COMMON-ION EFFECT
Adding or subtracting the conjugate base (acid) of an acid (base) affects the pH
of a solution by consideration of LeChâtelier’s principle.
- This is nothing new.
In other words, adding a common ion to solution decreases dissociation of acid
or base.
Example: A 1.0 M solution of HF at a pH of 1.58 Ka(HF) = 6.8 x 10-4. How
does the pH of the solution change when 42 g of NaF is added to the
solution? Assume the volume remains constant at 2.0 L.
First consider what happens to equilibrium when F- is added?
HF (aq)  H+ (aq) + F- (aq)
H   F 

Ka 
 HF
- Equilibrium will shift to the left.
- H+ will decrease.
- Since F- is a base, it is sensible that adding it will decrease H+ (increase pH).
Calculate initial concentration of F-.
Assume that x is small.
x(0.50)
 6.8  10 4
10
.
10
.  6.8  104 
x
 14
.  10 3
0.50

pH = 2.87
Example: 34.6 g of NH4Cl is added to 3.98 L of 0.0145 M solution of NH3.
Kb(NH3) = 1.8 x 10-5.
a) What is the pH of the original solution before the addition of the NH4Cl?
Assume that x is small.
x2
 18
.  105
0.0145
x2  0.014518
.  105   2.6  107
x = 5.1  10-4  pOH = 3.29  pH = 10.71
b) Calculate the pH of the solution after the addition of the NH4Cl.
Assume volume remains constant.
Calculate initial concentration of NH4+.
34.6 g

1 mol
.
 0162
.
M
53.4912 g 3.98 L
NH4+ (aq)  H+ (aq) + NH3 (aq)
 H  NH 

Ka 
 NH 

4
3
Assume that x is small.
x 0.0145
 5.6  1010
(0162
. )
0162
. 5.6  1010 
x
 6.26  10 9  pH = 8.20
0.0145
BUFFERS
- Buffers are solutions of conjugate acid/base pairs.
- Buffers change pH slowly when acid or base is added.
- Buffers take advantage of common-ion effect
Consider solution of 2.0 M HC2H3O2 and 1.0 M NaC2H3O2.
HC2H3O2 (aq)  H+ (aq) + C2H3O2- (aq)
- Because of common-ion effect, acid will dissociate very little.
When a strong acid is added to the solution, it neutralizes the base, C2H3O2-.
HCl (aq) + C2H3O2- (aq)  HC2H3O2 (aq) + Cl- (aq)
When a strong base is added to the solution, it neutralizes the acid, HC2H3O2.
NaOH (aq) + HC2H3O2 (aq)  NaC2H3O2 (aq) + H2O (l)
What happens to the solution if we add enough HCl to change to acetate
concentration from 1.0 to 0.9?
- Base is converted to acid. Therefore, acetic acid concentration must change
from 2.0 to 2.1.
before neutralization
0.1 M
1.0 M
2.0 M
0.0 M
HCl (aq) + C2H3O2- (aq)  HC2H3O2 (aq) + Cl- (aq)
0.0 M
0.9 M
2.1 M
0.1 M
after neutralization
- Now calculate pH
Ka 
 H   C2 H3O2 
 HC2 H3O2 

x(0.9  x) x(0.9)

 1.8 105
2.1

x
2.1

  
x  4.2  105
 pH = 4.38
Compare this pH to pH of 0.1 M HCl solution
 pH = 1.0
**The acetic acid/acetate solution has “buffered” the effects of addition of acid.**
Buffer Capacity
- Amount of acid or base that can be added to a buffer solution before its pH
changes appreciably.
Consider two 1.0 L ammonia/ammonium buffer solutions.
Buffer 1
1.0 M NH3
1.0 M NH4+
 H   NH 3
 NH 4    K 10.  5.6  1010

Ka 

H

K


a
NH 4 
 NH 3  a 10.




pH = 9.25
What happens when 0.1 mol of NaOH is added?
0.1 M
1.0 M
1.0 M
+
NaOH (aq) + NH4 (aq)  NH3 (aq) + H2O (l)
0.9 M
0.0 M

 H    Ka
 NH   K
 NH 

4
3
a
1.1 M
0.9
 4.6  10 10
11
.
pH = 9.34
- High concentration of conjugate acid/base pairs make for high buffer capacity.
Buffer 2
0.01 M NH4+
0.01 M NH3
 H  NH 

Ka 
 NH 
3

4

H   K
 NH   K
 NH 


4
a
a
3
0.01
 5.6  10 10
0.01
pH = 9.25
What happens when 0.1 mol of NaOH is added?
0.1 M
0.01 M
0.01 M
+
NaOH (aq) + NH4 (aq)  NH3 (aq) + H2O (l)
0.09 M
0.02 M
0M
- pH is dictated by excess of strong base.
pOH = -log(0.09) = 1.05  pH = 12.95
- Buffer 1 has much more capacity than Buffer 2.
- Capacity of buffer depends on concentration of conjugate acid/base pairs.
- high concentrations  high capacity
- low concentrations  low capacity
Henderson-Hasselbalch equation
- Relates pH of buffer solution to concentrations of acids and bases.
- No substantive difference from Ka expression, just in a more convenient form.
Derivation
Recall rules of logarithms
log ab  log a  log b
H A 


Ka
a log b  log b a
 HA



H   K HA
A 

a

  HA  

 log H     log K a
A   

  HA  
  HA  
pH   log K a  log    pK a  log  
 A  
 A  
  HA  
pH  pK a  log  
 A  
1
 A  

 pK a  log
  HA  
 [ base]
pH  pK a  log

 [acid ] 
Observations about the Henderson-Hasselbalch equation
- pH of buffer is approximately pKa of acid
- exactly equal when [base] = [acid]
- Can get same answers from ICE table since x is small.
- Equation neglects changes in acid or base when strong acid or base is
added.
- Equation is used to monitor pH of weak acid solution when titrated with a
strong base.
- When choosing buffer system for a particular pH, choose acid such that its pKa  pH
Example: What is the pH of a solution with 1.0 M NaC2H3O2 and 2.0 M HC2H3O2?
Ka(HC2H3O2) = 1.8  10-5
Note that answer agrees with answer to same problem done previously with ICE table.
Example: How many grams of NaH2PO4 and Na2HPO4 are needed to make 2.4 L
of a buffer with a pH of 7.40? The concentrations of the buffer species
should approximately 0.1 M. Ka (H3PO4) = 6.2  10-8 = Ka(H2PO4-).
2
H2PO4- (aq)  H+ (aq) + HPO42- (aq)
 [ base]
pH  pK a  log

 [acid ] 
 [ base]
log
.
  7.40  7.21  019
 [acid ] 
 [ base]
7.40  7.21  log

 [acid ] 
[ base]
 100.19  15
.
[acid ]
Since the pH of the buffer is dependent only on the ratio of acid to base, let us
allow the concentration of the acid to be 0.10 M and the concentration of the
base to equal to 0.15 M
Grams of acid, NaH2PO4
2.4 L 
010
. mol 119.98 g

 28.8 g
L
mol
Grams of base, Na2HPO4
2.4 L 
015
. mol 14196
. g

 511
. g
L
mol
The pH of blood serum is 7.40; thus, phosphate buffers are used in many
intravenous formulations.
ACID-BASE TITRATIONS
- We can measure amount of acid (base) in a solution by adding an equal
amount of base (acid).
- equivalence point – where amount of acid equals amount of base
**moles of acid = moles of base**
- How do we know when moles of acid = moles of base? Use indicator.
- indicator – substance which changes color when pH is changed.
indicator
methyl red
litmus
bromothymol blue
phenolphthalein
pH of color change
5.0
6.5
7.1
9.4
acidic color
red
red
yellow
colorless
basic color
yellow
blue
blue
pink
- endpoint – where indicator changes color.
- When doing titration, choose indicator whose endpoint is at approximately the
expected equivalence point.
Titration procedures
Deadstop titration
- A precise amount of acid (base) is measured with a pipet.
- Class A pipet tolerance = 0.001 mL
- Indicator is added.
- For titration of strong acid with strong base, phenolphthalein is often used, but
bromothymol blue is more accurate with experienced eyes.
- A precise amount of base (acid) is added slowly using buret until indicator
changes.
Potentiometric titration
- In a titration, a precise amount of acid (base) is measured with pipet.
- pH of acid (base) is monitored with a pH meter as base is added.
- Data of pH versus volume of base (acid) added is recorded and plotted to yield
titration curve.
Titration curves
- A titration curve is a plot of pH versus volume of base (acid) added.
- Curve yields equivalence point and pKa (pKb) of acid (base) titrated.
Strong acid – strong base titration curve
- Curve is a very sharp S shape.
- Equivalence point occurs at pH = 7
- moles of acid = moles of base
- nH+ = nOH- Near equivalence point, pH changes
over 6 units.
- drastic change allows flexibility
with choice of indicator.
- Note small slope at beginning and end
of curve
- pH changes very little until
equivalence point is reached.
- For strong base titrated with strong acid, curve is reversed.
Titration of 50.00 mL of Strong Base with 0.1000 M HCl
14
13
12
11
10
9
pH
8
7
6
5
4
3
2
1
0
0
25
50
75
volum e of HCl (m L)
100
125
Weak acid – strong base titration curve
- Curve is less sharp
- pKa is found at point halfway between starting point and equivalence point.
- At halfway point, half of acid is converted to conjugate base, thus
[HA] = [A-], thus pH = pKa.
 [ base]
pH  pK a  log
  pK a  log(1)  pK a
 [acid ] 
- The halfway point is where a buffer solution has been created.
- This halfway point is called an inflection point.
- At equivalence point, change in pH is not as dramatic as strong acid – strong
base titration.
- Need to be more careful about choosing indicator
- Slope at beginning and end of curve are slightly steeper.
- Indicates that H+ is reacting to smaller extent with A- than with OH-.
- The stronger the conjugate base, the steeper the slope.
- For weak base titrated with strong acid, curve is reversed with pKa at halfway
point between start and equivalence point.
- recall that pKb = pKw – pKa
- pH at equivalence point not necessarily 7.
- We have four regions where can calculate pH during a weak acid-strong base
titration.
i) Initial (calculate the pH of the weak acid)
ii) Buffer region (Henderson-Hasselbalch)
iii) Equivalence point (weak base solution)
iv) Excess titrant (calculate excess OH- conc.)
Example: Calculate the pH of the resultant solution after 25.00 mL of 0.1102 M
HF is titrated with the following volumes of 0.1004 M NaOH.
a) 0.00 mL
This is a weak acid solution.
b) 15.63 mL
Now we need to know where this point is on the titration curve.
Compare the amount of acid to the amount of base.
The amount of base is less than the amount of acid; therefore, this point is in the
buffer region.
Before considering the equilibrium, consider the acid-base reaction first.
HF (aq) + OH- (aq)  H2O (l) + F- (aq)
After the chemical reaction: n F  1.569 mmol
n HF remaining   n HF total   n HF used   n HF total   n OH  used 
 2.755 mmol  1.569 mmol  1.186 mmol
 [ base]
pH  pK a  log

 [acid ] 


 1.569 mmol 40.63 mL 

pH   log 6.8  10 4  log 
 1.186 mmol 40.63 mL 
 1.569 mmol 
pH  3.17  log 
  3.17  0.12  3.29
 1.186 mmol 
c) 25.00 mL
Where on the titration curve is this point?
Still in the buffer region (barely!)
After the chemical reaction: n F  2.510 mmol
d) 27.44 mL
We’re at the equivalence point! We now have a solution of sodium fluoride (a
weak base).
F- (aq) + H2O (aq)  HF (aq) + OH- (aq)
What is the fluoride concentration?
[ F ] 
moles of F 
25.00 mL  01102
.
M

 0.05254 M
total volume 25.00 mL  27.44 mL
Now we can treat solution as a weak base.
Assume x is small.
x2
 15
.  1011  x 2  0.0525415
.  1011 
0.05254
x = 8.8  10-7 = [OH-]
pOH = -log 8.8  10-7 = 6.06
pH = 14.00 – pOH = 7.94
If we are doing a deadstop titration, phenolphthalein (pKa = 9.4) would not be an
optimal indicator. We would get a better result with phenol red instead. (pKa = 7.9)
e) 30.34 mL
This point is in the excess titrant region.
After the chemical reaction: n F  2.755 mmol
n OH  remaining   n OH   total   n OH   used 
 3.046 mmol  2.755 mmol  0.291mmol
[OH  ] 
moles of OH  0.291 mmol

 0.00526 M
total volume
55.34 mL
pOH   log[OH ]   log  0.00526  2.28
pH  14.00  pOH  14.00  2.28  11.72
Polyprotic acids
Each proton of polyprotic acid has an equivalence point and a buffer region to
find pKa.
Titration of 25.00 mL of Diprotic Acid with 0.1000 M NaOH
13
12
11
10
pKa
2
9
8
equivalence
points
pH
7
6
5
4
3
2
pKa
1
1
0
0
25
50
75
100
125
150
volum e of NaOH (m L)
Note: pKa points and equivalence points are at inflection points
SOLUBILITY EQUILIBRIA
An untruth: Solid substances are either soluble or insoluble in solvent.
The truth: All ionic compounds are soluble though some have extremely small
solubilities.
The less soluble substances are in equilibrium with their dissociated ions.
MX (s)  M+ (aq) + X- (aq)
M   X  

Kc 
 K sp   M   X  
MX
Ksp – solubility product constant
In general for MAXB (s)  A Ma+ (aq) + B Xb- (aq)
Ksp   M a    X b 
A
B
Examples:
Ca(OH)2
Ag3AsO4
SrCO3
**Ksp tells us concentration of ions in a saturated solution.**
Example: Calculate pH of Ca(OH)2 solution. Ksp = 5.5 10-6
Ca(OH)2 (s)  Ca2+ (aq) + 2 OH- (aq)
Ksp   Ca 2  OH    55
.  106
2
Solubility – The amount of solute (in grams or moles) that can be dissolved in a
specific amount of solvent.
- If solubility is given, Ksp can be calculated.
- If Ksp is given, solubility can be calculated.
Example: The solubility of Ca3(PO4)2 is 2.2110-5 g/100 mL Calculate Ksp.
In a problem like this, consider the change in dissociation (“x” variable) to be the
molar solubility.
Thus we will need to convert the mass solubility to molar solubility.
Keep in mind that we’ll need to convert the units in the denominator of 100 mL to L.
(100 mL = 0.1 L)
2.21105 g 100 mL 1mol


 7.12 107 M
100 mL
0.1L 310.18g
Write Ksp expression for
Ca3(PO4)2 (s)  3 Ca2+ (aq) + 2 PO43- (aq)

Ksp   Ca 2   PO4 3
3

2
A form of calcium phosphate, hydroxyapatite Ca10(PO4)6(OH)2, is the major mineral
component of bone.
Example: The Ksp of PbCl2 is 1.1710-5. Calculate its solubility in g/100 mL.
PbCl2 (s)  Pb2+ (aq) + 2 Cl- (aq)
Remember: “x” is the molar solubility.
Ksp  Pb2 Cl    117
.  105
2
Ksp  x 2 x  4x3  117
.  105
2
117
.  105
 x  0.0143 M
4
0.0143mol 0.1L 278.11 g


 0.398 g /100 mL
L
100 mL
mol
x3 
FACTORS THAT AFFECT SOLUBILITY
Common ion effect
- Addition of an ion in a solubility product reduces solubility. (LeChâtelier’s
principle)
Consider BaCrO4 (s)  Ba2+ (aq) + CrO42- (aq)
- Addition of BaCl2 (for example) shifts equilibrium to left, thus reducing
solubility.
Example: Ksp(BaCrO4) = 1.1710-10. Calculate the molar solubility of BaCrO4
a) in a solution of 0.100 M BaCl2
BaCrO4 (s)  Ba2+ (aq) + CrO42- (aq)
K sp   Ba 2  CrO 4 2    117
.  10 10
Yikes! A quadratic equation
Relax. Assume x is small
Ksp   0100
.  x  117
.  1010
x  117
.  109 M  s
While we’re at it, let’s calculate mass solubility.
b) in a solution of 0.00100 M BaCl2
Initial
Change
Equil.
[Ba2+]
0.00100
+x
0.00100 + x
[CrO42-]
0
+x
+x
Ksp   0.00100 x  117
.  1010
x  117
.  107 M  s
In pure water,
1.17 107 mol 0.1L 253.32 g


 2.96 106 g /100 mL
L
100 mL
mol
s  1.08 105 M
BaCrO4 is more soluble in a more dilute solution of BaCl2.
pH of solution
- pH affects solubility of hydroxides since [OH-] is related to [H+].
- pH affects solubility of fluorides, oxalates, phosphates, etc…, that is, salts with
weak acid anions.
Calculate the Cu(II) concentration in a saturated solution of Cu(OH)2.
Ksp(Cu(OH)2) = 2.210-20.
a) when pH = 10.00
pH = 10.00  [H+] = 1.010-10  [OH-] = 1.010-4
Ksp   Cu2  OH    2.2  1020
2
Cu 2  
2.2 1020

2.2 1020
OH 
1.0 10
- In this case [Cu ] is also molar solubility.

2

4 2
2+
b) when pH = 5.00
pH = 5.00  [H+] = 1.010-5  [OH-] = 1.010-9
 2.2 1012 M
Ksp   Cu2  OH    2.2  1020
2
 Cu  
2
2.2  1020
 OH 
 2

2.2  1020
10.  10 
9 2
 2.2  102 M
Example: How does pH affect solubility of SrCO3?
SrCO3 (s)  Sr2+(aq) + CO32- (aq)
H+ (aq) + CO32- (aq)  HCO3- (aq)
Consider both equilibria together.
SrCO3 (s) + H+ (aq)  Sr2+(aq) + HCO3- (aq)
Increasing pH, decreases solubility.
Decreasing pH, increases solubility.
True in general for solubilities of salts of weak acids.
FRACTIONAL PRECIPITATION
Consider the use of Q (ion product) to determine precipitation.
MX (s)  M+ (aq) + X- (aq)
Q = [M+][X-]
If Q > Ksp
If Q = Ksp
If Q < Ksp
precipitation occurs
saturated solution
precipitate dissolves
Example: A sample has 2.410-5 M of Pu3+ in an acidic solution. To what pH
must the solution be adjusted to begin precipitation of the plutonium as
Pu(OH)3? Ksp = 2.010-20
Thus, the solution will not begin to precipitate until enough base is added to
adjust the pH to 8.97.
When two metal ions are in solution, they can be separated from one another if
their Ksp for a specific anion are very different.
Example: Consider a solution of 0.114 M Pb2+ (aq) and 0.214 M Bi3+ (aq).
Ksp(PbI2) = 7.110-9 Ksp(BiI3) = 8.110-19
a) What is the minimum concentration of I- needed to start precipitation of the Pb2+ ion?
b) What is the minimum concentration of I- needed to start precipitation of the Bi3+ ion?
c) What range of concentrations of I- can be used to precipitate only Bi3+?
1.6 10-6 M < [I-] < 2.5 10-4 M
Bismuth and lead are commonly found together in ores.
COORDINATION COMPLEXES
Lewis acid/base theory is useful to explain structure of metal ions in aqueous
solution (coordination complexes).
Consider Cu2+ ion.
CuSO4 (s)
+
very pale blue
H2O (l)

CuSO4 (aq)
bright blue
What has changed?
- In the solid, Cu2+ is surrounded by SO42- In the solution, Cu2+ is surrounded by H2O
- Cu2+ acts as Lewis acid.
- H2O acts as Lewis base.
H
H
O
H
H
O
O
Cu2+
H
H
O
H
Consider the following equilibrium
H
CoCl42- + 6 H2O  Co(H2O)62+ + 4 Cl-
Lewis acid
Lewis base H
H
O
Cl-
H
H
Cl-

Cl-
Co2+
O
O
Co2+
H
H
O
ClH
blue
pink
Ag+ + 2NH3  [Ag(NH3)2]+
Lewis acid
Lewis base
H
H
H
N
H
H
Ag+
N
H
H
AgCl (s) + 2NH3  [Ag(NH3)2]+ + Cl- (aq)
Ammonia is stronger Lewis base than chloride when silver ion is the Lewis acid.
Complexation of metal ions
- The addition of substances with strong Lewis base character can react with
certain metal ions to form soluble complexes.
- Formation of complexes increases solubility of “insoluble” salt.
Consider: Silver diammine complex
Ag+ + 2NH3  [Ag(NH3)2]+
Ag( NH 3 ) 2
Kf 
.  107
2  17

Ag NH 3

 


- Note that Kf (formation constant) tells us that formation of the complex is
overwhelmingly favored.
- The ammonia will stay bound to the silver ion until a stronger Lewis base is
added to the solution.
Example: How does the addition of ammonia affect the solubility of AgCl?
AgCl (s)  Ag+ (aq) + Cl- (aq)
Ag+ (aq) + 2NH3 (aq)  [Ag(NH3)2]+ (aq)
AgCl (s) + 2NH3 (aq)  [Ag(NH3)2]+ (aq) + Cl- (aq)
Kc = Ksp  Kf = (1.8  10-10)  (1.7  107) = 0.00306
- From LeChâtelier’s principle, addition of ammonia increases solubility.
Other substances that form complexes with metal ions
I-, Br-, SCN-, Cl-, NO3-, F-, OH-, H2O, CN-, CO
Amphoteric Metals
Metals generally react with acids to form metal ion solutions.
Ni (s) + 2H+ (aq)  Ni2+ (aq) + H2 (g)
However, some metals also react with bases to form metal ion solutions.
Al (s) + 4 OH- (aq)  Al(OH)4- (aq)
Amphoteric metals react in both acid and bases to form ions.
The oxides and hydroxides of amphoteric metals will dissolve in acid and base.
Amphoteric metals tend to be close to the metal/nonmetal boundary.
A partial list of amphoteric metals includes: Si, Al, Zn, Ga, Ge, As, Sn, Sb, Pb, etc…
[Read Section 17.7 for help in lab (will not be tested in lecture).]