ANALYTICAL CHEMISTRY

Lecture Note

Chem. 243

Chapter 5

Acids, Bases and Buffers

Chapter 5 —ACIDS DONATE HYDROGEN IONS TO BASES

•By controlling the acid-base conditions, the clinician can ensure that medications stay in solution.•If acid content of blood changes slightly, the patient dies

Dynamic Equilibrium (Exists in a system made up of at least 2 states of the matter when the populations of the 2 states are constant, even though the members of the system are constantly changing from one state to another; e.g., vapor pressure: From Chapter 7)

❍ Most chemical reactions are reversible

– Reactants combine to give products

– Products can fall apart to give reactants (products can break back down into starting materials)

reactants products

These two competing processes occur simultaneously

❍ This process is called a dynamic equilibrium: forward and reverse reactions continue to take place but the [reactants] and [products] remain constant

Example: Hemoglobin (Hb loads up on O2 in the lungs and dumps

O2 into the cells) Le Chatelier’s principle (Nature strives to attain/maintain equilibrium)

❍ Hemoglobin (Hb) is the protein in blood that carries oxygen from the lungs to the rest of the body. Hb is a quaternary aggregate of four polypeptide (i.e., protein) chains. Each polypeptide carries an organic heme group (the cyclic structure that carries a Fe2+ ion). Each iron (II) ion can carry one O2 molecule, so each Hb molecule can carry 4 O2 molecules.

❍ Hb combines reversibly with oxygen in the lungs

❍ In the lungs the concentration of O2 is high. The system (i.e., your body) “perceives” the increased O2 concentration as adding reactants; therefore, the equilibrium shifts towards the product: oxyhemoglobin, Hb(O2)4

❍ The opposite happens in cellular tissue (low [O2] is “perceived as removing a reactant: equilibrium shifts towards reactants)

422 OHb 4O Hb

The Equilibrium Constant

❍ A system is in a state of equilibrium when there is a balance between reactants and products– This balance is defined by thermodynamic

parameters (e.g., bond strengths and intermolecular forces between all the molecules in the system)

– The equilibrium constant (K) is the numerical description of that balance

Equilibrium Constant Expression

❍ For a model reaction:a A + b B ⇌ c C + d D

The equilibrium constant is given by:

❍ The molar concentrations of the products, raisedto their stoichiometric coefficients, are in the numerator

❍ The molar concentrations of the reactants, raisedto their stoichiometric coefficients, are in the denominatorStoichiometry: calculation of measurable (quantitative) relationships of the reactants and products in a balanced chemical reaction

ba

dc

BA

DCK

A Word About Subscripts

❍ The equilibrium constant (K) is often appended with a subscript❍ The subscript denotes the type of equilibrium reaction

– Keq is a generic equilibrium constant

– Ka is the equilibrium constant for the inonization of weak acids

– Kb is the equilibrium constant for the inonization of weak bases

– Ksp is the equilibrium constant for the solubility of sparingly soluble compounds

Meaning of K

❍ As K increases (K > 1), the reaction tends to increasingly favor products (the forward reaction becomes more favorable)

a A + b B c C + d D ❍ As K decreases (K < 1), the reaction tends to increasingly favor

starting materials (the reverse reaction becomes more favorable)

a A + b B c C + d D

ba

dc

BA

DCK

Example

❍ K for the reaction below is 56:

H2 + I2 2 HI⇌

❍ If equal amounts of hydrogen and iodine are placed into a reaction vessel and allowed to come to equilibrium, will there be more HI (hydrogen iodide) or H2?

Since K >1, the reaction is “product-favored”, so at equilibrium, there will be more HI

Units of K

❍ The units on K simply cancel out the units on the other side of the equation

❍ You have been reminded for the last few weeks about always including units

❍ However, the units on K are meaningless❍ You do not need to include units with K

Carbon Monoxide

Carbon monoxide (CO) kills by “tying up” the Hb molecule so that not enough O2 can be delivered to the tissues in the body

The K for Hb and CO is about 200 times larger that the K for Hg and O2 (the system favors the forward reaction towards the formation of carboxyhemoglobin: Hb(CO)4

Hb + 4 CO Hb(CO)4

There is less free Hb available to bind O2

Solids and Liquids

❍ The concentration of a pure solid or liquid is a constant value (or very nearly so)– Pure solids or liquids comprise a different phase from where the reaction

occurs. Physiologic reactions usually occur in aqueous media.❍ The concentration of a liquid solvent (e.g., water) is also nearly constant

(because the concentration of water is so much greater than the concentrations of any of the reacting species in aqueous solutions) so these constant values for concentration are included with the equilibrium constant value.

❍ Therefore, the concentrations of solids, liquids, and water (as a solvent) do not appear explicitly in the equilibrium constant expression

Example❍ Write the equilibrium constant expression for the partial ionization of a weak

base:

K = [NH4+] [OH-]

[NH3] [H2O]this can be rearranged to:

K [H2O] = [NH4+] [OH-]

[NH3]this can be rearranged to:

Kb = [NH4+] [OH-]

[NH3]*Water is the solvent and does not appear in the equilibrium constant expression

(aq) OH (aq) NH OH NH -423

Reversing a Reaction

❍ Since equilibrating reactions are, by definition, reversible, what happens to K when the chemical equation is reversed?

Kforward is the reciprocal of Kreverse

Kforward = 1

Kreverse

Le Chatelier’s Principle

❍ When a system in a state of dynamic equilibrium is disturbed, it will react to re-establish the equilibrium condition

❍ Nature likes being in an equilibrium state❍ Changing reaction conditions (concentrations, temperature,

etc.) moves the system away from being at equilibrium❍ The system will adjust (go forwards or backwards) until it

again is in an equilibrium state

Changing Concentration

❍ If you add products, the equilibrium will shift towards reactants

❍ If you remove products, the equilibrium will shift towards products

❍ The system will adjust to counteract whatever change you make on it

22 I H 2HI

22 I H 2HI

Example

❍ In which direction will the equilibrium shift if more calcium hydroxide is added?

Increasing reactants will shift the equilibrium to the products (to the right)

OH (s) CaCO (g) CO (aq) OHCa 2322

Example

❍ In which direction will the equilibrium shift if calcium hydroxide is removed?

Decreasing reactants will shift the equilibrium to the reactants (to

the left)

OH (s) CaCO (g) CO (aq) OHCa 2322

Example

❍ In which direction will the equilibrium shift if more calcium carbonate is added?

❍ Increasing volume will decrease pressure; therefore, not favoring the smaller number of gas particles. Since the gas (CO2) is in the reactants, an increase in volume will shift the equilibrium to the products (shift to the right)

OH (s) CaCO (g) CO (aq) OHCa 2322

Self-Ionization of Water

❍ The acid-base chemistry we will encounter occurs in aqueous media

❍ A tiny fraction of water molecules fragment or ionize into a hydrogen ion and a hydroxide ion

H2O H➔ + + OH-

❍ Thus, the is H+ represents the acid and OH- represents the base

Conjugate Acid-Base Pairs

❍ When an acid donates a proton, it is converted into its conjugate base

HA H➔ + + A—

acid conjugate base� يتحد مؤقتًا When a base accepts a proton, it is converted into

its conjugate acid

B + H+ ➔ BH+

base conjugate acid

21Lecture 3

HCl(g) → Cl-(aq)Conjugate

baseacid

Conjugate acid-base pairs differ by a proton.

When an acid donates a proton it becomes the conjugate base.

22Lecture 3

Conjugate acid-base pairs differ by a proton.

When a base accepts a proton it becomes the conjugate acid.

H3O+(aq)H2O (l) →Conjugate acidbase

23Lecture 3

Conjugate acid-base pairs differ by a proton.

baseacid

HCl(g) + → Cl-(aq) + H3O+(aq)H2O (l)

acidbase

24Lecture 3

Examples

❍ Give the conjugate bases for each of these acids (the charge of the conjugate base is always one lower than the charge of its conjugate acid)

HCl

H2SO4

HSO4-

H2O

Cl-

HSO4-

SO42-

OH-

Examples

❍ Give the conjugate acids for each of these bases (the charge of the conjugate acid is always one greater than the charge of its conjugate base)

OH-

NH3

HPO42-

H2O

H2O

NH4+

H2PO4−

H3O+

Example❍ Identify the conjugate acid-base pairs in this reaction:

NH3 + HC2H3O2 NH➔ 4+ + C2H3O2

2-

base conjugateacid

acidconjugate

base

acetic acid (gives vinegar its sour taste and pungent smell)

acetate ion

Amphoteric (Amphoprotic) Species

❍ An amphoteric species can behave as either an acid or a base

❍ H2O is an amphoteric species:

– Here water behaves as a(n)

base (water accepts a proton)

H+ + H2O H3O+

– Here, water behaves as a(n)

acid (water donates a proton)

H2O H+ + OH-

Diprotic Acids

❍ A diprotic acid has 2 hydrogen ions to donate. A diprotic acid can behave as an acid twice.

H2SO3 H➔ + + HSO3—

HSO3- ➔ H+ + SO3

2—

❍ The terms triprotic acid and polyprotic acid are self-explanatory

Acid and Base Strength

❍ A stronger acid is more determined to give its proton to some base

❍ A stronger base is more determined to take a proton from some acid

Acid Strengths

Ionization Equilibria of Weak Acids

❍ Weak acids do not ionize 100% in water❍ Weak acids establish an equilibrium:

HA ➔ H+ + A-

❍ And have a corresponding equilibrium constant expression:

HA

AHKa

Ionization Equilibria of Weak Bases

❍ Weak bases establish an equilibrium by accepting a proton from water:

B + H2O = BH+ + OH-

❍ And have a corresponding equilibrium constant expression:

BOHBH

Kb

Acid/Base Strength and Ka/Kb

❍ Stronger acids have larger Ka values

❍ Stronger bases have larger Kb values

Self Ionization of Water : Kw

❍ The ionization of water is an equilibrium:

H2O ↔ H+ + OH-

❍ The equilibrium constant is called Kw:

Kw = [H+][OH-] = 1.0x10-14 at 25°C

pKa and pKb

pKa = -log Ka

pKb = -log Kb

The Key Relationships

pH = -log [H+]

pOH = -log[OH-][H+][OH-] = 1.00 x10-14

pH + pOH = pKw = 14.00

Calculting pH: Strong Acids

❍ Strong acids are 100% ionized in water

– Well, at least as long as the concentration is less than 0.1M, this is mostly true

❍ Therefore, [H+] = cacid (formal concentration or molarity of the acid)

Example: [H+] pH➔

❍ What is the pH of a 0.025 M solution of HCl?

HCl is a strong acid so it fully dissociates.

The [H+] = formal concentration (molarity) of HCl.

pH = -log [H+] = -log 0.025 = 1.6

Example: pH [H➔ +]

❍ The pH of a dilute nitric acid solution is 4.59. What isthe[H+]? What is the formal concentration of nitric acid?

Nitric acid is a strong acid.

antilog of -4.59 = 2.6 x 10-5 M

To take the “antilog” of -4.59 enter -4.59 then click “inv” and “log” keys in your calculator

Calculating pH: Strong Bases

❍ In a strong base, [OH-] = cbase

Example

❍ What is the pH of a 0.025 M solution of NaOH?

NaOH is a strong base so:

pOH = -log [OH-] = -log [0.025M] = 1.6

Since pH + pOH i= 14

so pH = 14 – 1.6 = 12.4

Example

❍ The pH of a dilute KOH solution is 9.59. What is the [H+]?

Strong base so full dissociation of KOH into K+1 and OH- .

K+1 does not matter (it is a pure “spectator”). pH = -log[H+]

so [H+] is antilog of pH

antilog of -9.59 = 2.57 x 10-10

so [H+] = 2.57 x 10-10

Buffers❍ A pH buffer is a solution that resists changes in pH❍ A pH buffer contains a weak acid (HA) and its conjugate base

(A-); or a weak base (A-) and its conjugate acid.❍ If a strong base (OH-) is added to a buffered solution, the weak

acid in the buffer (HA) will react with OH- to give H2O and the weak base (A-). This results in converting a strong base (OH-) into a weak base (A-). As a result, the pH will slightly increase.

HA + OH- H➔ 2O + A-

If a strong acid acid (H+) is added to a buffered solution, the weak base in the buffer (A-) will react with the (H+) ion to give HA. This results in converting a strong acid H+ into a weak acid HA. As a result, the pH will slightly decrease.

A- + H+ HA➔

Buffer pH: The Henderson-Hasselbach Equation (the Buffer Equation)

❍ The pH of a buffer IS NOT NECESSARILY 7!

HA

ApKpH a

log

Example❍ Calculate the pH of a buffer solution that is 0.10 M in

acetic acid and 0.15 M in sodium acetate. The ka of acetic acid is 1.8 x 10-5.

❍ pka = -log ka = -log 1.8 x 10-5 = 4.74❍ [A-] = acetate ion = 0.15 M❍ [HA] = acetic acid = 0.10 M

pH = 4.74 + log 0.15 M 0.10 M

pH = 4.74 + log 1.5 = 4.74 + 0.18 = 4.92 pH = 4.92 is slightly higher than pka (on the “basic” side of pka since the concentration of the base (0.15 M) is higher than the concentration of the acid (0.10 M)

HA

ApKpH a

log

Example❍ Calculate the pH of a solution that is 0.25 molar in ammonia and

0.75 M in ammonium chloride. The pkb for ammonia is 4.74.

The Henderson-Hasselbach equation calls for the pka of the acid so 14.00-4.74 = 9.26 is the pka.

pH = 9.26 + log 0.25M 0.75 MpH = 9.26 + log 0.33 = 9.26 + (-0.48) = 8.78pH = 8.78 is slightly lower than pka (on the “acidic” side of

pka since the concentration of the base (0.25 M) is smaller than the concentration of the acid (0.75 M)

HA

ApKpH a

log

Special Example

❍ Calculate the pH of a buffer that is 0.10 M each inمهم sodium acetate and acetic acid. The ka of acetic acid is 1.8 x 10-5.

pka of acetic acid = 4.74

pH = 4.74 + log 0.10 M

0.10 M

pH = 4.74 + log 1

pH = 4.74 + 0

pH = 4.74

pH = pKa (true of any buffer system when the concentrations of the weak acid and its conjugate base are equal)

pKa is the pH at which the concentrations of ionized and unionized species in a system are the same

HA

ApKpH a

log

Buffers

Example: What is the pH of a buffer containing 0.12 M benzoic acid and 0.20 M sodium benzoate? Ka = 6.3 x 10-5.

Buffers

Example: What base to acid ratio is needed to make a pH 4.95 buffer using benzoic acid and sodium benzoate?

Ka = 6.3 x 10-5.

Preparation of Buffers

❍ Buffers can be prepared in three ways

– Weak acid plus its conjugate base or weak base plus its conjugate acid

– Excess weak acid plus strong base

– Excess weak base plus strong acid

Buffers

Buffer capacity– the amount of acid or base a buffer can neutralize

before the pH begins to change appreciably

The buffer capacity depends on the amount of acid and base from which the buffer is prepared.

Buffer Capacity

❍ To be an effective buffer,

pH = pKa ±1

the pH must be within one pH unit of the pKa of the weak acid

Thank you!Thank you!

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