1. Acid/Base and Solubility Equilibrium
CH1121

2. Bronsted-Lowry Acids and Bases
Acid-base reactions involve the transfer of H+ ions (protons) from one substance to another
Acids are proton donors
Bases are proton acceptors
Some substances are amphoteric and can act as a base or an acid depending on what they are being reacted with

3. Conjugate Acids and Bases
When an acid reacts with a base it loses a proton
What’s left of the acid is then known as the conjugate base
These two substances are known as a conjugate pair

4. Conjugate Acids and Bases
When a base reacts with an acid it gains a proton
When the base has gained a proton it is called the conjugate acid
These two substances are known as a conjugate pair

5. Conjugate Acids and Bases

7. pH Numerical scale that describes the acidity of a substance
Measurements are logarithmic and fall between 0-14 for most substances
Bases >7
Acids <7

8. pH

9. pOH Measures the basicity of substances
Just like pH is a measure of the concentration of H+ ions pOH is a measure of the concentration of OH- ions
When added together pH and pOH of a substance will be equal to 14

10. pH and pOH Calculated by the following equations: pH = -log[H+]
pOH = -log [OH-]
pH + pOH = 14.00
Keep in mind, as the concentration of H+ ions increases the pH will decrease

11. pH and pOH A solution at 25℃ has [OH-] = 6.7 x 10-3M.
What is the pH of the solution?
Is the solution acidic or basic?

12. pH and pOH
𝑝𝑂𝐻=− log 𝑂𝐻− =−log 6.7× 10 −3 =2.17 𝑝𝐻+𝑝𝑂𝐻=14 𝑝𝐻=14−2.17=11.83 pH >7 ∴ basic

13. pH
What is the concentration of H+ ions in a solution with a pH of 7.5?
7.5=− log 𝐻+
𝐻+ = 10 −7.5 =3.16× 10 −4 𝑀

14. H2O(l) + H2O(l) ⇋ H3O+(aq) + OH-(aq)
Water
Because water is amphoteric, in a container of many H2O molecules there is a constant process called self ionization
H2O(l) + H2O(l) ⇋ H3O+(aq) + OH-(aq)
This means there are ions present as well as water molecules
It is important to note that H3O+ and H+ are often used interchangeably

15. Water
If we write an equilibrium constant for water (Kw) it would look like this:
Kw=[H+][OH-]
Kw = 1.00 x 10-14
This value for Kw is for 25℃

16. Kw Like other equilibrium constants, Kw is temperature dependent.
Kw also applies to aqueous solutions of acids and bases

17. Kw
If at 25℃ the concentration of H+ ions in an aqueous basic solution is 2.55 x 10-5M what is the concentration of OH- ions?
[H+] [OH-] = Kw = 1.00 x 10-14
𝑂𝐻 − = 1.00× 10 − × 10 −5 =3.92× 10 −10 𝑀

18. Strong Vs. Weak Acids and Bases
Ionization (breaking into ions) is what determines the strength of acids and bases
If an acid/base completely dissociates into its ions it is said to be strong
If an acid/base only partially dissociates into its ions it is said to be weak

19. Common Bases Hydroxide ion indicates a strong base
KOH
NaOH
LiOH
Ba(OH)2
For this course use this rule

20. Strong Acids There are 7 strong acids All other acids are weak acids
H2SO4
HNO3
HClO3
HClO4
HCl
HBr HI
All other acids are weak acids

21. H3PO4(aq) ⇋ H+(aq) + H2PO4-(aq)
Acid Strength (Ka)
For weak acids we can write an equilibrium equation to compare strengths
Ka is just like all other equilibrium constants
For the reaction:
H3PO4(aq) ⇋ H+(aq) + H2PO4-(aq)
𝐾𝑎= [ 𝐻 + ][ 𝐻2𝑃𝑂4 − ] [𝐻3𝑃𝑂4]

22. Acid Strength (Ka) As Ka increases, the strength of the acid increases
The lower Ka the weaker the acid is
A weak acid will be closer to the basic end of the pH scale
A weak base will be closer to the acidic end of the pH scale

23. Titrations Mixing of 2 solutions
Titrant : known concentration, known volume
Analyte: unknown concentration, known volume
To find the unknown concentration of the analyte we use stoichiometry (unit 5 of CH1120)

24. Titrations

25. Titrations
Titration reveals that 11.6mL of 3.0M sulfuric acid are required to neutralize the sodium hydroxide in 25.00mL of NaOH solution. What is the molarity of the NaOH solution?

26. Titrations
H2SO4(aq) + 2NaOH(aq) ⇋ 2H2O(l) + Na2SO4(aq) 𝐿 3.0𝑚𝑜𝑙 𝐻2𝑆𝑂4 1𝐿 2𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1𝑚𝑜𝑙 𝐻2𝑆𝑂4 =0.069𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 𝐶= 𝑛 𝑉 = 0.069𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 0.025𝐿 =2.76𝑀 𝑁𝑎𝑂𝐻

27. Indicators
Dyes made up of weak acids and bases used to identify pH ranges
Protonated (acid) and unprotonated (base) forms take on different colours
Litmus paper, turns red for acid, blue for base
Red is protonated form
Blue is unprotonated form

28. Indicators

29. Acid Rain
Rain is naturally slightly acidic due to CO2 in the atmosphere mixing with rain water
pH = 5.5
Pollutants decrease the pH of rain water even further
Nitrogen oxides (general formula = NOx)
Sulfur oxides (general formula = SOx)

30. Acid Rain (Pollutant = NOx)
Naturally:
Volcanic eruptions
Forest fires
Lightning
Human Activity:
Internal combustion engines
Smelting
Thermal generation
Form H2NO3(aq) when combined with rain water

31. Acid Rain (Pollutant = SOx)
Naturally:
Volcanic eruptions
Forest fires
Decay
Human Activity:
Burning fossil fuels
Smelting
Deforestation
Form H2SO4(aq) when combined with rain water

32. Acid Rain
This creation of stronger acids lowers the pH of rain water below the normally slightly acidic level
Acidifies lakes and oceans
Dissolves nutrients in soil
Damages manmade structures
Increases the solubility of dangerous heavy metals
Levels as low as pH = 2.5

33. Acid Rain

34. Buffers
Solution that resists change to pH even with the addition of small amounts of strong acid or base
Weak acid and conjugate base
Weak base and conjugate acid
Buffer capacity refers to the amount of acid or base a buffer can neutralize before the pH begins to change a significant amount

35. Buffers Since buffers have both acidic and basic properties
Acidic part neutralizes bases that are added
Basic part neutralizes acids that are added

36. Buffers Common examples of buffers:
Carbonic acid (H2CO3)/bicarbonate ion (HCO3-)
Ammonia (NH3)/ammonium chloride (NH4+Cl-)
Acetic acid (CH3COOH)/sodium acetate (CH3COO-Na+)

37. Buffers

38. Solubility Equilibrium (Ksp)
Relates the solid form of a compound to the dissolved/ionized version of that solid
Calculated just like all other equilibrium constants
BaSO4(s) ⇋ Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-]
The smaller Ksp, the less soluble the substance is

39. Solubility Equilibrium (Ksp)
The Ksp for CaF2 is 3.9 x at 25℃. Assuming equilibrium is established between solid and dissolved CaF2, and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in mols per litre.

40. Solubility Equilibrium (Ksp)
CaF2(s) ⇋ Ca2+(aq) + 2F-(aq) Ksp = [Ca2+][F-]2 We can let “X” represent the concentration of CaF2 and use our mol ratio in our Ksp Ksp = [X][2X]2 Ksp = 3.9 x = 4X3 X = 2.1 x 10-4 The solubility of CaF2 is 2.1 x 10-4mols/L