Transcript Acid-Base Equilibria and Solubility Equilibria
Acid-Base Equilibria
Chapter 16
The
common ion effect
is 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.
Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid).
CH CH 3 3 COONa COOH ( (
s aq
) ) Na + (
aq
) + CH 3 COO (
aq
) H + (
aq
) + CH 3 COO (
aq
) common ion 16.2
A
buffer solution
is a solution of: 1. A weak acid or a weak base
and
2. 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.
Consider an equal molar mixture of CH 3 COOH and CH 3 COONa CH 3 COOH (
aq
) H + (
aq
) + CH 3 COO (
aq
) Adding more acid creates a shift left IF enough acetate ions are present 16.3
Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na 2 CO 3 /NaHCO 3 (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) CO 3 2 is a weak base and HCO 3 buffer solution is it conjugate acid 16.3
What is the pH of a solution containing 0.30
M
HCOOH and 0.52
M
HCOOK?
Mixture of weak acid and conjugate base!
Initial (
M
) Change (
M
) Equilibrium (
M
) HCOOH (
aq
) 0.30
-
x
0.30 -
x
K a for HCOOH = 1.8 x 10 -4 [H + ] [HCOO ] K a = [HCOOH] H + 0.00
+
x x
(
aq
) + HCOO x = 1.038 X 10 -4 pH = 3.98
0.52
+
x
0.52 +
x
(
aq
) 16.2
OR…… Use the Henderson-Hasselbach equation
Consider mixture of salt NaA and weak acid HA.
NaA HA ( (
s aq
) ) Na + (
aq
) + A (
aq
) H + (
aq
) + A (
aq
)
K a
= [H + ][A [HA] ] p
K a
= -log
K a
[H + ] =
K a
[HA] [A ] -log [H + ] = -log
K a
- log [HA] [A ] [A ] -log [H + ] = -log
K a
+ log [HA] pH = p
K a
[A ] + log [HA] Henderson-Hasselbach equation pH = p
K a
+ log [conjugate base] [acid] 16.2
What is the pH of a solution containing 0.30
M
HCOOH and 0.52
M
HCOOK?
Mixture of weak acid and conjugate base!
Initial (
M
) Change (
M
) Equilibrium (
M
)
Common ion effect
0.30 –
x
0.30
0.52 +
x
0.52
HCOOH p
K a
= 3.77
HCOOH (
aq
) 0.30
-
x
H 0.00
+
x
+ (
aq
) + HCOO 0.52
+
x
0.30 -
x x
0.52 +
x
[HCOO ] pH = p
K a
+ log [HCOOH] [0.52] pH = 3.77 + log [0.30] = 4.01
(
aq
) 16.2
HCl H + + Cl HCl + CH 3 COO CH 3 COOH + Cl 16.3
K b = Calculate the pH of the 0.30
M
NH 3 /0.36
M
NH 4 Cl 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?
Initial Change End NH 3 (
aq
) + H 2 O (
l
) [NH 4 + ] [OH ] [NH 3 ] 0.30
- x = 1.8 X 10 -5 0.30 - x NH 4 + (
aq
) + OH (
aq
) 0.36
+ x 0.36 + x 0 + x x 1.8 X 10 -5 = (.36 + x)(x) (.30 – x) 1.8 X 10 -5 0.36x
x = 1.5 X 10 -5 0.30 pOH = 4.82
pH= 9.18
16.3
Calculate the pH of the 0.30
M
NH 3 /0.36
M
NH 4 Cl 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 NH 4 + 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 NH 3 start (M) end (M) 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M
0.29
0.01
0.28
0.0
0.24
NH 4 + (
aq
) + OH (
aq
) H 2 O (
l
) + NH 3 (
aq
) 0.25
K a [H + ] [NH 3 ] = = 5.6 X 10 -10 [NH 4 + ] [H + ] 0.25
= 5.6 X 10 -10 0.28
[H + ] = 6.27 X 10 pH = 9.20
-10 16.3
Calculate the pH of the 0.30
M
NH 3 /0.36
M
NH 4 Cl 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?
NH 4 + (
aq
) H + (
aq
) + NH 3 (
aq
) pH = p
K a
+ log [NH 3 ] [NH 4 + ] p
K a
= 9.25
pH = 9.25 + log final volume = 80.0 mL + 20.0 mL = 100 mL [0.30] [0.36] = 9.17
start (M) end (M) 0.29
NH 4 + (
aq
) + OH (
aq
) H 2 O (
l
) + NH 3 (
aq
) 0.28
0.01
0.0
0.24
0.25
pH = 9.25 + log [0.25] [0.28] = 9.20
16.3
Chemistry In Action:
Maintaining the pH of Blood 16.3
Titrations
In a
titration
a solution of accurately known concentration is 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
Strong Acid-Strong Base Titrations
NaOH (
aq
) + HCl (
aq
) H 2 O (
l
) + NaCl (
aq
) OH (
aq
) + H + (
aq
) H 2 O (
l
) 100% ionization!
No equilibrium 16.4
Weak Acid-Strong Base Titrations
CH 3 COOH (
aq
) + NaOH (
aq
) CH 3 COOH (
aq
) + OH (
aq
) CH 3 COONa (
aq
) + H 2 O (
l
) CH 3 COO (
aq
) + H 2 O (
l
) At equivalence point (pH > 7): CH 3 COO (
aq
) + H 2 O (
l
) OH (
aq
) + CH 3 COOH (
aq
) 16.4
Strong Acid-Weak Base Titrations
HCl (
aq
) + NH 3 (
aq
) H + (
aq
) + NH 3 (
aq
) NH NH 4 4 Cl Cl At equivalence point (pH < 7): ( (
aq aq
) ) NH 4 + (
aq
) + H 2 O (
l
) NH 3 (
aq
) + H + (
aq
) 16.4
Acid-Base Indicators 16.5
pH 16.5
The titration curve of a strong acid with a strong base.
16.5
Which indicator(s) would you use for a titration of HNO 2 with KOH ?
Weak acid titrated with strong base.
At equivalence point, will have conjugate base of weak acid.
At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5
Finding 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 K b for conjugate base equilibrium • Strong Acid vs. Weak Base – Base is neutralized; Need K a equilibrium for conjugate acid • Weak Acid vs. Weak Base – Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7
Exactly 100 mL of 0.10 a 0.10
M M
HNO 2 are titrated with 100 mL of NaOH solution. What is the pH at the equivalence point ?
start (moles) end (moles) 0.01
HNO 0.0
2 (
aq
) 0.01
+ OH (
aq
) 0.0
NO 2 0.01
(
aq
) + H 2 O (
l
) Final volume = 200 mL NO 2 (
aq
) + H 2 O (
l
) [NO 2 ] = 0.01
0.200
= 0.05
M
OH (
aq
) + HNO 2 (
aq
) Initial (
M
) Change (
M
) 0.05
-
x
Equilibrium (
M
) 0.05 -
x K b
= [OH ][HNO 2 ] [NO 2 ] = 0.05 –
x
0.05
x
x
2 0.05-
x
= 2.2 x 10 1.05 x 10 -6 = [OH ] -11 0.00
+
x x
pOH = 5.98
pH = 14 0.00
+
x x
– pOH = 8.02
Complex Ion Equilibria and Solubility A
complex ion
is an ion containing a central metal cation bonded to one or more molecules or ions.
Co 2+ (
aq
) + 4Cl (
aq
) CoCl 4 2 (
aq
) Co(H 2 O) 6 2+ CoCl 2 4 16.10
16.10
Complex 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(NH 3 ) 4 2+ test for Cu 2+ (ammonia is used as a ions), and Ag(NH 3 ) 2 + .
Memorize
the common ligands.
Ligands
H 2 O NH 3 OH Cl Br CN SCN-
Common Ligands
Names used in the ion
aqua ammine hydroxy chloro bromo cyano thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)
Names
• Names: ligand first, then cation Examples: – tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ – diamminesilver(I) ion: Ag(NH 3 ) 2 + .
– tetrahydroxyzinc(II) ion: Zn(OH) 4 2 • The charge is the sum of the parts (2+) + 4(-1)= -2.
When Complexes Form
• Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H can form complex ions. 2 O) • The odd complex ion, FeSCN 6 2+, 3+ • Transitional metals, such as Iron, Zinc and Chromium, shows up once in a while • Acid-base reactions may change NH 3 into NH complex ion is formed. For example, Cu 2+ 4 + (or vice versa) which will alter its ability to act as a ligand.
• Visually, a precipitate may go back into solution as a + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2 . With excess ammonia, the complex, Cu(NH 3 ) 4 2+ , forms.
• Keywords such as "
excess
" and "
concentrated
" of any solution
may
indicate complex ions. AgNO HCl, the complex ion, AgCl 2 3 + HCl forms the white precipitate, AgCl. With excess, concentrated , forms and the solution clears.
Coordination 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.
Some Coordination Complexes
molecular formula
Ag(NH 3 ) 2 + [Zn(CN) 4 ] 2 [Ni(CN) 4 ] 2-
Lewis base/ligand
NH 3 CN CN-
Lewis acid donor atom
Ag + N Zn Ni 2+ 2+ C C [PtCl 6 ] 2 Cl [Ni(NH 3 ) 6 ] 2+ NH 3 Pt 4+ Ni 2+ Cl N 4 6 6
coordination number
2 4