Title: AcidBase Equilibria
1Acid-Base Equilibria
2The common ion effect is the shift in equilibrium
caused by the addition of a compound having an
ion in common with the dissolved substance.
Consider mixture of CH3COONa (strong electrolyte)
and CH3COOH (weak acid).
16.2
3What is the pH of a solution containing 0.30 M
HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
0.30
0.00
0.52
-x
x
x
0.30 - x
x
0.52 x
Ka for HCOOH 1.8 x 10 -4
x 1.038 X 10 -4
H HCOO-Ka
HCOOH
pH 3.98
16.2
4OR Use the Henderson-Hasselbach
equation Consider mixture of salt NaA and weak
acid HA.
pKa -log Ka
Henderson-Hasselbach equation
16.2
5What is the pH of a solution containing 0.30 M
HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base!
0.30
0.00
0.52
-x
x
x
0.30 - x
x
0.52 x
Common ion effect
0.30 x ? 0.30
4.01
0.52 x ? 0.52
HCOOH pKa 3.77
16.2
6- A buffer solution is a solution of
- A weak acid or a weak base and
- The salt of the weak acid or weak base
- Both must be present!
A buffer solution has the ability to resist
changes in pH upon the addition of small amounts
of either acid or base.
CH3COOH (aq) H (aq) CH3COO- (aq)
Adding more acid creates a shift left IF enough
acetate ions are present
16.3
716.3
8Which of the following are buffer systems? (a)
KF/HF (b) KCl/HCl, (c) Na2CO3/NaHCO3
(a) KF is a weak acid and F- is its conjugate base
buffer solution
(b) HCl is a strong acid
not a buffer solution
(c) CO32- is a weak base and HCO3- is it
conjugate acid
buffer solution
16.3
9Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl
buffer system. What is the pH after the addition
of 20.0 mL of 0.050 M NaOH to 80.0 mL of the
buffer solution?
NH4 OH- NH3
Kb
1.8 X 10-5
0.36
0
Initial
0.30
x
x
Change
- x
End
0.30 - x
0.36 x
x
(.36 x)(x) (.30 x)
1.8 X 10-5
0.36x 0.30
pH 9.18
1.8 X 10-5 ?
x 1.5 X 10-5
pOH 4.82
16.3
10Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl
buffer system. What is the pH after the addition
of 20.0 mL of 0.050 M NaOH to 80.0 mL of the
buffer solution?
final volume 80.0 mL 20.0 mL 100 mL
NH4 0.36 M x 0.080 L 0.029 mol / .1 L
0.29 M OH- 0.050 x 0.020 L 0.001 mol / .1
L 0.01M NH3 0.30 M x 0.080 0.024 mol /
.1 L 0.24M
0.01
0.24
start (M)
0.29
end (M)
0.28
0.0
0.25
H NH3 NH4
Ka 5.6 X 10-10
H 6.27 X 10 -10
pH 9.20
H 0.25 0.28
5.6 X 10-10
16.3
11Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl
buffer system. What is the pH after the addition
of 20.0 mL of 0.050 M NaOH to 80.0 mL of the
buffer solution?
pKa 9.25
9.17
final volume 80.0 mL 20.0 mL 100 mL
0.01
0.24
start (M)
0.29
end (M)
0.28
0.0
0.25
9.20
16.3
12Chemistry In Action Maintaining the pH of Blood
16.3
13Titrations
In a titration a solution of accurately known
concentration is added gradually added to another
solution of unknown concentration until the
chemical reaction between the two solutions is
complete.
Equivalence point the point at which the
reaction is complete
Indicator substance that changes color at the
endpoint (hopefully close to the
equivalence point)
Slowly add base to unknown acid UNTIL
The indicator changes color (pink)
4.7
14Strong Acid-Strong Base Titrations
100 ionization! No equilibrium
16.4
15Weak Acid-Strong Base Titrations
At equivalence point (pH gt 7)
16.4
16Strong Acid-Weak Base Titrations
At equivalence point (pH lt 7)
16.4
17Acid-Base Indicators
16.5
1816.5
19The titration curve of a strong acid with a
strong base.
16.5
20Which indicator(s) would you use for a titration
of HNO2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of
weak acid.
At equivalence point, pH gt 7
Use cresol red or phenolphthalein
16.5
21Finding the Equivalence Point(calculation method)
- Strong Acid vs. Strong Base
- 100 ionized! pH 7 No equilibrium!
- Weak Acid vs. Strong Base
- Acid is neutralized Need Kb for conjugate base
equilibrium - Strong Acid vs. Weak Base
- Base is neutralized Need Ka for conjugate acid
equilibrium - Weak Acid vs. Weak Base
- Depends on the strength of both could be
conjugate acid, conjugate base, or pH 7
22Exactly 100 mL of 0.10 M HNO2 are titrated with
100 mL of a 0.10 M NaOH solution. What is the pH
at the equivalence point ?
start (moles)
0.01
0.01
end (moles)
0.0
0.0
0.01
Final volume 200 mL
0.05
0.00
0.00
-x
x
x
0.05 - x
x
x
pOH 5.98
2.2 x 10-11
pH 14 pOH 8.02
0.05 x ? 0.05
x ? 1.05 x 10-6
OH-
23Complex Ion Equilibria and Solubility
A complex ion is an ion containing a central
metal cation bonded to one or more molecules or
ions.
16.10
2416.10
25Complex Ion Formation
- These are usually formed from a transition metal
surrounded by ligands (polar molecules or
negative ions). - As a "rule of thumb" you place twice the number
of ligands around an ion as the charge on the
ion... example the dark blue Cu(NH3)42 (ammonia
is used as a test for Cu2 ions), and Ag(NH3)2. - Memorize the common ligands.
26Common Ligands
27Names
- Names ligand first, then cation
- Examples
- tetraamminecopper(II) ion Cu(NH3)42
- diamminesilver(I) ion Ag(NH3)2.
- tetrahydroxyzinc(II) ion Zn(OH)4 2-
- The charge is the sum of the parts (2) 4(-1)
-2.
28When Complexes Form
- Aluminum also forms complex ions as do some post
transitions metals. Ex Al(H2O)63 - Transitional metals, such as Iron, Zinc and
Chromium, can form complex ions. - The odd complex ion, FeSCN2, shows up once in a
while - Acid-base reactions may change NH3 into NH4 (or
vice versa) which will alter its ability to act
as a ligand. - Visually, a precipitate may go back into solution
as a complex ion is formed. For example, Cu2 a
little NH4OH will form the light blue
precipitate, Cu(OH)2. With excess ammonia, the
complex, Cu(NH3)42, forms. - Keywords such as "excess" and "concentrated" of
any solution may indicate complex ions. AgNO3
HCl forms the white precipitate, AgCl. With
excess, concentrated HCl, the complex ion,
AgCl2-, forms and the solution clears.
29Coordination Number
- Total number of bonds from the ligands to the
metal atom. - Coordination numbers generally range between 2
and 12, with 4 (tetracoordinate) and 6
(hexacoordinate) being the most common.
30Some Coordination Complexes