Homework
Fundamentals of
Analytical Chemistry
Chapter 9
20, 21, 25, 27, 29
Acids, Bases, and Buffers A
Chapter 9(b)
Definitions
Focus in
Acid
Aqueous Solutions
Releases H+ ions in water (Arrhenius)
Proton donor (Bronsted
(Bronsted – Lowry)
ElectronElectron-pair acceptor (Lewis)
Characteristic cation for the solvent (solvent(solvent-system)
Base
Releases OH- in water (Arrhenius)
Proton acceptor (Bronsted
(Bronsted – Lowry)
ElectronElectron-pair donor (Lewis)
Characteristic anion for the solvent (solvent(solvent-system)
All aqueous solutions have OH-
KW
[H+][OH-]
More correct is H3O+
We will assume H+ and H3O+ are equivalent
• Even more correct – H9O4+
• Most common approach
= [H ][OH ] = 1.0 x 10
25OC)
(at 25 C)
Note that concentrations rather than activities
• Should use activities, but µ = 1.0 x 10-7
• Will ignore activity coefficients unless specifically
asked to use them in calculations!
Aqueous Solutions
10-14
Note H2O(l) is not part of the equilibrium
expression
• Activity of solvents by definition equal to 1
to)
have H+ ions.
When [H+]
H2O(l) H+ + OH2H2O(l) H3O+ + OH-
ions
All aqueous solutions (are considered
Aqueous Solutions
For any aqueous solution
this course is water as a solvent
Most common reference will be Bronsted –
Lowry definitions
If
= [OH-] = 1.0 x 10-7
‘Neutral’
Neutral’ solution
[H+] > [OH-]
Acidic
• [H+] > 10-7
• [OH-] < 10-7
Basic
• [H+] < 10-7
• [OH-] > 10-7
1
Typical Values
pH
range for H+ and OH- are
from greater than 10 to less than 10-15
Concentration
Difference between 1¢
1¢ and $100 trillion.
‘Compress’
Compress’ scale
(10-15)
pH = -log [H+]
Actually, pH = -log aH+
Compress scale, gives mostly positive values
Inverse relationship to [H+]
• As [H+] increases, pH decreases
[[-15]
scale more palatable by changing
sign
using logarithms
Range is then log (10) [1] to log
Make
Working definition:
Since
Range is then -1 to 15
acid has been defined as [H+] > 10-7
Acidic pH has values less than 7
Basic values have pH greater than 7
Error in
measurement of [H+] ≈ 5%
Since pH log function, error is ±0.02 pH units
• [0.05 / ln 10] = 0.0217 = ±0.02
Calculation of pH for a Solution
Must
first define the type of solution
Acid
And the last shall be first…
first…
None
• Strong
• Weak
Base
• Solution will still be neutral
• Note that the change in ionic strength d/n change
the activity from 1.0 x 10-7
• Kw = aH+ * aOH-
• Strong
• Weak
of the Above
Nothing in the solution to cause the hydrogen
ion concentration to change (NaCl
(NaCl))
pH = 7.00
Buffer
Acid Salt
None of the Above
pH for Acids
If
the solution is an acid
Species is a proton donor
Different levels of ‘donation’
donation’
• All (strong acid)
• Some (weak acid)
Strong acids
HCl,
HCl, HBr,
HBr, HI
HNO3, HClO3, HClO4
H2SO4
pH for Acids
Strong acids donate all protons
[H+] = CHA
What
is the pH for the following solutions
of HCl (a strong acid)
1.0 x 10-2
1.0 x 10-5
1.0 x 10-7
1.0 x 10-10
• Special situation – will discuss later
2
pH for Acids
HOW CAN WE ADD
AN ACID TO A
SOLUTION AND
MAKE THE
SOLUTION BASIC???
pH for Acids
Remember,
pH values greater than 7
mean that the solution is basic!
What happened?
pH for Acids
For ALL acidic solutions,
there are 2
sources of protons
pH for Acids
The acid
Water
As [H+]
approaches 10-7
When [H+]
is large, we can ignore
contribution from water.
‘Neutral’
Neutral’ [H+] = 10-7
LeChatlier’
LeChatlier’s Principle causes a shift away
from the production of H+ from the
autoionization of water
Equilibrium shift
decreases
[H+] from water
becomes more
significant.
What is the pH
for a solution that
is 1.0 x 10-7 M in
HCl?
HCl?
pH for Acids
When do
we have to account for addition
of hydrogen ion due to the dissociation of
water, and when can we safely ignore this
contribution?
When [H
[H+]calc < 3 x 10-7, must account for
water dissociation
Translates to a pH > 6.5
Most cases we can ignore!
[ ][ ]
[ ] [ ] + [H ]
Also, [H ]
= [OH ] = x
Then, 1.0 x10 = (1.0 x10 + x )( x )
K w = H + OH −
But H + = H +
+
water
+
acid
−
water
−14
−7
x 2 + 1.0 x10 −7 x − 1.0 x10 −14 = 0
x=
− 1.0 x10 −7 ±
(1.0 x10 )
−7 2
− 4(1)( −1.0 x10 −14 )
2(1)
x = 6.2 x10 −8
[H ] = 1.0 x10
+
−7
+ 0.62 x10 −7 = 1.62 x10 −7
pH = − log(1.62 x10 −7 ) = 6.79
pH for Acids
Weak acids
CANNOT assume complete dissociation!
For the weak acid HA H+ + AEquilibrium constant for this dissociation is Ka
Ka = [H+][A-] / [HA]
• Subscript a indicates an acid dissociation constant
• Unfortunately, we know none of the equilibrium
concentrations
• Must relate analytical concentration to equilibrium
concentration
3
pH for Acids
What do
Cation
we know?
Ka
CHA
balance equation
Since a solution must be electrically neutral,
the concentration or all of the positive charges
must equal the concentration of all of the
negative charges
What
H+
Anions
Charge
pH for Acids
A-, OH• Always have hydroxide ions in an aqueous
solution
Using the
• Remember, that as [H+] increases, [OH-]
decreases
ions do we have in solution?
pH for Acids
pH for Acids
We can assume that [A-] >> [OH-]
Reduced our expression from 3
• Therefore, [A-] + [OH-] ≈ [A-]
Substituting, [H+] ≈ [A-]
(1)
• True when [H+] > 3 x 10-7
• Will be true for all weak acids we work with!
the equilibrium
constant expression
Ka = [H+] [H+] / [HA]
Ka = [H+]2 / [HA]
variables
to 2!
Mass balance equation
Substituting (1) into
charge balance equation
[H+] = [A-] + [OH-]
The analytical concentration for a species that
dissociates is equal to the sum of the
concentration of it’
it’s identifiable pieces
For HA, CHA = [HA] + [A-]
Rearranging, [HA] = CHA – [A-]
Substituting (1), [HA] = CHA – [H+]
(2)
pH for Acids
pH for Acids
[ ]
[ ]
* (C − [H ]) = [H ]
+ 2
Substituting (1) and
(2) into the
equilibrium
expression
Ka = [H+]2 / CHA – [H+]
One equation, one
variable
Ka =
Ka
H
C HA − H +
+ 2
+
Works for all situations we
HA
rearranging ;
[H ] = K C − K [H ]
[H ] + K [H ]− K C = 0
Quadratic in [H ]
+ 2
+
a
HA
+ 2
a
+
a
a
HA
+
[H ] = − K
a
[H ] = − K
a
+
+
± K a2 − 4 * (− K a C HA )
2
will encounter,
but…
but…
What if CHA >> [H+]
• Then CHA - [H+] ≈ CHA
• Ka = [H+]2 / CHA
• Rearranging [H+]2 = KaCHA
• H + = K a CHA
[ ]
+ K a2 + 4 K a C HA
2
4
pH for Acids
Limitations for ‘short’
short’ equation
CHA >> [H+]
Method 1
pH for Acids
Method
• Do the problem the short way, test the assumption,
and redo the problem if assumption is wrong.
3
Assumption will be valid if relative error < 5%
If 100*Ka < CHA, the assumption will introduce
less than 5% error
First test, is 100*Ka < CHA
• YES, then you can use the ‘short’
short’ equation
• NO, then use the quadric based equation
• NOT SURE, then use the quadratic based
equation
Method 2
• ALWAYS use the quadratic based equation
(safest)
Will always work
pH for Bases
What
changes for bases?
[OH-] > [H+]
• Calculate [OH-]
• Must convert to pH somehow
Method
1
pH for Bases
Method
[H+] * [OH-] = 1.0 x 10-14
[H+] = 1.0 x 10-14 / [OH-]
pH = -log [H+]
pH for Bases
Strong bases
Soluble hydroxides
• Group I hydroxides
• Ba2+, Sr2+
Soluble oxides
• Group I oxides
M2O 2M + + O2O2- + H2O 2OH[OH-] = 2*CSALT
2
[H+] * [OH-] = 1.0 x 10-14
log ([H+] * [OH-]) = log (1.0 x 10-14)
-log ([H+] * [OH-]) = -log (1.0 x 10-14)
-log [H+] + -log [OH-] = 14.00
pH + pOH = 14.00
14.00 – pOH = pH
pH for Bases
Strong base –
we know [OH-]
Calculate pH using Method 1 or 2
Ignore contribution due to dissociation of
water unless:
•
•
•
[OH-]calc < 3 x 10-7
pHcalc < 7.5
‘Mirrors’
Mirrors’ strong acid rules.
Others which are of no analytical importance
5
pH for Bases
Weak bases
We can
B + H2O HB+ + OHKb = [HB+][OH-] / [B]
[OH-]
for weak bases
[OH ] = − K
−
b
pH for Bases
use the second equation only
when CB > 100Kb
Similar to situation with weak acids
When in doubt, use the quadratic form
+ K b2 + 4 K bC B
2
OR
[OH ] =
−
K bC B
Conjugate Acid/Base Pairs
A
conjugate acid/base pair is a pair of
chemical species that differ by the
presence/absence of one proton (H+.) The
conjugate acid has one more proton than
the conjugate base.
HA / AHB+ / B
Do
not confuse conjugate base (or acid)
with base (or acid).
acid).
Conjugate Acid/Base Pairs
Conjugate
base of a strong acid IS NOT
basic!
Donates protons completely – doesn’
doesn’t ‘want
any’
any’ back
Conjugate
base of a weak acid IS basic
Will ‘take’
take’ protons from water
• Not really, but easier to picture
Conjugate
acid of a weak base IS acidic
Will donate protons
pH for Weak Bases
Appendix 3
Ka values only
• Acetic acid (HC2H3O2)
• Ammonium ion (NH4+)
How do we get Kb for a weak base?
• What is the relationship between Ka and Kb for a
conjugate acid/base pair?
For HA ↔ H + + A − ; K a =
[ H + ][ A − ]
[ HA]
For A− + H 2O ↔ HA + OH − ; K b =
∴ K a * Kb =
[ HA][OH − ]
[ A− ]
[ H + ][ A − ] [ HA][OH − ]
*
[ HA]
[ A− ]
K a * K b = [ H + ][OH − ] = K w = 1.0 x10 −14
∴ for conjugate acid / base pairs :
Kb =
K w 1.0 x10 −14
=
Ka
Ka
6
Misc.
Buffers
Water is a ‘leveling solvent’
solvent’
Two
Strongest acid in water is H3O+
• Acids stronger than H3O+ transfer protons
completely to water to form the hydronium ion.
• Relative to acidic strength, all strong acids are the
same strength
Strongest base in water is OH• Bases stronger than OH- react completely with
water to form the hydroxide ion (O2-)
properties
Resistant to ∆pH upon the addition of a small
amount of strong acid/base
Maintain pH upon dilution
Mixture
of a (significant amount of a) weak
conjugate acid/base pair
HC2H3O2/C2H3O2NH4+/NH3
Buffers
Resistant
to ∆pH
Assume [ HA] = C HA ; [ A− ] = C A−
Strong acid reacts with the base
Strong base reacts with the acid
Maintains pH upon
Buffers
[ H + ][ A− ]
Ka =
[ HA]
Then K a = [ H + ]
C A−
C HA
C −

∴ log K a = log  [ H + ] A 
C HA 

C − 
log K a = log [ H + ] + log A 
 C HA 
rearranging :
dilution
pH of a buffer is controlled by the ratio of the
weak acid to the weak base
C − 
− log [ H + ] = − log K a + log A 
 C HA 
This is the Henderson − Hasselbalch equation
Buffers
What if CHA ≠ [HA]
‘Strong’
Strong’ weak acid
• Dissociation of acid
causes CHA > [HA]
Buffers
Use when Ka≥10-3
Analagous equation
for ‘strong’
strong’ weak
bases
Will not derive (or use)
From the Henderson − Hasselbalch equation :
[H + ] = Ka
What
+
C HA − [ H ]
C A− + [ H + ]
+
[ H ] * (C A− + [ H + ]) = K a * (C HA − [ H + ])
[ H + ]2 + C A− [ H + ] = K a C HA − K a [ H + ]
• Maximum pH = pKa + log (10) = pKa + 1
• Minimum pH = pKa + log (0.1) = pKa – 1
rearranging :
+ 2
+
+
[ H ] + C A− [ H ] + K a [ H ] − K aC HA = 0
∴[ H + ]2 + (C A− + K a )[ H + ] − K a C HA = 0
+
[H ] =
is …a significant concentration…
concentration…
For a buffer, the ratio of base to acid must be
between 0.1 and 10
Using these limits
For a buffer, pH = pKa ± 1
− (C A− + K a ) + (C A− + K a ) 2 + 4 K aC HA
2
7
Buffers
Buffer capacity (β
(β)
Moles of a strong acid or a strong base that
causes 1.00 L of the buffer to undergo a 1.00
unit change in pH
Function of concentration of components and
also the ratio of the components
‘Best’
Best’ buffer when ratio of base to acid is 1
• pH = pKa + log (1)
• pH = pKa gives the ‘ideal’
ideal’ buffer
Preparing Buffers
Mix
weak acid and a soluble salt of the
conjugate base
HC2H3O2 + NaC2H3O2
Mix
weak base and a soluble salt of the
conjugate acid
NH3 + NH4Cl
Partial neutralization
HC2H3O2 + NaOH when NaOH is the limiting
reactant
NH3 + HCl when HCl is the limiting reactant
8