Chapter 7 acids and bases

Chapter 7
Acid-Base Equilibria

Acid-Base Theories
• Arrhenius theory (Nobel Prize)
An acid is any substance that ionizes (partially or
completely) in water to give hydrogen ions, H+ (which
associate with the solvent to give hydronium ions, H30+)
HA + H2O H3O+ + A-
• A base: any substance that ionizes in water to
give hydroxyl ions OH-. Weak (partially ionized)
bases generally ionize as follows:
B + H2O BH+ + OH-
• strong bases such as metal hydroxides (e.g.,
NaOH) dissociate as
M(OH)n Mn+ + nOH-

THEORY OF SOLVENT SYSTEMS
• solvent ionizes to give a cation and an anion; for example,
2H20 H30+ + OH-
2NH3 NH4
+ + NH2
-
• According to this theory:
Acid is defined as a solute that yields the cation of the solvent
Base is a solute that yields the anion of the solvent.
• Thus, NH4C1 is a strong acid in liquid ammonia similar to HCl in
water:
HCl + H2O → H3O+ + CI-

• NaNH2 is a strong base in ammonia (similar to
NaOH in water);
• both of these compounds ionize to give the
solvent cation and anion, respectively.
• Ethanol ionizes as follows:
• 2C2H5OH C2H5OH2
+ + C2H5O-
• Hence, sodium ethoxide, NaOC2H5, is a strong
base in this solvent.
THEORY OF SOLVENT SYSTEMS

BRONSTED-LOWRY THEORY
• The theory of solvent systems is suitable for
ionizable solvents, but it is not applicable to
acid-base reactions in nonionizable solvents
such as benzene or dioXane.
• In 1923, Bronsted and Lowry separately
described what is now known as the
Bronsted-Lowry theory.
• This theory states that an acid is any substance
that can donate a proton, and a base is any
substance that can accept a proton.
• Thus, we can write a "half-reaction”
Acid = H+ + Base
The acid and the base are called “Conjugate pairs”

LEWIS THEORY
• Acid is a substance that can accept an electron
pair
• Base is a substance that can donate an electron
pair. The latter frequently contains an oxygen or
a nitrogen as the electron donor.
• Thus, non-hydrogen-containing substances are
included as acids
• Examples of acid-base reactions in the Lewis
theory are as follows:
• H+ (solvated) + :NH3 → H:NH3
+

Acid-Base Equilibria in Water
• When an acid or base is dissolved in water, it
will dissociate, or ionize.
• The amount (degree) of ionization is
dependent on the strength of the acid.
• A "strong" electrolyte is completely
dissociated, while a "weak" electrolyte is
partially dissociated.

Dissociation of weak acids
Thermodynamic acidity constant, o
aK
In dilute solutions, the activity of water remains
essentially constant, and is taken as unity at standard
state

Thermodynamic autoprotolysis/Self-ionization
constan

Molar equilibrium constant/Acid dissociation
constant, Ka
The product of the
hydrogen ion
concentration and the
hydroxyl ion
concentration in aqueous
solution is always equal
to 1.0x10-14
[H+] [OH-] = 1.0 X 10-14
Except when the hydrogen ion
concentration from the acid is very small,
10-6 M or less, any contribution to [H+]
from the ionization of water can be
neglected.

Example
A 1.0 X 10-3 M solution of hydrochloric acid is
prepared. What is the hydroxyl ion concentration?

The pH scale
• It is more convenient to compress the acidity
scale by placing it on a logarithm basis.
• The pH of a solution was defined by Sorenson
as
pH = -log [H+]
pOH = -log [OH-]

Example
1. Calculate the pOH and the pH of a 5.0 X 10-2
M solution of NaOH
2. Calculate the pH of a solution prepared by
mixing 2.0 mL of a strong acid solution of pH
3.00 and 3.0 mL of a strong base of pH
10.00.

Example
• The pH of a solution is 9.67. Calculate the
hydrogen ion concentration in the solution.
[H+] = 10-pH

pH and acidity and alkalinity
• When [H+] = [OH-], the solution neutral.
• If [H+] > [OH-], the solution acidic.
• If [H+] < [OH-], the solution is alkaline.
• in pure water at 25°C [H+] = [OH-],
pH= pOH = 7
• A solution of pH < 7 is acidic
• A solution of pH > 7 is alkaline (or basic)

Comments on basicity and acidity
• Is there a pH of negative value?
• How about the pH value of a solution that
contains 1x10-8 M HCl?
• If the concentration of an acid or base is
much less than 10-7M, then its contribution
to the acidity or basicity will be negligible
compared to the contribution from water.

Example
• Calculate the pH and pOH of a 1.0 X 10-7 M
solution of HCl.
Since the hydrogen ions contributed from the ionization of
water are not negligible compared to the HCl added,

Comments
• The calculation in this example is more academic
than practical because carbon dioxide from the air
dissolved in water exceeds these concentrations.
• Since carbon dioxide in water forms an acid,
extreme care would have to be taken to remove
and keep this from the water, to have a solution of
10-7 M acid .
• We usually neglect the contribution of water to
the acidity in the presence of an acid since its
ionization is suppressed in the presence of the
acid.

pH and temperature
• Does pH change with temperature? If yes, why?
• K w = 5.5x10-13 at 100oC
•The pH of blood at body temperature (37°C) is 7.35 -7.45.
•This value represents a slightly more alkaline solution relative to
neutral water than the same value would be at room temperature
• At 37 oC, Kw = 2.5 X 1014 and pK w, = 13.60.
•The pH (and pOH) of a neutral solution is 13.60/2 = 6.80.
•The [H+] = [OH-] 1.6 X 10-7 M
• Since a neutral blood solution at 37°C would have pH 6.8

pH and temperature
• Since a neutral blood solution at 37°C would have pH 6.8.
• A blood pH of 7.4 is more alkaline at 37°C by 0.2 pH units than
it would be at 25°C.
• This is important when one considers that a change of 0.3 pH
units in the body is
extreme.
• The HCl concentration in the stomach is about 0.1 to 0.02 M.
• Since, the pH at 0.02 M would be 1.7.
• It will be the same regardless of the temperature since the
hydrogen ion concentration is the same and the same pH
would be measured at either temperature.
• But, while the pOH would be 14.0 - 1.7 = 12.3 at 25°C,
it is 13.6 - 1.7 = 11.9 at 37°C.

pH of the weak acids and bases
• pH and pOH are determined readily from the
concentration of the acid or base.
• Weak acids (or bases) are only partially
ionized.
• Most organic acids and bases, as found in
clinical applications, are weak.
• The ionization constant can be used to
calculate the amount ionized and, from
this, the pH.

• If x less than about 10 or 15% of the acid is ionized,
the expression may be simplified by neglecting x
compared with C (10-3 M in this case).
• The simplification applies if Ka is smaller than about
0.01 C, that is, smaller than 10-4 at C = 0.01 M, 10-3
at C = 0.1 M, and so forth.
• Under these conditions, the error in calculation is
5% or less (results come out too high
• If CHA > l00Ka , x can be neglected compared to CHA.
AaCkHx
In
 
][
General

pH of the weak bases
The basicity constant K, for ammonia is 1.75 X 10-5 at 25°C. (It is
only coincidental that this is equal to Ka for acetic acid.) Calculate
the pH and pOH for a 1.00 X 10-3 M solution of ammonia

pH of salts of weak acids and bases
• Salts of weak acids or weak bases are completely
ionized
• NaOAc → Na+ + OAc-
• NH4Cl → NH4
+ + Cl-
• BHCl → BH+ + Cl-
• Anions of salts of weak acids are Bronsted bases
• Oac- + H2O HOAc - + OH-
• Cations of salts of weak bases are Bronsted acids
• BH+ + H2O B + H3O+
• Anions and cations of strong acids or strong bases are
neutral

pH of salts of weak acids
This ionization is known as hydrolysis and its equilibrium constant
may be called Hydrolysis constant or the basicity constant
of the salt
Thus for a conjugate acid-base, always:
Kw = Ka x Kb

pH of salts of weak acids, HA
--
A
2
A
2
a
w
C
x
x-C
x
k
k
bK CA- = Csalt
This equation holds
only if:
-
A
bA
Ctocomparedneglected
becanxandk100C 

pH of salts of weak bases, BHA
BHA → BH+ + A-
B + H2O BH+ + OH-
][
]][[BH
B
OH
Kb



x-CC-CC saltBABHBH

x-Csalt x x
saltsalt
33
salt
2
salt
2
C
]][H[H
C
]][[
C
x
x-C
x 

OHOH
Ka
b
w
a
K
KH
K 

saltsalt C
][
C
]][H[H 2
salt.C][H
b
w
K
K

 BHsalt CC

Buffers (Buffer solutions)
• A buffer is a solution that resists change in pH when a
small amount of an acid or base is added or when the
solution is diluted.
• A buffer solution consists of a mixture of a weak acid and
its conjugate base or a weak base and its conjugate acid at
predetermined concentrations or ratios.
• That is, we have a mixture of a weak acid and its salt or a
weak base and its salt.

pH of a buffer of weak acid solution containing its salt
Henderson-Hasselbalch equation
HA H+ + A-
[Acid]
[Salt]
logpKa pH

Example
Calculate the pH of a buffer prepared by adding
10 mL of 0.10 M acetic acid to 20 mL of 0.10M
sodium acetate.
[Acid]
[Salt]
logpKa pH

Example
Calculate the pH of a solution prepared by adding
25 mL of 0.10 M sodium hydroxide to 30 mL of
0.20 M acetic acid.
left
formed
a
[Acid]
[Salt]
logpK pH
left
formed
a
[Acid]
[Salt]
logpK pH

Buffering mechanism
• If the solution is diluted, the ratio remains constant, and so
the pH of the solution does not change
• If a small amount of a strong acid is added, it will combine
with an equal amount of the A- to convert it to HA.
• The change in the ratio [A -]/[HA] is small and hence the
change in pH is small.
• If a small amount of a strong base is added, it will combine
with part of the HA to form an equivalent amount of A -.
• Again, the change in the ratio is small.
HA H+ + A-

Buffering capacity
• The amount of acid or base that can be added without
causing a large change in pH is governed by the Buffering
capacity .
• Buffering capacity is determined by the concentrations of
[HA] and [A -].
– The higher their concentrations, the more
acid or base the solution can tolerate.
• The buffer capacity (buffer intensity, buffer index) of a
solution is defined as

Buffering capacity
• The buffer capacity is a positive number.
• The larger it is, the more resistant the solution is to
pH change.
• the buffering capacity is governed by the ratio of
HA to A -.
• The buffering capacity is maximum when the ratio
is unity.
• That is, when the pH = pKa

Example
• A buffer solution is 0.20 M in acetic acid and in sodium acetate.
Calculate the change in pH upon adding 1.0 mL of 0.10 M
hydrochloric acid to 10 mL of this solution.
pK a = 4.76

pH of a buffer of a weak base solution containing its salt
• A mixture of a weak base and its salt acts as a buffer in
the same manner as a weak acid and its salt.
Consider an equilibrium between a base and its conjugate
acid BH+ (same as before). Then write a Ka expression for
the conjugate (Bronsted) acid.

pH of a weak base solution containing its salt
Since pH= pK w -pOH

Buffering mechanism for a buffer of weak base solution and its salt
• When a strong acid is added, it combines with some of the
base B to form the salt BH+.
• Conversely, when a strong base is added, it combines with
BH+ to form B.
• Since the change in the ratio will be small, the change in pH
will be small.
• Again, the buffering capacity is maximum when
pH = pKa ; that is pH = 14 – pKb or pOH = pKb

Example
Calculate the volume of concentrated ammonia and the weight of
ammonium chloride you would have to take to prepare 100 mL of
a buffer at pH 10.00 if the final concentration of salt is to be 0.200 M.
Given that the molarity of concentrated ammonia is 14.8 M
[NH3] = 1.16 M

Example
How many grams ammonium chloride and how many milliliters 3.0M
sodium hydroxide should be added to 200 mL water and diluted to 500
mL to prepare a buffer of pH 9.50 with a salt concentration of 0.10 M?
NH4Cl + NaOH → NH3 + NaCl + H2O
Find the concentration of NH3 formed from the equation, [NH3] = 0.18 M
Total # mmoles of NH4Cl added =
#mmoles of NH4Cl left (final) +# mmoles of NH3formed(final)
Total # mg NH4Cl added = Total # mmoles NH4Cl x MM (NH4Cl )
Volume of NaOH added = # mmoles of NaOH added /M (NaOH)
= # mmoles of NH3 formed/M (NaOH)

Polyprotic acids and their salts
• Polyfunctionaly acids or bases are substances that have
more than one ionizable proton or hydroxide ion.
• These substances ionize stepwise, and an equilibrium
constant can be written for each step.
The stepwise K, values of polprotic acids get progressively smaller
as the increased negative charge makes dissociation of the next
proto difficult.

Polyprotic acids and their salts
• The overall ionization is the sum of these individual steps
and the overall ionization constant is the product of the
individual ionization constants:
• In order to make precise pH calculations, the contributions of
protons from each ionization step must be taken into account.
• Exact calculation is difficult and requires a tedious iterative
procedure

Buffer calculations for polyprotic acids
• The anion on the right side in each ionization step can be
considered the salt (conjugate base) of the acid from which
it is derived.
• That is, in, H2P04
- is the salt of the acid H3P04
• HPO4
2- is the salt of the acid H2P04
–
• PO4
3- is the salt of the acid HPO4
2-
So each of these pairs constitutes a buffer system.

Buffers of orthophosphoric acid
• Orthophosphate buffers can be prepared over a wide pH
range.
• The optimum buffering capacity of each pair occurs at a pH
corresponding to its pKa.
• The HPO4
2- /H2P04
- couple is an effective buffer system
in the blood

Example
• The pH of blood is 7.40. What is the ratio of
[HPO4
-]/[H2P04
-] in the blood (assume/ 25°C)?

Example
Calculate the pH of a 0.100 M H3P04 solution.
• The pH of a solution of H3P04 can be calculated same as
any weak monoprotic acid.
• The H+ from the first ionization step effectively suppresses
• the other two ionization steps, so that the H+ contribution
from them is negligible compared to the first ionization.
• The quadratic equation must be solved because Kal is
relatively large.
Approximation is not acceptable

Fractions of Dissociation at a given pH:  Values
• Consider the following equilibria for H3PO4
• At any given pH, all the four phosphoric acid species will coexist
in equilibrium with one anotother.
• By changing the pH, the equilibria shift, and the relative
concentrations change.
• It is possible to derive general equations for calculating the
fraction of the acid that exists in a given form, from the given
hydrogen ion concentration.

• For a given total analytical concentration of
phosphoric acid, C we can write
43POH
• The a's are the fractions of each species present at equilibrium.
• Note that the subscripts denote the number of dissociated protons
or the charge on the species

Example
Calculate the equilibrium concentration of the different species
in a 0.10 M phosphoric acid solution at pH 3.00 ([H+] = 1.0 X 10-3 M).

Salts of polyprotic acids
• Salts of acids such as H3P04 may be acidic or ba ic.
• The protonated salts pose both acidic and basic properties
(H2P04
-, HPO4
-)
• The unprotonated salt is simply a Brensted base that
hydrolyzes (P04
3-).
Amphoteric Salts.
• H2P04
- possesses both acidic and basic properties.
• That is, it is amphoteric. It ionizes as a weak acid and it
also is a Brensted base that hydrolyzes:
43
-
42 POHH  
POH
43
-
42 POHH  
POH

Salts of polyprotic acids
• The solution of H2P04
- could be either
alkaline or acidic, depending on which
ionization is more extensive.
• Since Ka2 for the first ionization is nearly 105
greater than Kb for the second ionization, the
solution in this case will obviously be acidic.

[H+] in a solution of amphoteric salts, H2P04
-
43
-
42 POHH  
POH
OR
• [OH-] will be counted only if pH is around 7.
• In the case of acidic solution [H+] is negligible therefore no need to include [OH-}

Multiplying each side of the equation by [H+], collecting the
terms containing [H+]on the left side, and solving for [H+]2
In the above equation substitute values from equilibrium
constants in the right hand side of the equation

If the dissociation of H2O is neglected, the Ka1Kw will be very small and can
be removed from the numerator of the equation above
In most cases [HA-] or [H2PO4
-]>> Ka1; thus Ka1 can be
neglected from the denominator

Consequently, [H+] for a solution of NaH2PO4 (H2PO4
- )
This equation is valid on the basis that the stepwise dissociation
constants differ by factors of 100 or more
• (pH = 4.54) thus, pH is approximately independent
of the salt concentration
• This would be the approximate pH of a NaH2P04 solution.
• Similarly, HPO4
2- (Na2HPO4) is both an acid and a base.
• The K values involved here are Ka2 and Ka3 of H3P04
• Again, stepwise dissociation constants differ by factors of 100 or more
• pH of a Na2HP04 solution is 9.72

Unprotonated Salt
• Unprotonated phosphate Na3PO4, (PO4
3-) is a fairly strong
Brensted base in solution and ionizes as follows:
• The constant Ka3 is very small, and so the equilibrium lies
significantly to the right.
• Because Ka3 << Ka2, hydrolysis of HPO4
- is suppressed by
the OH- from the first step.
• Thus, pH of P04
3- can be calculated just as for a salt of a
monoprotic weak acid
• Because Kb is relatively large, the quadratic equation must s
be solved, that is, P04
3- is quite a strong base.

Example
• Calculate the pH of 0.100 M Na3P04.

Physiological buffers
• The pH of the blood in a healthy individual remains
remarkably constant at 7.35 to 7.4
• This is because the blood contains a number of buffers that
protect against pH change due to the presence of acidic or
basic metabolites.
• From a physiological viewpoint, a change of ±0.3 pH unit is
extreme.
• Acid metabolites are ordinarily produced in greater
quantities than basic metabolites, and carbon dioxide is the
principal one.
• The buffering capacity of blood for handling CO2 is estimated
to be distributed among various buffer systems as follows:
– Hhemoglobin and oxyhemoglobin, 62%
– H2P04
-/HP04
2- 22%
– plasma protein, 11%
– bicarbonate, 5%.

Buffers for Biological and Clinical Measurements
• Many biological reactions of interest occur in the pH range of
6 to 8.
• A number, particularly specific enzyme reactions that might
be used for analyses may occur in the pH range of 4 to 10 or
even greater.
• The proper selection of buffers for the study of biological
reactions or for use in clinical analyses can be critical in
determining whether or not they influence the reaction.
• A buffer must have the correct pKa, near physiological pH so
the ratio of [A -]/[HA] in the Henderson-Hasselbalch
equation is not too far from unity, and it must be
physiologically compatible.

Phosphate Buffers
• Biological systems usually contain some phosphate already,
and phosphate buffers will not interfere in many cases.
• By choosing appropriate mixtures of H3P04/H2P04
-, H2P04
-
/HPO4
2-, or HPO4
2- I P04
3-
• phosphate will precipitate or complex many polyvalent
cations, and it frequently will participate in or inhibit a
reaction.
• It should not be used, for example, when calcium is present
if its precipitation would affect the reaction of interest.

Tris Buffers
• It is that prepared
fromtris(hydroxymethyl)aminomethane and its
conjugate acid (the amino group is protonated).
• It is a primary standard and has good stability, has
a high solubility in
physiological fluids, is nonhygroscopic, does not
absorb CO2 appreciably, does not precipitate
calcium salts, does not appear to inhibit many
enzyme systems, and is compatible with
biological fluids.

Chapter 7 acids and bases

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