1. Acid Base Chemistry

2. Acid-Base definitions
Arrhenius
Produces H+
Produces OH-
Bronsted-Lowery
Proton donor
Proton acceptor
Lewis
Electron acceptor
Electron donor
Amphoteric: can act as an acid or base (H2O)

3. Conjugate acid-Base pairs
Conjugate acid-base pairs consists of two substances related to each other by the transfer of a proton.
HNO2(aq) + H2O(l) NO2-(aq) + H3O+(aq)
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

4. Quiz Identify the conjugate acid-base pairs in the following reaction:
HBr + NH NH4+ + Br-
HCO3- + HCl H2CO3 + Cl-
What is the conjugate base of H2S?
What is the conjugate acid of NO3-?

5. Acid Strength
Strong acid—completely ionizes, equilibrium lies far to the right.
Weak Acid—partially ionizes, equilibrium lies to the left.
NOTe*--do not confuse with dilute or concentrated acids.

6. Strong acids Memorize this list of strong acids:
Generally, the more oxygen atoms, the stronger the acid. Why?
HCl
HBr HI
HNO3
HClO4
H2SO4

7. WEAK ACIDS
Degree of ionization depends on attraction between anion (conjugate base) and the hydrogen ion, relative to attractions of these ions to water.
Acid ionization constant (Ka)
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
The smaller the constant, the weaker the acid.

8. Write the simple ionization reaction for the following acids
Hydrochloric acid
Acetic Acid
The ammonium ion
The anilinium ion (C6H5NH3+)

9. Autoionization of water and pH
Water is amphoteric
H2O(l) H+(aq) + OH-(aq)
Ion product constant for water: Kw
Kw = [H+][OH-] = 1.0x10-14
[H+]= [OH-] = √Kw = 1.0x10-7 (neutral solution)
Acidic solution—contains an acid that creates additional H3O+
Basic solution—contains base that creates additional OH-.

10. Summary
Neutral solution: [H3O+]= [OH-] = 1.0x10-7 Acidic solution: [H3O+] > [OH-] Basic solution: [H3O+] < [OH-]

11. USING KW IN CALCULATIONS
Calculate the [OH-] at 25oC for each solution and determine if the solution is acidic, basic, or neutral:
A.) [H3O+] = 7.5x10-5 M
B.) [H3O+] = 1.0x10-7 M
C.) [H3O+] = 1.5x10-9 M

12. pH Scale Way to specify the acidity of a solution. pH = -log[H3O+]
Example: A solution with [H3O+] = 1.0x10-3 M
pH = -log[1.0x10-3] = 3.00
Note*--Sig figs: only numbers to the right of the decimal point are significant in logarithms.

13. Relation of pH to acidity
[H3O+]
Solution is:
= 7
1.0X10-7
NEUTRAL
> 7
>1.0X10-7
Basic
< 7
<1.0X10-7
Acidic

14. pH Scale

15. Calculating pH from [H3O+] OR [OH-]
Calculate the pH of each solution at 25oC and indicate whether the solution is acidic or basic.
[H3O+] = 1.8x10-4 M
[OH-] = 1.3x10-2 M

16. Calculating [H3O+] from pH
Calculate the [H3O+] of a solution with a pH of 4.80 (use correct sig figs in your answer)

17. pH and Other p Scales Notice that p = -log; therefore pX = -logX
pOH = -log[OH-]
14 = pH + pOH
pKa = -logKa = another way to quantify strength of an acid.
The smaller the pKa, the stronger the acid.

18. Finding the [H3O+] and pH of strong and weak acids
For strong acids, [H3O+] = concentration of strong acid.
For weak acids, must use Ka to find concentrations at equilibrium.
Remember ICE tables??

19. Practice Find the pH of a 0.100 M solution of HCN.
Ka for HCN is 4.9x10-10

20. Practice Find the pH of a 0.200 M HNO2 solution.
Ka for HNO2 = 4.6x10-4

21. Finding equilibrium constant from pH
A M weak acid solution has a pH of Find the Ka for the acid.

22. pH of Mixtures of Acids
Strong and Weak acid: completely ignore the ionization of the weak acid. The pH is equivalent to that of the strong acid.
Two weak acids: If Ka’s are similar, find concentration of H30+ from both and then add them together. If different, use the largest one and assume any more formation of H3O+ is hindered.

23. Base solutions Strong Bases: completely dissociate in water
Memorize these: LiOH, NaOH, KOH, Sr(OH)2, Ca(OH)2, Ba(OH)2
Completely dissociate in water.

24. Weak Bases
Analogous to weak acid
Can find pH, pOH, Kb, and [OH-]

25. Different bases at the same concentration:
base with larger Kb is the stronger base
Stronger base will have:
higher [OH–]
higher pH
higher % dissociation
Same base at 2 different concentrations:
Kb is the same, so base strength is the same
solution with higher concentration will have:
lower % dissocation

26. Acid-Base properties of ions and salts
Salts are produced when acids and bases react.
Neutral salts: Formed from strong acids and bases.
Acidic salts: Strong acid reacts with weak base.
Basic salts: Weak acid reacts with strong base.

27. Practice
Predict whether an aqueous solution of the following salts is acidic, basic, or neutral. Prove with appropriate equations.
NaC2H3O2
NH4NO3
Al2(SO4)3

28. Practice
Calculate the pH of a 0.30 M solution of NaF. The Ka of HF is 7.2x10-4.

29. Practice
Find the pH of a M NaCHO2 solution.

30. Polyprotic acids Ionizes in successive steps Example: Sulfurous acid
H2SO3(aq) H+(aq) + HSO3-(aq) Ka1 = 1.6x10-2
HSO3-(aq) H+(aq) + SO32-(aq) Ka2 = 6.4x10-8
Note*--Ka2 is much smaller than Ka1.

31. Finding pH of polyprotic acids
First ionization step is much larger than any sequential step. Inhibits any more formation of H3O+ in the second step. Find the pH of a M ascorbic acid (H2C6H6O6). Ka1 = 8.0x10-5 Ka2 = 1.6x10-12

32. Buffers Resist pH change by neutralizing added acid or base.
Contains either:
Weak acid and its conjugate base
Weak base and its conjugate acid
Example: Carbonic acid in blood

33. Concept check Which one of the following solutions is a buffer?
A M HNO3 AND M HCl
B M HNO3 AND M NaNO3
C M HNO2 AND M NaCl
D M HNO2 AND M NaNO2

35. Calculating pH of a buffer Solution
Consider a solution of sodium acetate and acetic acid.
Common Ion effect: Same ions in buffer solution cause acid to ionize less (Le Châtelier’s principle)
Practice: A buffer solution contains M acetic acid and M sodium acetate. Determine the [H3O+], pH, and % ionization in this
solution. For HC2H3O2, Ka = 1.8 x 10–5

36. Henderson-Hasselbalch equation
pH = pKa + log[base]/[acid]
Calculate the pH of a buffer solution that is M in benzoic acid (HC7H5O2) and M sodium benzoate (NaC7H5O2). For benzoic acid, Ka = 6.5x10-5

37. Calculating pH Changes in Buffer solutions
Two parts:
Stoichiometry calculation: calculate how the addition changes the relative amounts of acids and bases.
Equilibrium calculation: calculate pH based on new amounts of acid and conjugate base.

39. Calculating the pH change in a buffer solution
A 1.0 L buffer solution contains M acetic acid and M sodium acetate. The value for Ka for acetic acid 1.8x Calculate the pH of the buffer solution. Calculate the new pH after addition of mol of solid NaOH to the buffer. For comparison, calculate the pH after adding mol of solid NaOH to 1.0 L of pure water.

40. Buffer Effectiveness
Most effective with the following characteristics:
Relative concentrations of weak acid and conjugate base do not differ by more than a factor of 10.
Concentrations of acid and conjugate base are high.
Range of the system is one pH unit on either side of the pKa