1. Equilibrium Acids and bases Titrations Solubility
AP Review Big Idea 6
Equilibrium
Acids and bases
Titrations
Solubility

2. Equilibrium Constant – Keq
Most chemical reactions are reversible
aA + bB cC + dD
Keq = [C]c [D]d = products
[A]a [B]b reactants
Keq has no units
Solids and pure liquids are not included in equilibrium

3. HC2H3O2(aq) H+(aq) + C2H3O2-(aq)
[H+] [C2H3O2-] [HC2H3O2]
Keq =

4. 2H2S(g) + 3O2 (g) 2H2O (g) + 2SO2 (g)
You can make an equilibrium expression for gases too. Keq = [H2O]2 [SO2]2 [H2S]2 [O2]3 KP = P2H2O P2 SO2 P2H2S P3 O2
You can use equilibrium expression if you have concentrations of the gases
You can use equilibrium expression if you have pressures of the gases

5. Ksp Used for dissociation of a (very) slightly soluble solid
Product is always a solid … so not part of equilibrium expression
CaF2 (s) Ca2+ (aq) + 2F- (aq)
Ksp =
[Ca2+ ] [F-]2

6. Le Chatelier
When stress is placed on equilibrium, equilibrium will shift to relieve the stress and re-establish equilibrium
Note: heat is added need to know if exo or endothermic rxn.
Exo = heat is a product
Endo= heat is a reactant
Pressure only effects gases. When pressure is increases = shift to side with LEAST # gas moles

7. Q Q is a particular point in the reaction Set up same as Keq
We compare Q to Keq to see if this point in the reaction favors products (forward), favors reactants(reverse) or is at equilibrium
If Q< Keq favors reactants(reverse)
If Q> Keq favors products (forward)
If Q= Keq at equilibrium

8. Keq and multistep problem
If A + B C Keq = K1 and C D + E Keq = K2 Then A+ B D + E Keq = K1K2

9. Use ICE Box

10. NH3 H2O NH4+ OH- 0.0124M 0M 0M - 4.62x10-4M +4.62x10-4M +4.62x10-4M
K = [NH4+] [OH-]
[NH3]
K = [4.62x10-4] [4.62x10-4]
[ ]
K = x 10 -5

11. Common Ion Effect
A solid is thrown in some water. The solid is only very slightly soluble in water so an equilibrium is established AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq) So using le Chatilier, what would happen if I add more Ag+ ? More Cl- ?
Product side would be too high so ….Shift to the left
Product side would be too high so ….Shift to the left

12. Common Ion Effect AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq)
You can add more of the products, its called common ion effect.
It does effect the way equilibrium will shift.
If I add more [Cl- ]then the equil. will shift to the left and make more reactant while [Ag+ ] down

13. Common Ion Effect 1.6 x 10 -10 = [Ag+ ] [Cl-]
AgCl(s)
H2O
Ag+
Cl-
AgCl (s) + H2O (l) Ag+ (aq) + Cl- (aq)
If Ksp = 1.6 x , what is [Ag+ ] and [Cl-] ?
Ksp = 1.6 x = [Ag+ ] [Cl-]
1.6 x = (x) (x)
1.6 x = x2
1.3 x = x = [Ag+ ] =[Cl-]
So what happens if I add .1mol NaCl to 1L of AgCl solution?
NaCl is completely soluble and will completely dissociate to Na+ and Cl-
0.1mol of Cl- is a significant amount of Cl- compared to [Cl-]=1.3 x 10 -5
1.6 x = [Ag+ ] [Cl-]
1.6 x = [Ag+ ] [.1M]
[Ag+ ] = 1.6 x
0.10
[Ag+ ] = 1.6 x 10 -9
[Ag+ ] has decreased drastically 0M 0M +x +x x x

14. Gibbs Free Energy and Equilibrium
DG = -RT lnK (K is equilibrium constant) If DG = negative, K must be >1 so product are favored If DG = positive, K must be <1 so reactant are favored

15. Acids and bases Acids donate H+ Bases accept H+ (Ex. NaOH or NH3)
pH = -log[H+]
pOH = -log[OH-]
pH = 7 neutral
pH < 7 acid
pH> 7 base

16. HCl HBr HI H2SO4 HNO3 HClO3 HClO4 They completely ionize. 100%
Strong acids
HCl HBr HI H2SO HNO HClO3 HClO4
They completely ionize. 100%
Can find pH using pH = -log[H+]
Strong bases
LiOH NaOH KOH RbOH CsOH Ca(OH)2 Sr(OH)2 Ba(OH)2
and then 14 = pOH +pH

17. Weak acids Weak bases HA H+(aq) + A- (aq) B HA+(aq) + OH- (aq)
Only a small amount ionizes
Need to set up equilibrium
HA H+(aq) A- (aq)
Ka = [H+] [A- ]
[HA ]
To find pH -ice box!
Only a small amount ionizes
Need to set up equilibrium
B HA+(aq) OH- (aq)
Kb = [HA+] [OH- ]
[B ]
To find pH -ice box

18. Find pH of .2M HC2H3O2, with Ka = 1.8 x 10-5 HC2H3O2 H+ + C2H3O2-E
.2M 0M 0M -x +x +x
.2-x x x
1.8 X 10-5 = x2 .2
x = [H+] = 1.9 x 10 -5
pH = -log (1.9 x 10 -5)
pH = 2.7

19. Buffers and Salt solutions

20. Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4
NaF(aq) Na F-
step 1: identify major species
Na+, F-, H2O
Step 2: write dominate reaction
F-(aq) + H2O (l) HF(aq) + OH-(aq)
expect something basic
If from weak acid then stronger conjugate base)
(from strong base)
(from weak acid)

21. Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4
F-(aq) + H2O (l) HF(aq) + OH-(aq)
Since F- is a stronger conjugate base we need Kb
Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4
Step 3: calculate Kb, using Ka and Kw
Ka x Kb = Kw
Acid Conj base
Kb=Kw/Ka (for HF)
Kb = 1.0x10-14/7.2x10-4
Kb= 1.4 x

22. Calculate the pH of a 0.30M NaF solution. The Ka for HF =7.2x10-4
Step 4 make icebox
F H2O (l) HF OH-
0.30 -x +x +x
0.30-x x x
Use short cut 5%
X is significantly smaller than .3 so x is negligible and
.3M-x is about equal to .3M
Kb = 1.4x10-11 = (x)(x)
0.30
x= 2.0x10-6= [OH-]

23. x= 2.0x10-6= [OH-] pOH = -log [OH-] = -log(2.0x10-6) = 5.69 14=pH +pOH
14 – pOH = pH
= 8.31 = pH

24. Buffer…. Weak acid + its conjugate base Or
Weak base + its conjugate acid

25. Which of following pairs make a buffer?
A. NH3, NH4Cl
B. RbOH, HBr
C. KOH, HF
D. NaC2H3O2, HCl

26. Calculate the pH of a buffer solution that contains 0
Calculate the pH of a buffer solution that contains 0.15M of acetic acid and 0.20mol of sodium acetate in 1L. (Ka =1.8 x 10-5)
Solve by icebox and then solve by Henderson-Hasselbach
HC2H3O H C2H3O2-
.15M
.2M -x +x +x
.15-x +x
.2 +x
5% siort cut
K= x(.2)
____
.15
1.8x10 -5= x(.2)
____
.15
x= 1.35x and x also = [H+]
pH =-log[H+]
pH =-log(1.35x10 -5)
pH = 4.87

27. Yes- Henderson-Hasselbalch equation
Is there an easier way to calculate pH of a buffered system?
Yes- Henderson-Hasselbalch equation
pH = pKa + log[A-]/[HA]
pH =pKa + log[base]/[acid]
Is useful for calculating pH of solutions when the HA and A- are known
pH =pKa + log[base]/[acid] look back at our problem
Calculate the pH of a buffer solution that contains 0.15M of acetic acid and 0.20mol of sodium acetate in 1L. (Ka =1.8 x 10-5)
pH = -log(1.8x10 -5) + log[.2]/[.15]
pH =
pH = same as by ice box method just faster 

28. Equivalence pt. During strong acid/strong base titration is always pH7
Equivalence pt. happens during titration. It is when moles H+ = moles OH-
Equivalence pt. During strong acid/strong base titration is always pH7 7 7
Volume Base added
Volume Acid added
Titration curve of strong acid with strong base
Titration curve of strong base with strong acid
Type 1: strong acid-strong base titration

29. Weak acid- Strong base HF + OH- H2O + F-
(A) Maximum buffering where [HA] / [A-]= 1
(B) Equivalence point (above pH = 7) and [HA] =0
(C) Basic
(D) ½ way point and pH =pKa
No calculation needed C A B D
Definition of buffer = weak acid + its salt (conj. base)
Or weak base + its salt (conj.acid)

30. Weak Base- Strong Acid NH3 + HCl NH4+ +Cl-
Equivalence pt