&sz=43&hl=en&start=2&um=1&tbnid=J4lede1Vryg6gM:&tbnh=130&tbnw=130&prev=/images?q=Filter&um=1&hl=en&sa=G

What color is the juice after filtering?

What color is the juice after filtering? The color of pH 6, 7, or

8

Colors of cabbage juice at various pH values

Diffusion, Osmosis and Osmotic

Pressure

Diffusion

• Molecules are in continuous random motion

(Brownian motion)

• Evident mostly in liquids and gases whose

molecules are free to move

• Greater the concentration of molecules greater

the likelihood of collision and movement to

chamber with low concentration

Diffusion

Osmosis

• Diffusion of water through the semi permeable

membrane from a solution of lower concentration

towards a solution of higher concentration

Osmosis and Osmotic Pressure

•Osmosis: The selective passage of solvent

molecules through a porous membrane from a

dilute solution to a more concentrated one.

•Osmotic pressure (π or ∏): The pressure

required to stop osmosis.

π = iMRT (Approximate Form)

R = 0.08206 (Latm)/(molK)



• Isotonic: Solutions have equal concentration

of solute, and so equal osmotic pressure.

•Hypertonic: Solution with higher

concentration of solute.

•Hypotonic: Solution with lower

concentration of solute.


OSMOREGULATION

Osmoregulation is the means by which cells keep the concentration of cell

cytoplasm or blood at a suitable concentration.

What is the osmotic pressure of a 1.00 M solution of sucrose at 25°C?

Problem 1.

π = iMRT

R = 0.08206 (Latm)/(molK) i = 1 T = 25°C = 298 °K M = 1 mol/l Π = 1 x 1 mol/l x 0.08206 (latm)/(molK) x 298 °K Π = 24.45 atm

Lysozyme is an enzyme that breaks bacterial cell walls. A solution containig

0.150 g of this enzyme in 210 ml of solution has an osmotic pressure of

0.00125 atm at 25 degrees Celsius. What is molar mass of lysozyme.

π = iMRT

M = π/iRT

M = n/V

n/V = π/iRT

n/ 0.21 l = 0.00125 atm/ 0.08206 (latm)/(molK) x 298 °K

n =0.21l x 0.00125 atm x 0.08206 (latm)/(molK) x 298 °K

n = 0.0000107mol

n = m/M

M = m/n

M = 0.15g/0.0000107mol

M = 14018,6 g/mol

Problem 1.

Buffer capacity

• A buffer solution resists change in pH upon addition of

any compound that tend to alter hydrogen ion

concentration. Buffer capacity is defined as the

magnitude of the resistance of a buffer to pH changes.

Other names for it include buffer index, buffer value,

and buffer efficiency.

• To have a better understanding, lets assume that we have a

1 liter acetate buffer solution containing 0.2 moles acetic

acid and 0.2 moles sodium acetate. According to the buffer

equation , the pH of this solution is equal to pKa value of

acetic acid which is 4.76.

Problem 1.

http://1.bp.blogspot.com/-cyvGYshSJ_U/UeHXNcX3rTI/AAAAAAAABMs/PiCTMMcO-i4/s1600/buffer.jpg

Concentration

Mol/l

CH3COOH CH3COO-

Initial 0.2 0.2

Final 0.2 – 0.02 = 0.18 0.2 + 0.02 =0.22

Furthermore, lets assume that 0.02 moles of KOH are introduced into this

solution without significantly changing the volume.

CH3COOH + OH- ↔ CH3COO- + H2O

Accordingly, 0.02 moles of acetic acid will change into acetate. Thus, acetic acid

concentration becomes equal to 0.2-0.02 = 0.18 M and acetate concentration

becomes equal to 0.02+0.02 = 0.22.

http://1.bp.blogspot.com/-S2JLWlaXrRs/UeHYwi0De0I/AAAAAAAABM8/IlPr_18kh_E/s1600/buffer2.jpg

Thus, for an addition of 0.02 moles of a strong base the pH changed by 4.85-

4.76 = 0.09 units. If another portion (0.02 moles) of the base is added this will

lead to a solution pH = 4.94 and a pH change = 0.09 units (try the calculation).

A third portion (0.02 moles) of the base will lead to a solution pH = 5.03 and a

pH change = 0.09 units. The buffer capacity, A, can now be approximately

calculated using the following equation:

Buffer capacity (A) is a ratio of acid or base added (to 1 litre of a buffer) to change

its pH:

Where, Δn is the molar amount of the base added to the buffer to introduce a

pH change of ΔpH. Accordingly, the buffer capacity of the above mentioned

acetate buffer is 0.22.

• To have a better understanding, lets assume that we have a

1 liter acetate buffer solution containing 0.2 moles acetic

acid and 0.2 moles sodium acetate. According to the buffer

equation , the pH of this solution is equal to pKa value of

acetic acid which is 4.76.

Problem 2.

http://1.bp.blogspot.com/-cyvGYshSJ_U/UeHXNcX3rTI/AAAAAAAABMs/PiCTMMcO-i4/s1600/buffer.jpg

Concentration

Mol/l

CH3COOH CH3COO-

Initial 0.2 0.2

Final 0.2 + 0.02 =0.22 0.2 – 0.02 = 0.18

Furthermore, lets assume that 0.02 moles of HCl are introduced into this

solution without significantly changing the volume.

CH3COO- + H+ ↔ CH3COOH

Accordingly, 0.02 moles of acetate will change into acetic acid.

pH = 4.76 + log 0.18/0.22

pH = 4.76 + (-0.087)

pH = 4.673

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