 Last
week, we discussed the Brønsted – Lowry
concept of acids and bases
 This model is not limited to aqueous solutions; it
can be extended to reactions in the gas phase!
 According


to this model:
Acids are proton (H+) donors
Bases are proton (H+) acceptors

Result of this type of acid/base reaction is a
conjugate acid-base pair


There is an important connection between the
strength of an acid and that of its conjugate base
A strong acid has an equilibrium that lies far to
the right; thus, almost all the original HA is
dissociated at equilibrium

A strong acid yields a weak conjugate base


A weak conjugate base has a low affinity for a proton
Conversely, a weak acid has an equilibrium that
lies far to the left; thus, most of the HA
originally placed in the solution is still present as
HA

A weak acid yields a strong conjugate base

A strong conjugate base has a high affinity for a proton

There is also a connection between the strength
of a base and that of its conjugate acid
A
strong base has an equilibrium that lies far
to the right; thus, almost all the original BOH
is dissociated at equilibrium

A strong base yields a weak conjugate acid
 Conversely,
a weak base has an equilibrium
that lies far to the left; thus, most of the B
originally placed in the solution is still
present as B

A weak base yields a strong conjugate acid
 During
a Brønsted – Lowry acid/base
reaction, a salt is formed
 Recall,
a salt is an ionic compound that
dissolves and completely dissociates in water
into its ions, which move about
independently
 Salts do not always produce neutral solutions
when dissolved in water

Under certain conditions, these ions behave as
acids or bases

Look at the salt and ask yourself:
Which acid and which base reacted to form it?
 Is the acid strong or weak?
 Is the base strong or weak?


Embrace the fact that STRONG WINS and is a
spectator

The remaining ion of the salt reacts with water
If you predict a basic solution, YOU MUST WRITE:
H2 O ↔ OH−
 If you predict an acidic solution, YOU MUST
WRITE:
H2 O ↔ H +

 We
just discussed how the conjugate base of
a strong acid has virtually no affinity for
protons in water

Thus, when anions such as Cl- and NO3- are
placed in water, they do not combine with H+
and have no effect on pH
 Cations
such as K+ and Na+ from strong bases
have no affinity for H+, nor can they produce
H+, so they too have no effect on the pH
 So, salts that consist of the cations of strong
bases and the anions of strong acids have no
effect on [H+] when dissolved in water
 Salts
that are formed from the cation of a
strong base and the conjugate base (anion)
of a weak acid will produce a basic aqueous
solution
 Salts
that are formed from the conjugate
acid (cation) of a weak base and the anion of
a strong acid will produce an acidic aqueous
solution
 #1(a),
#1(c)
 Can
predict whether the solution will be
basic, acidic, or neutral by comparing Ka
value for the acidic ion with the Kb value for
the basic ion

Ka = Kb


Ka > Kb


Neutral solution
Acidic solution
Ka < Kb

Basic solution
Type of Salt
Examples
Comment
pH of Solution
Cation is from
strong base; anion
is from strong acid
KCl, KNO3, NaCl,
NaNO3
Neither acts as an
acid or a base
Neutral
Cation is from
strong base; anion
is conjugate base
from weak acid
NaC2H3O2, KCN,
NaF
Anion acts as a
base; cation has
no effect on pH
Basic
NH4Cl, NH4NO3
Cation acts as an
acid; anion has no
effect on pH
Acidic
NH4C2H3O2, NH4CN
Cation acts as an
acid; anion acts as
a base
Neutral if Ka = Kb
Acidic if Ka > Kb
Basic if Ka < Kb
Cation is
conjugate acid of
weak base; anion
is from strong acid
Cation is
conjugate acid of
weak base; anion
is conjugate base
of weak acid
 The
most important application of acid-base
solutions containing a common ion is for
buffering

A buffered solution is one that resists a change
in its pH when either hydroxide ions or protons
are added

HUGE EXAMPLE OF A BUFFERED SOLUTION– Our blood!
 Blood can absorb the acids and bases produced in
biologic reactions without changing its pH


Vital because cells can survive only in a very narrow pH
range!
Buffering system in blood involves HCO3- and H2CO3
 Defined
as the shift in equilibrium caused by
the addition of a compound having an ion in
common with the dissolved substance
 The presence of a common ion suppresses
the ionization of a weak acid or a weak base

In other words, a mixture of NaC2H3O2 and
HC2H3O2 is less acidic than a solution of HC2H3O2
alone
CH3COONa (s)
Na+ (aq) + CH3COO- (aq)
CH3COOH (aq)
H+ (aq) + CH3COO- (aq)
Common
Ion
A

buffer solution can consist of:
A weak acid and its salt


HF and NaF
OR a weak base and its salt

NH3 and NH4Cl
 How
do we calculate the pH of a buffer
solution?

Solving these types of problems is very similar to
the common ion problems from last chapter!
 All

groups work on #18 (a) and #19 (a)
Be prepared to share out your solutions with the
rest of the class!
 Remember,
buffered solutions resist pH
change with addition of a STRONG acid or
base
 In a buffered solution, the pH is governed by
the ratio [HA]/[A-]

As long as this ratio remains virtually constant,
the pH will remain virtually constant

For a weak acid
buffer (pH < 7)



Added strong acid (H+)
reacts with conjugate
base
Added strong base (OH-)
reacts with weak acid
For a weak base
buffer (pH > 7):


Added strong acid (H+)
reacts with weak base
(B)
Added strong base (OH-)
reacts with conjugate
acid (BH+)

The acid dissociation equilibrium expression can be
rearranged to provide a useful equation when calculating pH:
[ H ][ A ]
[ HA]
Ka 
 [ H ]  Ka
[ HA]
[A ]
[ HA]
  log[ H ]   log Ka  log
[A ]
pH = pK a + log

A−
HA
= pK a + log
base
acid
This log form of the expression for Ka is called the
Henderson-Hasselbach equation and is useful for calculating
the pH of solutions when the ratio [HA]/[A-] is known

So, how do we go about calculating the pH of a
buffered solution once a strong acid or base is
added?

Solution #1




Do a “Before Reaction” – “Change” – “After Reaction”
table using MOLES ONLY for stoichiometry calculations to
determine new concentrations
 Assume reaction with H+/OH- goes to completion
Set-up an ICE table with new concentrations to
determine equilibrium concentrations
Calculate pH from [H+] using Ka expression
Solution #2


Do a “Before Reaction” – “Change” – “After Reaction”
table for stoichiometry calculations to determine new
concentrations
 Assume reaction with H+/OH- goes to completion
Use Henderson-Hasselbach equation to solve for pH
 All

groups work on #18 (c) and #18 (d)
Be prepared to share out your solutions with the
rest of the class!
 Buffer
capacity represents the amount of H+
or OH- the buffer can absorb with a
significant change in pH
 A good buffer can withstand the addition of
relatively large amounts of H+ or OH- without
changing pH!
 The most effective buffer occurs when the
ratio of [A-] to [HA] or [B] to [BH+] is 1

If this ratio is 1, then according to the
Henderson-Hasselbalch equation:
pH = pKa

Work backwards

Choose a weak acid whose pKa is close to the desired
pH


You can:
 Calculate [H+] from pH
 Substitute this value with Ka value of weak acid into
the equation:
[HA]
+
H = Ka
[A− ]
 Choose the ratio [HA]/[A-] that is closest to 1
OR:
 Substitute pKa (-log Ka) and pH values into the
Henderson-Hasselbach Equation


This will give a ratio of [A-]/[HA]
Then, convert ratio to molar quantities
 All

groups work on #17 (a) and (b)
Be prepared to share out your solutions with the
rest of the class!