16.1 Common Ion Effect
Buffer Solutions
The resistance of pH change
Dr. Fred Omega Garces
Chemistry 201
Miramar College
1
Common Ion Effect
January 13
Common Ion Effect
Ionization of an electrolyte, i.e., salt, acid or base is decreased when a
common ion is added to that solution.
i) What is the % ionization for 0.100 M acetic acid ? (Pure)
HC2H3O2 + H2O D
C2H2O2- + H3O+
Ka =1.8•10-5 M
Solving the iCe problem:
ka = =1.8•10-5 M = [C2H3O2-] [H3O+] /0.10 M g [H3O+]=1.34•10-3 M
% α = (1.34•10-3 / 0.10 ) * 100 = 1.34 %
pH = 2.87
ii) What is % α if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? (Buffer)
HC2H3O2 + H2O D
C2H2O2- + H3O+ Ka =1.8•10-5 M
i
0.100
Lots
0.100
1•10-7
C
-x
-x
+x
+x
Lots
0.100+x
1•10-7+x
[c] 0.100-x
ka = 1.8•10-5 M = [0.100+ x ] [x] /( 0.100 -x) ∼ [0.100] [x] /( 0.100 )
x = [H3O+]= 1.8•10 -5 M pH = 4.74
% α = (1.8•10-5 / 0.10 ) * 100 = 0.0180 %
Ionization % decrease in presence of common ion !!
2
Common Ion Effect
January 13
Common Ion Effect Equation
Consider the previous problem in which a common ion is in the same solution.
or
i
HC2H3O2 +
0.100
H2 O
Lots
C2H3O2- +
0.100
H3 O +
1•10-7
C
-x
-x
+x
+x
[c]
0.100-x
Lots
0.100+x
1•10-7+x
[c]
[HC2H3O2 ]
Lots
ka =
[C2H3O2- ] [H3O+]
D
[C2H3O2- ]
rearrange the equation
Ka =1.8•10-5 M
[H3O+ ]
[H3O+] =
ka •
[HC2H3O2]
[HC2H3O2 ]
[C2H3O- ]
Taking the - log of both side -
- log [H3O+]
= - log (ka • [HC2H3O2 ] / [C2H3O2-] )
or
pH
= -log ka -
let Ca =
[HC2H3O2]
and Cb =
pH = pKa or
pH = pKa +
log( [HC2H3O2] / [C2H3O2 - ] )
[C2H3O2
-
] ) therefore
log Ca / Cb
log Cb / Ca
This is the Henderson Hasselbach Equation:
pH = pKa + log Cb / Ca or
3
pOH = pKb + log Ca / Cb
Common Ion Effect
January 13
Henderson-Hasselbach Equation
pH of a solution can be calculated using a useful equation:
pH = pKa + log [A-] / [HA]
Where HA & Aare the weak acid and its
conjugate and Ka is for HA
Similarly,
pOH = pKb + log [HA] / [A-]
Where HA & Aare the weak base and its
conjugate and Kb is for A-
4
Common Ion Effect
January 13
Henderson-Hasselbach Equation: Example
Consider the common ion effect problem and lets see how the Henderson-Hasselbach
equation can be used to simplify this problem.
What is pH if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ?
HC2H3O2 + H2O D
C2H2O2- + H3O+
Ka =1.8•10-5 M
i
0.100
Lots
0.100
1•10-7
C
-x
-x
+x
+x
[c]
0.100-x
Lots
0.100+x
1•10-7+x
Using the Henderson-Hasselbach equation:
pH = - log (4.3•10-7 ) + log (0.100 / 0.100)
pH = 4.74 + log 1
pH = 4.74 + 0
pH = 4.74
Note: When a common ion is present in the same solution, the strategy to
solve the problem requires a Buffer Type of calculation.
5
Common Ion Effect
January 13
Henderson-Hasselbach Equation and Buffer Problems (sRF)
A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (Ka = 1.8 •10-5).
i)
What is the pH of the buffer?
ii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL
of the buffer.
iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL
of water.
Note: HCl = 0.100 M • 1.00mL = 0.1 mmol.
C2H3O2- =0.100 M •100mL = 10 mmol and HC2H3O2 =0.200 M •100mL = 20 mmol
i) pH = pKa + log Cb/Ca = -log(1.8•10-5) + log ( 0.10 / 0.20) g pH = 4.44
C2H3O2-
ii)
+
H3 O +
HC2H2O2
D
+
s
10mmol
0.1 mmol
20 mmol
Lots
R
-0.1
-0.1
+0.1
-
f
9.9
0
20.1
Lots
[c] 9.9/101
0
20.1/101
VT = 101 mL
pH = -log (1.8•10-5)+log [(9.9/101) / [20.1/101)] = 4.74 - 0.31 →
pH (initial) = 4.44,
6
H2 O
pH (final) 4.43,
pH = 4.43
ΔpH (change) = -0.01
Common Ion Effect
January 13
...Continue: Henderson-Hasselbach Equation
and Buffer Problems
...continue
A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (Ka = 1.8 •10-5). Reger
14.19
iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is
added to 100.0 mL of water.
Note: HCl = 0.100 M • 1.00mL = 0.100 mmol.
HCl +
iii)
H 2O
D
+
Cl-
s
0.100 mmol
1•10-7M
-
R
-0.100mmol
+0.100mmol
-
f
0
0.100 mmol
[c]
0
0.100mmol / 101 mL
[H3O+] = 9.9•10-4 M
pH (initial) = 7.00 ,
7
H3 O +
g
pH = 3.00
pH(final) 3.00, ΔpH(change) = -4.00
Common Ion Effect
January 13
Essential Feature of Buffer Systems
A buffer solution exhibits very small change in pH changes when
H3O+ and OH- is added. A buffer solution consists of relatively
high concentration of the components of a conjugate weak acidbase pair. The buffer-components concentration ratio
determines the pH, and the ratio and pH are related by the
Henderson-Hasselbalch equation. A buffer has an effective
range of pKa + 1 pH unit.
8
Common Ion Effect
January 13
Blood Buffer System
Buffer - A solution whose pH is resistant to change
Your body uses buffers to maintain the pH of your blood
Blood pH 7.35 - 7.45
Buffer system in body -
1. Proteins
2. Phosphates HPO42- / H2PO4- :
3. Carbonates H2CO3 / HCO3 - :
Reaction:
H3O+ + HCO3- D
H2CO3 +
H2CO3 g H2O + CO2
9
Common Ion Effect
1.6 / 1
10 / 1
H2O
(exhale)
January 13
Acidosis
Blood pH i 7.35 (ACIDOSIS)
Depression of the acute nervous symptom. Or respiratory center in the
medulla of the brain is affected by an accident or by depressive drugs.
Symptoms:
•Depression
of the acute
nervous
system
•Fainting
spells
•Coma
•RIP
10
Causes:
1. Respiratory Acidosis
Difficulty Breathing
(Hypo-ventilation)
Pneumonia, Asthma
anything which diminish
CO2 from leaving lungs.
2. Metabolic Acidosis
Starvation or fasting
Heavy exercise
Common Ion Effect
Mechanism:
1. Respiratory Acidosis
CO2 doesn’t leave lungs which
result in the build up of
H2CO3 in the blood
2. Metabolic Acidosis
If body doesn’t have enough
food then Fatty acids (Fat)
are used.
Fatty Acids g Acidic.
Furthermore, exercise leads
muscle to produce lactic acid.
January 13
Alkalosis
Blood pH h 7.45 (ALKALOSIS)
Hyperventilation during extreme fevers or hysteria. Excessive ingestion of
basic antacids and severe vomiting
Symptoms:
•Over
simulation of
the nervous
system
•Muscle
cramps
•Convulsion
•Death
11
Causes:
1. Respiratory Alkalosis
Heavy rapid breathing
(hyperventilation).
Results from - fear,
hysteria, fever,
infection or reaction
with drugs.
2. Metabolic Alkalosis
Metabolic irregularities
or by excess vomiting
Common Ion Effect
Mechanism:
1. Respiratory Alkalosis
Excessive loss of CO2
lowers H2CO3 and raise
HCO3- level (Can be
remedied by breathing in
a bag)
2. Metabolic Alkalosis
Vomiting removes excess
acidic material from
stomach. (pH of
stomach equals one).
January 13
Buffer System at Work
Buffer - System that resists change in pH when H3O+ or OH- is added.
Buffer solution may be prepared by a weak acid and its conjugate base.
How it Works:
A-
H3 O +
g HA
Buffer
H2 O
Remember pH = Conc. of H3O+
Acidosis
Your blood
Rxn: HCO3-
Excess
H3O+ + HCO3-
D
g
H3 O +
Alkalosis
12
Excess
OH-
H2CO3
H2CO3 + H2O
CO2 + H2O
OH- + H2CO3
D HCO3- + H2O
Common Ion Effect
January 13
Equation / Concept
Summary
The following
summary lists the
important tools
needed to solve
problems dealing
with acid-base
equilibria.
1
[H+] [OH-] = Kw
2
3
p X = - log X
pH + pOH = 14.00
HA ! H++ A[H+ ] [A − ]
Ka =
[HA]
4
5
6
7
8
9
10
11
13
Function
Permits the calculation of [H+] or [OH-] when the other is known.
This equation is the basis of the p-scale.
This equation shows the relationship between the pH and the pOH
This is the Mass Action Equation for the ionization of a weak acid in
water. This equation yields the ka given the equilibrium
concentration of all specie. The equation also yields the [H3O+]
given the initial concentration of the weak base [HA] and the ka.
This is the Mass Action Equation for the ionization of a weak base
B + H2O ! HB + OHin water. This equation yields the kb given the equilibrium
[HB] [OH− ]
Kb =
concentration of all specie. The equation also yields the [OH-]
[B]
given the initial concentration of the weak base [B] and the kb.
Percent ionization (α)
The percent ionization can be calculated from the initial
amount ionized
concentration of the acid (or base) and the change in the
α=
× 100%
concentration of the ions. Given the percent ionization (α) and the
initial amount
pH, the ka (or kb) can be determined.
Ka • Kb = Kw
This equation relates Ka and Kb for conjugate pairs in aqueous
solution,
ID of the solute as :
Identification of the function of the solute leads to the correct Mass
i) only a weak acid
Action expression and thereby leading to the correct equilibrium
ii) only a weak base
law. This is a critical first step to solve any acid-base equilibria
iii) a mixture of a weak acid
and its conjugate base
Identification of acidic
Identification of function of cation and anion of a salt lead to pH of
cations and basic anions
the salt solution. Given the ka or kb of the conjugates of these ions
leads to the calculation of the pH or pOH
Assumption which
In order to simplify the math calculation of a Mass Action
simplifies Mass Action
expression, assumption can be made base on the ka or kb value.
Reactions when H+ or OH- Understanding the buffer reaction permits the determination of the
are added to a buffer
effect of a strong acid or strong base on the pH of the solution.
solution.
Adding H+ lowers the [A-] and raises [HA], adding OH- lowers
[HA] and raises [A-].
Common Ion Effect
January 13