Chapter 16: Applications of
Aqueous Equilibria
Renee Y. Becker
Valencia Community College
1
Acid-base Neutralization Reaction
Acid-base neutralization reaction
1.
Products are water and a salt
2.
Four types
a)
Strong Acid-Strong Base
b) Weak Acid-Strong Base
c)
Strong Acid-Weak Base
d) Weak Acid-Weak Base
2
Strong Acid-Strong Base
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
1. Because HCl is a strong acid and NaOH is a
strong base they are both strong electrolytes, they
dissociate nearly 100%
2.
3.
Complete ionic equation
H+ + Cl- + Na+ + OH-  H2O(l) + Na+ + Cl-
Net ionic equation
H3O+ + OH-  2 H2O(l)
3
Strong Acid-Strong Base
4.
Kn = 1 / [H3O+] [OH-] = 1 / Kw = 1 / 1 x 10-14 =
1 x 1014
a)
5.
Kn is the equilibrium constant with respect
to neutralization
1 x 1014 is a large # and means that
for a strong acid-strong base reaction
proceeds essentially 100% to
completion
pH = 7
4
Weak Acid-Strong Base
HF(aq) + NaOH(aq)  H2O(l) + NaF
1. Because HF is a weak acid-weak electrolyte it does
not dissociate well and will not be ionized on the
reactant side
2. Complete ionic equation
HF + Na+ + OH-  H2O + Na+ + F-
3. Net Ionic equation
HF + OH-  H2O + F5
Weak Acid-Strong Base
4. To obtain the equilibrium constant, Kn we
need to multiply known equilibrium constants
for reactions that add to give the net ionic
equation for the neutralization
HF(aq) + H2O(l)  H3O+(aq) + F-(aq)
H3O+(aq) + OH-  2 H2O(l)
Ka = 3.5 x 10-4
1/Kw = 1 x 1014
Net: HF + OH-  H2O(l) + F-
Kn = Ka (1/Kw)
= 3.5 x 10-4 (1 x 1014) = 3.5 x 1010
6
Weak Acid-Strong Base
For any weak acid-strong base reaction
1. Kn = Ka (1/Kw)
2. 100% completion because of OH- strong
affinity for protons
3. The pH will be > 7, due to basicity of
conjugate base
7
Strong Acid-Weak Base
NH3(aq) + HCl(aq)  NH4+(aq) + Cl1. A strong acid is completely dissociated into
H3O+ and A- ions
2. Neutralization reaction is a proton transfer
3. Net ionic equation
H3O+ + NH3  H2O + NH4+
8
Strong Acid-Weak Base
5. Just as before we can obtain the equilibrium
constant for the neutralization reaction by
multiplying known equilibrium constants for
reactions that add to give the net ionic equation
NH3 + H2O  NH4+ + OHH3O+ + OH-  2 H2O
Kb = 1.8 x 10-5
1/Kw = 1 x 1014
Net: H3O+ + NH3  H2O + NH4+
Kn =Kb(1/Kw)
= 1.8 x 109
9
Strong Acid-Weak Base
For any strong acid-weak base reaction
1. Kn = Kb(1/Kw)
2. 100% completion because H3O+ is a
powerful proton donor
3. The pH will be < 7, due to the conjugate
acid
10
Weak Acid-Weak Base
CH3CO2H + NH3  NH4+ + CH3CO21.
Both acid and base are largely undissociated
2. Neutralization reaction is a proton transfer
from the weak acid to the weak base
3. The equilibrium constant can be obtained by
adding equations for the acid dissociation, the
base protonation and the reverse of the
dissociation of water
11
Weak Acid-Weak Base
CH3CO2H + H2O  H3O+ + CH3CO2NH3 + H2O  NH4+ + OHH3O+ + OH-  2 H2O
Ka = 1.8 x 10-5
Kb = 1.8 x 10-5
1/Kw = 1 x 1014
Net: CH3CO2H + NH3  NH4+ + CH3CO2Kn = Ka(Kb)(1/Kw) = 3.2 x 104
For any weak acid-weak base reaction
1.
Kn = Ka(Kb)(1/Kw)
Kn is smaller so the reaction does not go to completion
12
Example 1: The Common-Ion Effect
Calculate the pH of a solution prepared by
dissolving .10 mol acetic acid and .10 mol
sodium acetate in water, and then diluting
the solution to a volume of 1.00 L
13
Buffer Solutions
1. A weak acid and it’s conjugate base
2. Resist drastic changes in pH
3. If a small amount of OH- is added, the
pH increases but not by much because
the acid component of the buffer
neutralizes the OH4. If a small amount of H3O+ is added the
pH decreases but not by much because
the conjugate base in the buffer neutralizes the
added H3O+
14
Buffer Solutions
5. Examples
a) CH3CO2H + CH3CO2b) HF + Fc) NH4+ + NH3
d) H2PO4- + HPO42-
6. Very important in biological systems
(blood is a buffer) (H2CO3 + HCO3-)
15
Example 2:
Addition of OH- to a buffer, we add 0.01 mol
solid NaOH to 1.00 L of a 0.10 M acetic
acid-0.10 M sodium acetate solution. What
is the pH?
Because solutions involving a strong acid or
base go to nearly 100% completion we must
account for the neutralization before we can
calculate H3O+
16
Example 3:
Addition of H3O+ to a buffer, we add 0.01 mol
HCl to 1.00 L of a 0.10 M acetic acid-0.10
M sodium acetate solution. What is the pH?
17
Buffer capacity
1. Measure of the amount of acid or base that a
solution can absorb without a significant change
in pH
2. Measure of how little the pH changes with the
addition of a given amount of acid or base
3. Depends on how many moles of weak acid and
conjugate base are present
4. The more concentrated the solution (acid & conj.
base), the greater the buffer capacity
5. The greater the volume (acid and conj. Base), the
greater the buffer capacity
18
Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation
pH = pKa + log [base]/[acid]
1.
Tells us how the pH affects the % dissociation of a weak acid
2. Also tells us how to prepare a buffer solution with a given pH.
a) Pick a weak acid that has a pKa close to the desired pH
b) Adjust the [base]/[acid] ratio to the value specified by
the HH equation
The pKa of the weak acid should be within ± 1 pH unit of desired
pH
19
Example 4:
I want to prepare a buffer solution with a pH of 7.00 and
one of a pH of 9.00 which of the following pairs of
weak acid-conj. bases should I use?
CH3CO2H + CH3CO2-
Ka = 1.8 x 10-5
pKa = 4.74
HF + F-
Ka = 3.5 x 10-4
pKa = 3.46
NH4+ + NH3
Ka = 5.56 x 10-10
pKa = 9.25
H2PO4- + HPO42-
Ka = 6.2 x 10-8
pKa = 7.21
20
Example 5:
Use the HH equation to calculate the pH of a
buffer solution prepared by mixing equal
volumes of 0.20 M NaHCO3 and 0.10 M
Na2CO3 (Ka = 5.6 x 10-11 for HCO3-)
21
Example 6:
Give a recipe for preparing a NaHCO3-Na2CO3
buffer solution that has a
pH = 10.40
22
pH Titration Curves
1. A plot of the pH of the solution as a function of the
volume of the added titrant
2. A solution of a known concentration of base or
acid is added slowly from a buret to a second
solution with an unknown concentration of acid or
base
3. Progress is monitored with a pH meter or by color
of indicator
4. Equivalence point is the point at which
stoichiometrically equivalent quantities of acid and
base have been mixed together
23
pH Titration Curves
5.
Endpoint is when the color of the acid-base indicator
changes
6.
There are four important types of titration curves
a)
b)
c)
d)
Strong Acid-Strong Base
Weak Acid-Strong Base
Weak Base-Strong Acid
Polyprotic Acid-Strong Base
We will only be calculating for Strong acid-strong base titrations
but you are responsible to be able to recognize and label the
titration curves for all
24
Strong Acid-Strong Base
25
Weak Acid-Strong Base
26
Strong Acid-Weak Base
27
Polyprotic Acid-Strong Base
28
Strong Acid-Strong Base Titrations
Titration of a strong acid (50 mL of a 0.02 M HCl) by a
strong base (0.030 M NaOH)
There are four main calculations for this type of titration
1.
2.
3.
4.
Before any Base has been added
Before the equivalence point
At the equivalence point
After the equivalence point
29
1. Before any base has been added
1. Since HCl is a strong acid the initial
concentration of H3O+ = initial molarity
= 0.02 M
pH = -log[0.02] = 1.70
30
2. Before the equivalence point
Let’s say we have added 10 mL of 0.03 M NaOH
The added OH- ions will neutralize some of the H3O+ ions
Moles of H3O+ ions = M*V = 0.02 * .05 L = .001 mol H3O+
Moles of OH- ions = M*V = 0.03 * .01L = .0003 moles
Molarity of H3O+ after addition of NaOH
=[Moles H3O+ - moles OH-] /total volume,L
M = (.001 - .0003) / (.05 L + .01 L) = 1.2 x 10-2 M = [H3O+]
pH = - log[1.2 x 10-2] = 1.92
31
3. At the equivalence point
At the equivalence point the pH = 7
To find the volume of NaOH would give you
the equivalence point use equation:
M*V = M*V
.02(50 mL) = .03(x mL)
33.3 mL NaOH
32
4. After the equivalence point
After the equivalence point you have neutralized all
of the acid and now you have excess base, 45 mL
NaOH
Find the moles of acid & base
Moles of acid = M*V = .001 Moles H3O+
Moles of base = M*V = .03 * .045 L = .00135
moles OH33
4. After the equivalence point
[OH-] = (moles of base – moles of acid)/ total volume, L
= (.00135 - .001) / (.05 L + .045 L) = 3.7 x 10-3 M
Kw = [H3O+] [OH-]
[H3O+] = 1 x 10-14 / 3.7 x 10-3 = 2.7 x 10-12
pH = -log [ 2.7 x 10-12] = 11.57
34
Solubility Equilibria
Solubility Product Constant, Ksp
Same as Kc, Kp, Kw, Ka, & Kb Prod / reactant
Coefficients are exponents, omit solids and pure
liquids
CaF2(s)  Ca2+(aq) + 2 F-(aq)
Ksp = [Ca2+][F-]2
35
Example 7: Measuring Ksp and Calculating Solubility from Ksp
A saturated solution of Ca3(PO4)2 has [Ca2+] =
2.01 x 10-8 M and
[PO43-] = 1.6 x 10-5 M.
Calculate Ksp for Ca3(PO4)2
36
Factors that Affect Solubility
1. The Common ion effect
MgF2(s)  Mg2+(aq) + 2 F-(aq)
If we try dissolve this in a aqueous solution of NaF the
equilibrium will shift to the left. This will make MgF2 less
soluble
2. Formation of Complex ions
Complex ion: An ion that contains a metal cation bonded to
one or more small molecules or ions, NH3, CN- or OHAgCl(s)  Ag+ + ClAg+ + 2 NH3  Ag(NH3)2+
Ammonia shifts the equilibrium to the right by tying up Ag+
ion in the form of a complex ion
37
Factors that Affect Solubility
3. The pH of the solution
a) An ionic compound that contains a basic anion becomes
more soluble as the acidity of the solution increases
CaCO3(s)  Ca2+ + CO32H3O+ + CO32-  HCO3- + H2O
Net: CaCO3(s) + H3O+  Ca2+ + HCO3- + H2O
Solubility of calcium carbonate increases as the pH decreases
because the CO32- ions combine with protons to give HCO3- ions.
As CO32- ions are removed from the solution the equilibrium
shifts to the right to replenish the carbonate
PH has no effect on the solubility of salts that contain anions of
strong acids because these anions are not protonated by H3O+38
Precipitation of Ionic Compounds
Ion Product (IP)
Same as Ksp but at some time, t, snapshot like Qc,
reaction quotient
CaF2(s)  Ca2+ + 2 FIP = [Ca2+][F-]2
If IP > Ksp solution is supersaturated and precipitation will occur
If IP = Ksp the solution is saturated and equilibrium exists
If IP< Ksp the solution is unsaturated and ppt will not occur
39
Example 8:
Will a precipitate form on mixing equal volumes of
the following solutions?
a) 3.0 x 10-3 M BaCl2 and 2.0 x 10-3 M Na2CO3
(Ksp = 2.6 x 10-9 for BaCO3)
b) 1.0 x 10-5 M Ba(NO3)2 and 4.0 x 10-5 M Na2CO3
40