1. Strength of Acids and Bases
The terms weak and strong are used to compare the strengths of acids and bases
The terms dilute and concentrated are used to compare the concentration of solutions
They do not mean the same thing!
The combination of strength and concentration determines the behavior of the acid or base.

2. Strength of Acids and Bases
Strength is Determined by how much they ionize (dissociate) in water
Strong – ionize 100% in water
Weak – only partially ionize in water
Complete dissociation - all HCl compounds have separated into H+ and Cl- ions
Partial dissociation - some HNO2 compounds have separated into H+ and NO2- ions, while some remain together 2

3. Strength of Acids and Bases
Strong Acids:
Only HNO3, HI, HBr, HCl, H2SO4, HClO4
Remember: NO, I Brought Claude SOme ClOthes
All other acids are weak, and remain in equilibrium
Ex: HNO2 ↔ H+ + NO2-
Strong Bases: Group I or II metals
Ex: NaOH, Mg(OH)2…etc.
All other bases are weak, and remain in equilibrium
NH3 + H2O ↔ NH4+ + OH- 3

4. Conjugate Strength Strong Acid = weak conjugate base
Strong Base = weak conjugate acid
Weak Acid = strong conjugate base
Weak Base = strong conjugate acid
Example:
HCl + NH3 → Cl NH4+
(strong acid) (weak base) (weak c. base) (strong c. acid) 4

5. Strong Acid/Base Calculations
If the acid/base is strong, concentrations of reactants will match the concentration of the products
Ex: What is the [H+] in a 0.2M HCl solution? HCl is a strong acid; it ionizes completely.
HCl  H Cl-
What is the [OH-] in a 0.03 M NaOH solution. NaOH is a strong base and it completely ionizes.
NaOH  Na+ + OH-
0.2M
0.0M
[H+] = 0.2M
[OH-] = 0.03M

6. Weak Acids and Bases are Equilibriums
Since weak acids and bases only partially ionize in water, an equilibrium is established between the original acid (or base) and its ions.
Example: HCHOOH  H1+ + CHOOH1-
Acetic acid partially ionizes in to H1+ and
CHOOH1-.

7. Equilibrium Expressions for Acids
Equilibrium expressions relate concentrations of reactants to those of the products
In this case the acid and its ions
HA (aq) ↔ H+ (aq) + A- (aq)
Ka = [H+][A-]
[HA]
Ka = equilibrium constant for a acid (capital K)
Numerical value of the ratio of product concentrations to reactant concentrations
[ ] = concentration in M (mol/L)
Order = coefficient becomes exponent behind the brackets

8. Acid Equilibrium Expression Examples
Write the equilibrium expressions for the following reactions:
HF (aq) ↔ H+ (aq) + F- (aq)
Ka = [H+][F-] / [HF]
H2CO3 (aq) ↔ 2H+ (aq) + CO32- (aq)
Ka = [H+]2[CO32-] / [H2CO3]

9. Practice HNO2 (aq) ↔ H+ (aq) + NO2- (aq)
[HNO2] = 0.045M [H+]=[NO2-] = 0.045M
Write the equilibrium expression.
Ka = [H+][NO2-] / [HNO2]
Find Ka
0.045
H2SO3 (aq) ↔ 2H+ (aq) + SO32- (aq)
Ka = [H+]2[SO32-] / [H2SO3]
If Ka = 1.44, the [H+] = 0.2M, and [SO32-] = 0.1M, what is [H2SO3]?
2.8x10-3M

10. Weak Acid Calculation
A weak acid, HA, is found to be 25% ionized in water. Find the Ka for a 0.4M solution of HA.
HA  H A-
0.4M
0.0M
0.0M
M= 0.3M
0.1M
0.1M
Ka = [H+][A-]
[HA]
= [0.1][0.1]
[0.3]
= 0.033

11. Acid Equilibrium Constant (Ka)
The value of Ka tells you whether the products or reactants are favored.
The higher the value of Ka; the stronger the acid.
More products favored, so there is more ionization.

12. Comparing Ka Which acid is stronger?
HCOOH (formic acid) = 1.8 x 10-4
H2CO = 4.4 X 10-7
HC2H3O = X 10-5
Formic Acid
Which acid has the stronger conjugate base?
H2CO3

13. Equilibrium Expressions for Bases
Equilibrium expressions relate concentrations of reactants to those of the products
In this case the base and its ions
BOH (aq) ↔ B+ (aq) + OH- (aq)
Kb = [B+][OH-]
[BOH]
Kb = equilibrium constant for a acid (capital K)
Numerical value of the ratio of product concentrations to reactant concentrations
[ ] = concentration in M (mol/L)
Order = coefficient becomes exponent behind the brackets

14. Practice AgOH (aq) ↔ Ag+ (aq) + OH- (aq)
[AgOH] = 0.52M [Ag+]=[OH-] = 0.52M
Write the equilibrium expression.
Kb = [Ag+][OH-] / [AgOH]
Find Kb
0.52
Cu(OH)2 (aq) ↔ Cu2+ (aq) + 2OH- (aq)
Kb = [Cu2+][OH-]2 / [Cu(OH)2]
If Kb = 0.5, the [Cu2+] = 0.12M, and [OH-] = 0.24M, what is [Cu(OH)2]?
1.4x10-2M

15. Kb Calculations
A weak base, BOH, is prepared by dissolving 1.5 moles in 3 liters of solution. At equilibrium, the [OH] is 0.01 M. Find the Kb.
BOH  B+ + OH-
2x10-4

16. Additional KbCalculations
The Kb for NH3 is 1.8x Find the pH for a 0.5M NH3 solution.
The ionization constant (Kb) is less than 5% (5x10-2) of the original [base] so the amount of base that ionizes is negligible and can be ignored.
NH H2O  NH OH-
0.0M
0.5M
0.0M x x
0.5-x
Kb = [NH4+][OH-]
[NH3]
= [x][x]
[0.5-x]
= x2
[0.5]
= 1.8x10-5
X2 = 9.0x10-6
[OH-] = X = 3.0x10-3M

17. Additional Kb Calculations
Now that you have the [OH-], you can find pOH which gives you pH.
pOH = - log 3.0x10-3M
pOH = 2.52
pH = 14 – pOH = 11.5

18. Base Equilibrium Constant (Kb)
The value of Kb tells you whether the products or reactants are favored.
The higher the value of Kb; the stronger the base.
More products favored, so there is more ionization.

19. Acid-Base Reactions
The reaction of an acid and a base to produce a salt and water is called a neutralization reaction.

20. Neutralization Reactions
Sodium hydroxide (base) and hydrochloric acid (acid) react to form sodium chloride (salt) and water.

21. Neutralization Practice
Neutralizations are double displacement reactions!
Practice predicting products (always a salt and water):
Example #1
__Ca(OH)2 + __H3PO4 → ?
__Ca(OH)2 + __H3PO4 → __Ca3(PO4)2 + __H2O
3Ca(OH)2 + 2H3PO4 → Ca3(PO4)2 + 6H2O
Example #2
__Fe(OH)2 + __HBr → ?
__Fe(OH)2 + __HBr → __FeBr2 + __H2O
Fe(OH)2 + 2HBr → FeBr2 + 2H2O 21

22. Acid-Base Titrations
Titration is used in determining the concentration of an unknown acid or base.
An indicator is added to the standardized acid (acid of known concentration).
The unknown base is added slowly with a burette until the solution is neutralized and reaches the equivalence point (where [H+] = [OH-]).
The equivalence point is not necessarily where pH = 7
Adding one more drop of base changes the color of the solution to pink. This is called the endpoint.

23. Titration Calculations
MaVa(H+)= MbVb(OH-)
Note: pH at the equivalence point is not always 7.
Strong Acid + Strong Base, pH=7
Strong Acid + Weak Base, pH<7
Weak Acid + Strong Base, pH>7
Example: It took 75mL of NaOH to neutralize 50mL of 2M HCl. What is the concentration of the NaOH?
MaVa(#H+) = MbVb(#OH-)
(2M)(50mL)(1) = (x)(75mL)(1)
x = 1.33M NaOH

24. Titration Practice
It took 20mL of Ca(OH)2 to neutralize 25mL of 0.05M HCl. What is the concentration of the base?
MaVa(#H+) = MbVb(#OH-)
(0.05M)(25mL)(1) = (x)(20mL)(2)
x = M