1. CHAPTER 15
AP CHEMISTRY

2. COMMON ION EFFECT
If you have HC2H3O2(aq) <=> H+ (aq) + C2H3O2-(aq)
What would happen if you added the salt NaC2H3O2?
Addition of C2H3O2- would cause the system to shift to the left
This is called the COMMON-ION EFFECT
Suppose that we add 8.20 g, or mol, of sodium acetate, NaC2H3O2, to 1 L of a M of acetic acid, HC2H3O2. What is the pH of the resultant solution?
HC2H3O2(aq) <=> H+ (aq) + C2H3O2-(aq)
Initial
Change X X X
Equilibrium X X X
X(0.100+X)/( X)
0.100X/ = 1.8 X X= 1.8 X = [H+]
pH = 4.74
Page 699 example

3. BUFFER
Prepared by adding both the weak acid HB and its conjugate base B- to water
[H+] = ka[HB] = ka nHB
[B-] nB-
Effect of adding strong acids or bases to a buffer
HB(aq) + OH-(aq) ---> H2O(l) + B-(aq)
B-(aq) + H+(aq) ---> HB(aq)
H+ and OH- ions are consumed so the pH is not affected
Buffered solutions are most effective when concentrations of H+ and base B- are about the same

4. CONTINUED
Henderson-Hasselbach equation is used in biology to calculate the pH of the buffer
pH = pka+ log [base]
[acid]
What is the pH of a buffer mixture composed of 0.12 M lactic acid (HC3H5O3, ka = 1.4 X 10-4 ) and 0.10 M sodium lactate.
HC3H5O3 <=> H+ + C3H5O3-
-log(1.4 X 10-4 )= 3.85
Log(C3H5O3-/HC3H5O3) = log(0.10/0.12) =
pH = = 3.77
How many moles of NH4Cl must be added to 1.0L of 0.10 M NH3 to form a buffer whose pH is 9.00?
NH3 + H2O <=> NH4+ + OH-
1.8 X 10-5 = [NH4+] [OH-] = 5 for pOH
[NH3]
1.8 X 10-5 [NH3]/[OH-] = [NH4+]
1.8 X 10-5(0.10)/(1 X 10-5)
0.18mol

5. ACID-BASE TITRATION Read and study pages 701-703,706-708, 709-714
Titration curve
Graph of pH to concentration
strong acid - strong base page 715
H+(aq) + OH-(aq) <=> H2O(l) k = 1/kw = 1/(1 X 10-14)
pH = 7, change rapidly as you near the end point
When mL of M HCl is titrated to M NaOH, the pH increases, find the pH of the solution after the following volumes of M NaOH has been added a) mL b) mL
nH+ = L X 1.000molH+/1L = mol H+
nOH- = L X 1.000mol OH-/1L = mol OH-
The OH- and H+ will form water but in this problem there would still be H+ left over
n H+ excess = mol
The volume has changed it now has 100 mL, so the H+ concentration would be 1 X 10-5mol H+/0.100L = 1 X 10-4 M H+
pH = 4

6. CONTINUED Weak acid-strong base
HA(aq) + OH-(aq) <=> A-(aq) + H2O(l) k =1/kb
pH > 7 at end point, change occurs slowly at end point
Strong acid-weak base
H+ (aq) + A-(aq) <=> HA(aq) k = 1/ka
pH < 7, at end point, change occurs slowly at end point
Titration of polyproptic acids
When neutralization steps are separated the substance will show a titration curve with multiple equivalence points

7. INDICATORS HIn(aq) <=> H+(aq) + In-(aq) In = indicator
HIn and In- have different colors, the wavelengths that bounce back are different when hydrogen is attached
Color of a solution depends on the concentration ratio
[HIn] = [H+]
[In- ] ka
Endpoint occurs when [H+] = ka
About pH 5 for methyl red, pH 7 for bromothymol blue, and pH 9 for phenolphthalein
Read pages 732,

8. SOLUBILITY EQUILIBRIA
If Q(ion product) is compared with the ksp, the following can be said
Q > ksp ppt occurs until Q = ksp
Q = ksp equilibrium (saturation has occured)
Q < ksp solid dissolves until Q = ksp
AgCl(s) <=> Ag+ (aq) + Cl- (aq)
Slightly soluble ksp found on page 736
ksp = [Ag+][Cl-]
Will a precipitate form when 0.100L of 3.0 X 10-3M Pb(NO3)2 is added to 0.400L of 5.0 X 10-3M Na2SO4?
Find moles first
3.0 X 10-4mol Pb(NO3)2 = Pb2+
2.0 X 10-3mol Na2SO4 = SO42-
Add the two volumes together to get new volume of 0.500L
4.0 X 10-3M SO42- and 6.0 X 10-4M Pb(NO3)2
Q = (4.0 X 10-3)(6.0 X 10-4) = 2.4 X 10-6
If the ksp = 1.6 X 10-8 then a ppt would form

9. CONTINUED
What is [Pb2+] in a solution at equilibrium with PbCl2 (ksp = 1.7 x 10-5) if [Cl-] = 0.020M?
PbCl2 <=> Pb2+ + 2Cl-
ksp= [Pb2+][Cl-]2
0.042M Pb2+
the solubility of slightly soluble salts containing basic anions with increase as the [H+] increases
Looking at the solubility of a salt
In pure water it is the same as the reaction
PbCl2(s) <=> Pb2+(aq) + 2Cl-(aq)
[Pb2+] = s [Cl-] = 2s
[s][2s]2 = 4s3 = 1.7 x 10-5
s = 1.6 x 10-2M Pb2+ and 2(1.6 x 10-2) M Cl-

10. CONTINUED In a solution containing a common-ion 0.100 M HCl
[Pb2+] = s [Cl-] = 2s = 0.10
s(0.10)2 = 1.7 X 10-5
s = 1.7 X 10-3M Pb2+
Read pages 739, 742,
Complex ions
Metal ions with Lewis bases bonded to it page 754
Amphoterism
Behavior of water molecules around a metal ion
Read pages , 757
Read chapters 18, 19, and 22 notes are due the first day of fourth term