1. Acids and Bases

2. Conjugate Acid-Base Pairs
A Bronsted acid donates a proton
A Bronsted base accepts a proton
The conjugate base of a Bronsted acid is what remains after the acid gives up its proton
The conjugate acid of a Bronsted base is what is produced after the base accepts a proton
ex.; CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq) CH3COOH/ CH3COO- is a conjugate acid-base pair et H3O+/ H2O is another conjugate acid-base pair

3. Conjugate Acid-Base Pairs
Example: What are the conjugate acid-base pairs in the following reactions? (a) NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) (b) H3O+(aq) + OH-(aq)  2 H2O(l)
Solution: (a) NH4+/ NH3 is a conjugate acid-base pair H2O/ OH- is a conjugate acid-base pair (b) H3O+/H2O is a conjugate acid-base pair H2O/ OH- is a conjugate acid-base pair
N.B. In each case, the first is the acid and the second is the base.

4. The Acid-Base Properties of Water
We have just seen that water can donate a proton (and become OH-) or accept a proton (and become H3O+): Thus water can be both an acid and a base
Water can autoionize: H2O(l)  H3O+(aq) + OH-(aq)
The equilibrium constant, at 25oC, for the autoionization of water (using H+ for simplicity):
The concentrations of H+ and OH- are related to each other
If one is high, the other is low as their product is constant
In a neutral solution, at 25ºC, [H+] = [OH-] = 1.0 x 10-7 M

5. pH: A Measure of Acidity
A more practical measurement for the concentration of H+ is pH (a unitless quantity): pH = -log[H+]
In an acidic solution: pH < 7.00
In a basic solution: pH > 7.00
In a neutral solution: pH = 7.00
The pOH scale is less common: pOH = -log[OH-]
The sum of pH + pOH is always constant: pH + pOH = 14.00

6. pH: A Measure of Acidity
Example: Calculate the pH of a solution of HNO3 in which the concentration of hydrogen ions is 0.76 M.
Solution: pH = -log[H+] = -log(0.76) = 0.12
Example: The pH of a certain fruit juice is Calculate the concentration of H+ ions.
Solution:

7. Strong Acids and Strong Bases
A strong acid is a strong electrolyte that ionizes completely in water
ex.; HCl, HNO3, H2SO4
A weak acid is a weak electrolyte that only partially ionizes in water
ex.; HF, CH3COOH, NH4+

8. Strong Acids and Strong Bases
A strong base is a strong electrolyte that ionizes completely in water
ex.; Any alkali metal hydroxide (ex.; NaOH)
A weak base is a weak electrolyte that only partially ionizes in water
ex.; NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

9. Strong Acids and Strong Bases
In a conjugate acid-base pair, if an acid is strong, its conjugate base is very weak, and vice versa
The H3O+ ion is the strongest acid that can exist in an aqueous solution
A stronger acid will react with H2O ex.; HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
The OH- ion is the strongest base that can exist in an aqueous solution
A stronger base will react with H2O ex.; O2-(aq) + H2O(l)  2 OH-(aq)

10. Strong Acids and Strong Bases
Example: Calculate the pH of a solution of Ba(OH)2 with a concentration of 1.5 x 10-2 M.
Solution: For each Ba(OH)2, we have two OH If we have 1.5 x 10-2 M of Ba(OH)2, we have 3.0 x 10-2 M of OH The pOH is therefore -log(3.0 x 10-2) = 1.52.
The pH is thus =

11. Weak Acids and Their Ionization Constants
The dissociation of a weak acid is not complete HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
The equilibrium constant for this reaction is the ionization constant of the acid, Ka
When the acid is stronger, its Ka is larger
We can calculate the concentrations of each species in equilibrium with the methods that we have seen in the chapter on chemical equilibrium

12. Weak Acids and Their Ionization Constants
To calculate equilibrium concentrations, we make ​​the following approximations
The concentration of H+ (1.0 x 10-7 M) before adding the acid to the pure water is negligible
The amount of acid that dissociates is negligible, to a first approximation
Once we find x (the quantity of acid that dissociates), we verify whether x is less than ~ 5% of the initial quantity of acid
If yes, this is the value of x and we can calculate all of the concentrations
If no, we need to solve for x, without making the second approximation

13. Weak Acids and Their Ionization Constants
Example: Calculate the concentrations of H+, of A-, and of non-ionized HA in a solution of 0.20M HA. The value of Ka for HA is 2.7 x 10-4.
Solution: [H+] = x, [A-] = x, and [HA] = x  Verify our approximation: Our approximation is acceptable, therefore [H+] = 7.3 x 10-3 M [A-] = 7.3 x 10-3 M [HA] = x 10-3 = 0.19 M

14. Weak Acids and Their Ionization Constants
Example: What is the pH of a M solution of a monoacid, HA, for which the value of Ka = 5.7 x 10-4?
Solution: [H+] = x, [A-] = x, et [HA] = x  Verify our approximation: We cannot make the approximation that [HA]  M

15. Weak Acids and Their Ionization Constants
Solution: [H+] = x, [A-] = x, et [HA] = x
The second solution is not acceptable
Thus, [H +] = 8.06 x 10-3 M, and the pH = -log(8.06 x 10-3) = 2.09 or

16. Weak Acids and Their Ionization Constants
Example: A M of HA has a pH equal to Calculate the value of Ka for this acid.
Solution: If pH = 3.44, [H+] = = 3.63 x 10-4 M [A-] = [H+] = 3.63 x 10-4 M [HA] = x 10-4 = M

17. Percent Ionization The percent ionization is defined by
For an acid that donates a single proton where [HA]o is the initial concentration of the acid
The percent ionization decreases when the acid becomes more concentrated

18. We dissolved 1. 22 g of a monoacid, HA, in water to produce a 25
We dissolved 1.22 g of a monoacid, HA, in water to produce a 25.0 mL solution. The initial pH of this solution is To neutralize this acid, we need 17.7 mL of a M NaOH solution.
a) What is the molar mass of HA?
b) What is the value of the dissociation constant of HA?

19. Diprotic and Polyprotic Acids
A diprotic or polyprotic acid can donate two or more protons per molecule
The ionization of these acids happens in steps
Each step has an ionization constant
The ionization constant becomes smaller in each step

20. Diprotic and Polyprotic Acids
Example: Calculate the concentrations of C2H2O4, of C2HO4-, of C2O42-, and of H+ in a solution of 0.20 M oxalic acid (C2H2O4). For oxalic acid, Ka1= 6.5 x 10-2 and Ka2= 6.1 x 10-5.
Solution: Looking at the first equilibrium Verify our approximation:
We cannot make the approximation that [C2H2O4]  0.20 M.

21. Diprotic and Polyprotic Acids
Solution:
The second solution is not acceptable
Therefore [H+] = M [C2HO4-] = M [C2H2O4] = 0.11 M or

22. Diprotic and Polyprotic Acids
Solution: We must also find [C2O42-]. We make the assumption that this second dissociation does not affect [C2HO4-] and [H+] for the first dissociation.
Verify our approximation:
Our approximation is acceptable, therefore [C2H2O4] = 0.11 M [C2HO4-] = M [C2O4-] = 6.1 x 10-5 M [H+] = M

23. Weak Bases and Their Ionization Constants
We treat weak bases the same way we treat weak acids
ex.; NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
Kb is the ionization constant of a base

24. Weak Bases and Their Ionization Constants
Example: Calculate the pH of a 0.26 M solution of methylamine (CH3NH2). Kb = 4.4 x 10-4 for methylamine CH3NH2(aq) + H2O(l)  CH3NH3+(aq) + OH-(aq)
Solution: [CH3NH3+] = x, [OH-] = x, and [CH3NH2] = x  Verify our approximation: Our approximation is acceptable, therefore pOH = -log(0.011) = 1.96, and therefore the pH = = 12.04

25. The Relationship Between the Ionization Constants in Conjugate Acid-Base Pairs
For a weak acid
For its conjugate base
The product of the two ionization constants gives
Kwater

26. The Relationship Between the Ionization Constants in Conjugate Acid-Base Pairs
The fact that KaKb = Kwater is not surprising because the sum of the two reactions is
The stronger an acid becomes, the weaker its conjugate base becomes, and vice versa
Example: For acetic acid, Ka = 1.8 x What is the value of Kb for the acetate anion (its conjugate base)?
Solution:

27. The Acid-Base Properties of Salts
The hydrolysis of a salt is the reaction between an anion and/or a cation derived from the salt, and water
For a salt where the cation is the conjugate acid of a strong base and the anion is the conjugate base of a strong acid, the salt ions do not react with water and the pH  7.0
ex.; NaNO3 is produced by reacting NaOH and HNO3 solutions Upon its dissociation: the ions produced do not interact with water. We say that the salt is neutral.

28. Salts Producing Basic Solutions
ex.; When sodium acetate is dissolved in water The Na+ cation does not react with water, but the CH3COO- anion does react with water
The solution becomes basic
We say that this is a basic salt

29. Salts Producing Acidic Solutions
ex.; When ammonium chloride is dissolved in water The Cl- anion does not react with water but the NH4+ cation partially dissociates
The solution becomes acidic
We say that this is an acidic salt

30. Salts Where Both Cation and Anion Hydrolyze
If the cation and the anion react in water, it is the relative strengths of the base and the acid of the salt which will determine the pH of the solution
If Kb > Ka: basic solution
If Kb < Ka: acidic solution
If Kb  Ka: nearly neutral solution
For an amphoteric ion, the largest ionization constant dominates. For example,
Therefore a sodium bicarbonate solution is basic

31. We dissolve g of a monoacid, HA, in water to produce a solution of 25.0 mL. To neutralize this acid, one needs 36.2 mL of a M aqueous solution of NaOH. What is the molar mass of HA? If the pH at the equivalence point is 11.22, what is the value of Kb for A- (aq)?

32. Acidic, Basic and Amphoteric Oxides
Oxides often react with water
In general, metal oxides react with water to produce a base (thus the oxide is basic)
Examples:

33. Acidic, Basic and Amphoteric Oxides
In general, the oxide of a non-metal reacts with water to give an acid (therefore the oxide is acidic)
Examples:
The first reaction explains why rain is naturally acidic (pH  5.5), and the second and third reactions explain the phenomenon of anthropogenic acid rain

34. Acidic, Basic and Amphoteric Oxides
The oxide of a transition metal in a higher oxidation state is often an acid
ex.; permanganic acid :
ex.; chromic acid