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10/15/2014
1
More Acid and Base Chemistry
Common-ion effect In the last chapter, we calculated the [H3O
+] of a 0.123M HClO as 6.02x10-5M. The percent dissociation for this solution would be:
0.0489% x1000.123
6.02x10 5
But what would happen if we place 0.123moles of HClO in a 1.000L container that already has 0.100moles of the conjugate base NaClO. Will this change anything?
I
E C
nitial
hange
quilibirum
0.123M 0M
-x +x
0.123M - x x
HClO(aq)+ H2O(l)⇌ H3O
+(aq) + ClO-(aq)
0.100M
+x
0.100+x
[HClO]
]][ClOO[H3a
K
x-0.123
x)(x)(0.1002.95x10 8
Hypoclorite ion already in solution before acid added
Assume these “x’s” will be very small
0.123
(x)(0.100)2.95x10 8
0.100x3.63x10 9
83.63x10x
0.0000295% x1000.123
3.63x10 8
Note the change in percent dissociation. The percent dissociation decreased because there
were already ClO- ions in solution. The reaction did not proceed as far toward the reactants to
reach equilibrium. This is the common-ion effect. Common ions in solution decrease
dissociation of weak acids and bases.
10/15/2014
2
Common-ion effect What would happen if the solution had a pH of 1.000 before adding 0.123M HClO?
0.0489% x1000.123
6.02x10 5
I
E C
nitial
hange
quilibirum
0.123M 0.100M
-x +x
0.123M - x x
HClO(aq)+ H2O(l)⇌ H3O
+(aq) + ClO-(aq)
0M
+x
0.100+x [HClO]
]][ClOO[H3a
K
x-0.123
x)(x)(0.1002.95x10 8
Hydromiun ion already in solution before acid added
Assume these “x’s” will be very small
0.123
(0.100)(x)2.95x10 8
0.100x3.63x10 9
83.63x10x
0.0000295% x1000.123
3.63x10 8
Both products of dissociation of HClO, ClO- and H3O
+, caused a decrease in the dissociation of the acid. This is the common
ion effect.
How would ionization be affected if both H3O+
and ClO- were 0.100M initially? Hint: think about Qa
Buffers Buffers have the ability to stabilize pH, even with the addition of acids or bases. Buffers are
composed of weak conjugate bases and acids in equilibrium.
Add HCl
Plain ole water
pH = 7 pH = 0
Cl- H+
ΔpH = -7
NH4+
NH3
Add HCl
NH3 /NH4+ buffer solution
NH4+
NH4+
NH4+
NH3
NH3
NH3
NH4+
NH3 NH4+
NH4+
NH4+
NH3
NH3
NH4+
pH = 9.25
Cl-
pH = 9.03
ΔpH = -0.22
HCl + H2O → H3O+ + Cl-
HCl + NH3 → NH4+ + Cl-
HCl adds H3O+ to the solution with its
reaction with water.
HCl reacts with NH3 to give more NH4+
ions. H3O+ can only be added to the
solution through the NH3/NH4+
equilibrium
Through the reaction, 1 mol of H3O+ is
added to the water.
Through the reaction, only 3.7x10-10 mol of H3O
+ is added to the water.
10/15/2014
3
Calculating pH of buffer solutions
Initial Change Equilibrium
5.0M 0M
-x +x
5.0M - x x
NH4+(aq)+ H2O(l)⇌
H3O+(aq) + NH3(aq)
3.0M
+x
3.0M+x
Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH3 of and 5.0 moles of NH4+. Ka for NH4
+ is 5.8x10-10.
][NH
]][NHO[H
4
33a
K x-5.0
x)(x)(3.05.8x10 10
5.0
(x)(3.0)5.8x10 10
109.7x10x
pH = -log[H3O+] = -log(9.7x10-10) = 9.01
Because the [NH3] +
[NH4+] =
8.0M, this is an 8.0 M
buffer
acid] [conjugate
base] [conjugatelogppH a K
Henderson – Hasslebalch Formula •Change in acid/base concentrations by dissociation negligible. [acid/base]initial ≈ [acid/base]equilibrium
But instead of another I.C.E. table, we can use…
5.0M
3.0Mlog9.24pH pH=9.02
Let’s do the same calculation above, but using H-H
Adding Strong Acid/Base to a Buffer Solution
What happens when we add 0.50L of 2.00M HCl to the solution
First, we need to determine how adding HCl changes the NH3/NH4+ buffer. To do this, we will use a change table. It is best
to use moles not M in this table (you will see why later).
Let’s take our first buffer problem from the last page: “Calculate the pH of a buffer solution in a 1.0L container with 3.0 moles NH3 of and 5.0 moles of NH4
+. Ka for NH4+ is 5.8x10-10.”
H+ (aq) + NH3(aq) → NH4
+(aq) Note that the HCl (acid) will react with the NH3 (base) in the solution.
1.0mole 3.0moles 5.0moles The strong acid or base will always react completely.
Remember, 100% dissociation.
0mole 2.0moles 6.0moles
Before:
-1.0mole -1.0mole +1.0mole
After:
Change:
acid] [conjugate
base] [conjugatelogppH a K
6.0mol
2.0mollog9.24pH
pKa = -log Ka = -log(5.8x10-10) = 9.24
Since both NH3 and NH4+ are in the same
volume, we can use moles instead of concentration. If you divide both HA and A- by the same volume, the ratio stays the
same, pH=8.76
10/15/2014
4
A More Realistic Buffer Solution Problem Calculate the pH of a buffer solution in a 100.mL container with 0.100M of each component of a benzoic acid /
sodium benzoate buffer. Ka of benzoic acid is 6.4x10-5.
acid] [conjugate
base] [conjugatelogppH a K
0.100M
0.100Mlog19.4pH 19.4pH
What is the pH after adding 10.00mL of 0.500M NaOH?
OH- (aq) + HA(aq) → A- (aq)
0.00500mol 0.0100mol 0.0100mol
0mol 0.0150mol
Before:
After:
Change: -0.00500mol -0.00500mol +0.00500mol
0.00500mol
0.00500mol
0.0150mollog19.4pH 67.4pH
A-
A-
HA
A-
Since there are still appreciable amounts of both HA and A-, it still a buffer solution.
Thus, H-H can be used
Buffer Solution
10/15/2014
5
A-
A-
HA
A-
A-
A- A-
A-
A- A- OH-
HA H+ HA Na
+ Cl-
Cl- Na+
A More Realistic Buffer Solution Problem What is the pH after adding a total of 20.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer
solution? Ka 6.4x10-5.
OH- (aq) + HA(aq) → A- (aq)
0.0100mol 0.0100mol 0.0100mol
0mol 0.0200mol
Before:
After:
Change: -0.0100mol -0.0100mol +0.0100mol
0mol
Is this even a buffer solution any more? NO! Only A-! Then I cannot use H-H!
A-
A- A-
A-
Weak Base Solution
0.167M 0M
-x +x
F- (aq)+ H2O(l)⇌ OH-(aq) + HF(aq)
0M
+x
Initial Change
00.167M - x x x Equilibrium
--
A M167.00.1000L L02000.0
A mol0200.0
102
10x6.10.167
x 610x2.5x
6- 10x2.5x ][OH
)012.5log(14pOH41 pH -6x
72.8 pH
NaOH mol0100.0NaOH L 1
NaOH 0.500molNaOH L0200.0 HA mol0100.0HA L 1
HA 0.100molHA L1000.0
10/15/2014
6
A More Realistic Buffer Solution Problem What is the pH after adding a total of 30.00mL of 0.500M NaOH to the original (100.0mL 0.100M each) buffer
solution?
OH- (aq) + HA(aq) → A- (aq)
0.0150mol 0.0100mol 0.0100mol
0.00500mol 0.0200mol
Before:
After:
Change: -0.0100mol -0.0100mol +0.0100mol
0mol
Which base should have a greater effect on pH?
A-
A- A-
A-
Strong Base Solution
OH-
--
OH M0385.00.03000 L1000.0
OH mol00500.0
1.4150.0385 -logpOH
1.415000.41pH
12.585pH
How a buffer reactions to addition of strong base or strong acid
adding base
adding acid
Mol of conjugate base in 1mol total buffer
Note that our theoretical buffer has a Ka of 1x10-7 and thus a pKa of 7. acid] [conjugate
base] [conjugatelogppH a K
Assume we have a buffer with 1M total buffer strength. Let’s start with 0.50M of each HA and A-.
10/15/2014
7
When does a buffer stop “buffering”?
10:1
1:10
1:1 base:acid
Mol of conjugate base in 1mol total buffer
No longer buffering
A good rule of thumb: A buffer is relatively effective ± 1 pH point of the acid’s pKa. After that point the buffer is exghausted and can no longer
buffer in that direction.
Strong Acid / Strong Base Titration problems (determining the pH at different points of a titration) are very similar to buffer questions. Use change tables, and see what’s left.
Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.
OH- (aq) + H+ (aq) → H2O
7.50x10-5 mol 1.00x10-4 mol Before:
After:
Change: -7.50x10-5 mol -7.50x10-5 mol
0mol 2.5x10-5 mol H+
H+ Cl-
Cl-
Cl- Cl- Na+ Na+
H M0022.00.00150L L01000.0
H mol.5x102 -5
2M)-log(0.002pH 66.2pH Strong Acid Solution
Determine the pH during of10.00mL of 0.0100M HCl before
the titration. 0M)-log(0.010pH
000.2pH H+
H+
H+
H+
Cl-
Cl- Cl- Cl-
What type of solution is this?
Strong Acid Solution
10/15/2014
8
Strong Acid / Strong Base Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.
OH- (aq) + H+ (aq) → H2O
1.0x10-4 mol Before:
After:
Change:
0mol
1.0x10-4 mol
-1.0x10-4 mol
0mol
-1.0x10-4 mol Cl-
Cl-
Cl-
Cl- Na+ Na
+
Na+ Na+
00.7pH
Neutral Solution
Strong Acid / Strong Base
OH- (aq) + H+ (aq) → H2O
1.00x10-4 mol Before:
After:
Change:
2.5x10-5 mol
1.25x10-4 mol
-1.00x10-4 mol
0mol
-1.00x10-4 mol
Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HCl.
---5
OH M0020.00.00250L L01000.0
OH mol.5x102
70.10M)-log(0.002pOH
70.100.14pH
30.12pH
Cl- Cl-
Cl-
Cl-
Na+ Na+
Na+
Na+ OH-
Strong Base Solution
10/15/2014
9
Acid-Base Titrations Strong Acid/Strong Base
Adding Base to Acid
Vol of NaOH added (mL)
Equivalence Point: The amount of base added equals the amount of acid present before the titration.
The acid/base reaction is complete. For strong
acid/strong base titration the pH is 7.
After
During Before
H+
H+
H+
H+
Cl-
Cl- Cl- Cl-
Before Titration
H+
H+ Cl-
Cl-
Cl- Cl-
During Titration
Na+ Na+
add NaOH
Cl-
Cl-
Cl-
Cl-
Equivalence Point
Na+ Na+
Na+ Na+
Cl- Cl-
Cl-
Cl-
Past Equivalence Point
Na+ Na+
Na+
Na+ OH- add
NaOH add
NaOH
Acid-Base Titrations Weak Acid/Strong Base
Equivalence Point: The amount of base added equals the amount of acid present before the titration.
The acid/base reaction is complete. For strong
acid/strong base titration the pH is >7.
After
During (buffer zone)
Before
Vol of NaOH added (mL)
½ volume of equivalence point pH = pKa
[HA] = [A-]
Eq. Pt.
HF
Before Titration During Titration
add NaOH
Equivalence Point Past Equivalence
Point
add NaOH
add NaOH
F- F- Na+
Na+ HF HF
HF
HF HF F-
F- Na+
Na+ F- F-
Na+
Na+ F-
F- Na+
Na+
F- F-
Na+
Na+
OH- Na+
10/15/2014
10
Acid-Base Titrations
Adding Acid to Base
Equivalence Point: The amount of acid added equals the amount of base present before the titration.
The acid/base reaction is complete. For strong
acid/strong base titration the pH is 7.
Strong Acid/Strong Base
After
During
Before
Na+ Na+
Na+ Na+ OH- OH- OH-
OH- add HCl
Na+ Na+
Na+ Na+ OH- OH-
Cl-
Cl-
add HCl
Cl-
Cl-
Cl-
Cl-
Equivalence Point
Na+ Na+
Na+ Na+
Before Titration During Titration
add HCl
Cl-
Cl-
Cl-
Cl- Na+ Na
+
Na+ Na+ Cl-
Past Equivalence Point
H+
Strong Acid/Weak Base
Acid-Base Titrations
After
During (buffer zone)
Before
Equivalence Point: The amount of acid added equals the amount of base present before the titration.
The acid/base reaction is complete. For strong
acid/weak base titration the pH is
10/15/2014
11
Strong Acid / Weak Base Titrations
Determine the pH during a titration after 1.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.
Determine the pH during of10.00mL of 0.0100M HF before the titration. Ka = 6.3x10-4
Determine the pH during a titration after 2.00mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.
Determine the pH during a titration after 2.50mL of 0.0500M NaOH has been added to 10.00mL of 0.0100M HF.
Polyprotic acid Titrations
Since there are multiple protons to remove from a polyprotic acid, there are multiple equivalence points – one for each proton.
The the pH of the equivalence point of an amphoteric species (on that can either
donate or accept an proton)
2
p p species amphotericof pH 1nn
aa
KK
H3PO4 H2PO4- HPO4
2- PO43-
⇌ ⇌ ⇌
pKa1 = 2.12 pKa2 = 7.20 pKa3 = 12.38
- 2- 3-
10/15/2014
12
Solubility Product – Ksp In CHM151, we talked about ionic compounds that were “insoluble” in water. In reality, those
compounds do dissolve, just a very. very small amount.
For example, based on our rules, AgBr is insoluble. How some very small amount does dissociate into ions. The represent this we have another K expression called Ksp.
The solubility of AgBr is represented by the equation:
Since solids do not appear in any K expression…
Ksp = [Ag+][Br-] and Ksp = 5.4x10
-13
Ag+ Br-
So, how do we determine the concentration of Ag+ and Br- in the solution?
0M
+x
x
0M
+x
x
Initial Change Equilibrium
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
132 5.4x10x
132 5.4x10x
M7.3x10x 7
[Ag+] = [Br-] = 7.3x10-7M This is called the “molar
solubility”
Solubility Product – Ksp (more complicated example)
Mg2+
OH-
OH-
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
How do we deal with sparingly soluble salts that do not have cations and anions in a 1:1 ratio?
Ksp = [Mg2+][OH- ]2
0M
+x
x
0M
+2x
2x
Initial Change Equilibrium
(x)(2x)2 =1.2x10-11
(x)(4x2) = 4x3 = 1.2x10-11
Ksp = 1.2x10-11
3
11
4
1.2x10x
x = 1.44x10-4M
[Mg2+] = x = 1.4x10-4 M
[OH-] = 2x = (2)1.44x10-4 M = 2.9x10-4M
10/15/2014
13
Solubility Product – Ksp
Ag+ Br-
Ag+ Br-
Add solid AgBr
Remember LeChatelier’s Principle holds that the addition of a solid should not shift equilibrium. Does this hold true? YES! Assuming there is already some solid in the solution, the concentration of ions does not change.
Ag+ Ag+
Ag+ Add water
Initially, as the water is added, the concentration of ions is decreased. Based on LeChatelier’s Principle the reaction should shift toward the product’s side. Thus, in the end, more AgBr is dissolved and the concentration of ions is the same.
Br-
Br- Br-
In all fours cases, the [Ag+] 7.3x10-7 M and [Br-] is 7.3x10-7 M
Factors that effect solubility of sparingly soluble salts
Common-ions
pH (acid/base reactions
Acids will increase the solubility of a sparingly soluble salts with a basic ion (OH-, F- ,etc.) through an acid/base reaction.
Fe2+ OH-
OH-
Add HCl
Fe2+ Cl-
Cl- Fe2+ Cl-
Cl-
OH- Fe2+ OH-
Equilibrium reestablished
Common ions will decrease solubility of a sparingly soluble salt. Same as common ion effect.
Formation of Complex ions (Lewis acid/base reactions
Cu2+
Add a lot of NH3
OH-
OH-
10/15/2014
14
Common ion of a sparingly soluble salt
Let’s say we have 0.50L of 0.50M NaOH. Into this solution we drop in a large chunk of Mg(OH)2. Determine the concentrations of all ions in solution at equilibrium.
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
0M
+x
x
0.50M
+2x
0.50+2x
Initial Change Equilibrium
Ksp = [Mg2+][OH- ]2
Ksp = 1.2x10-11
1.2x10-11 = (x)(0.50)2 = 0.25x
1111
4.8x100.25
1.2x10x
[Mg2+] = x = 4.8x10-11 M
[OH-] = 0.50+2x = 0.50 +(2)4.8x10-11 ≈ 0.50M OH-
Na+
OH- Na+
Add Mg(OH)2
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