1. Lets walk through them all together:
Remember they don’t have notes GO SLOWLY, noooo slower .
Lets walk through them all together:
Atkins and Jones

3. Strong/Strong Titration Calculations
If 100. mL of a Mg(OH)2 solution is titrated with 25.0mL of 2.00 M HCl…..
What is the pH at the equivalence point?
How many moles of Mg(OH)2 were initially present?
What was the concentration of Mg(OH)2?

4. Strong/Strong Titration Calculations
A student weighs g of an unknown strong acid into a beaker with 10mL of water. It is titrated to equivalence with 50.0mL of 1.0M NaOH. What was the molecular mass of the acid.

5. Conjugate Acid/Base pairs
Acids lose a proton to become the conjugate base.
Bases gain a proton to become it’s conjugate acid
All acid/base reactions have both an acid and a base in them? So what is the base and conjugate acid in the first reaction? What is the acid and conjugate base in the second reaction?
Conjugate Base
Acid
Conjugate
Acid
Conjugate
Base
Base
Acid

6. Important Relations H2O ⇄ H+ +OH− [H+]=[OH−]=1x10-7 @25oC
Relations between [H+], [OH-], Kw, pH and pOH
H2O ⇄ H+ +OH−
[H+]=[OH−]=1x @25oC
Kw= [H+][OH−]=1x @25oC
Taking the log of both sides:
Can you just memorize this without understanding it?
What happens if the temperature changes?

7. Kw Example Problem [H+]=10−6.8 Kw= [H+][OH−]
If water is heated and placed under pressure it has a pH of 6.8. Find the Kw of water under these conditions. Is it acidic, basic or neutral?
𝑛𝑒𝑢𝑡𝑟𝑎𝑙
[H+]=[OH−]
[H+]=10−6.8
Kw= [H+][OH−]
kw=[10-6.8] [10-6.8]=2.5x10-14

8. Weak and Strong Acid and Bases: Compare and Contrast
For strong acids and bases, we assume they completely ionize.
Weak acids and bases ionize to a very limited extent.
Equilibrium lies far to the left
Equilibrium lies far to the right
How do I know which is strong and weak?
Very small Ka
Very Large Ka
Memorize the strong, the rest are weak

9. What happens to Ka as the strength of the acid increases?vs
The larger the Ka the stronger the acid.
Lets think about the relation of acid strength to Ka. **click** I have two examples drawn out. In one case very little of the molecules are ionized, and in the next much more. Exaggerated a bit for the sake of understanding the question.
Which has a higher concentration of H+ *pause*** The one that dissociated more. That’s your stronger acid. If you have a hard time remembering that, just think about what a strong acid is, its something that dissociates nearly completely, so therefore, the more something dissociates the stronger the acid.
How does that affect Ka, you may want to pause and think about this for a bit. If you need a hint, look at the ka equation **click**. As the strength of the acid increases what happens to each of the species in the equation. How does this effect the final number. ***pause***
Since as stronger acid has more H+ and A- and less HA, the ka will increase. **click** therefore the stronger the acid the larger the ka. Because of this we can now rank the strength of weak acids simply by knowing their ka. The larger the Ka, the stronger the acid.

10. What happens to Kb as the strength of the base increases?
If a base is stronger, which way does the equilibrium shift?
It shifts to the right, so what happens to the amount of each product?
So for a stronger base, Kb is higher!!!
It would be nice if we could judge the strength of a base by the kb. So lets look at how a bases strength’s affects its kb so we’ll have a rule to abide by when trying to determine how strong a weak base is.
We’ll approach this just like we did the ka and acid strength a few slides back. If we have a stronger base, what does that mean about the equilibrium shift. *pause** **click*** since a stronger base means it shifts to the right, to create more OH-, then lets look at how that effects each species in the reaction. *click*. For the original base we’ll have less since more of it has ionized. We’ll have more of the protonated base and Oh. **click*** since both of the ones that are increasing are in the numerator, and the one decreasing is in the denominator, the Kb must go up. Meaning that we can now rank how strong a base is by looking at the kb.

11. Ranking: Strong Base Weak Base Weak acid Strong acid HCl NaOH NH(CH3)2
Rank the following solutions in order of increasing acidity. Assume the concentration for each is the same, and that all are within solubility limits.
HCl
NaOH
NH(CH3)2
NH2(CH3)2+
CH3COOH
Ca(OH)2
Strong
Base
Weak
Base
Weak
acid
Strong
acid

12. Ranking: Strong Weak acid Weak Strong Base Base acid NH2(CH3)2+ NaOH
Rank the following solutions in order of increasing acidity. Assume the concentration for each is the same, and that all are within solubility limits.
Strong
Base
Weak
Base
Weak acid
Strong
acid
𝑘 𝑎 = 𝑘 𝑤 5.9𝑥 10 −4
=1.7𝑥 10 −11
NH2(CH3)2+
NaOH
NH(CH3)2
HCl
Ca(OH)2
CH3COOH
Ka=1.8x10-5

13. Acid/Base Conceptual Understanding Questions:
In what range must the pH of a 0.17M solution of a weak acid fall?
pH<7
pH>-log(0.17) pH=
What must be true about the [H+] of a weak acid
Neutral gives [H+]=[OH-]=10-7
Acidic has more [H+] so: >10-7
What must be true about the [OH-] of a weak acid solution?
Acidic has less [OH-] so: <10-7

14. Percent Ionization: Pictorial Representation
We’ll do some real examples in a moment
Original Acid: 14
Dissociated Acid: 3
% ionization= 3/14*100%= 21.4%

15. 𝐾 𝑎 = 𝐻 + 𝐴 − 𝐻𝐴 = 8𝑥 10 −4 8𝑥 10 −4 0.8−8𝑥 10 −4 = 8.00x 10 −7
Example: The percent dissociation of a M aqueous monoprotic weak acid is 0.10%. What is the Ka value for the acid?
HA ⇄ H A−
%𝐷𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛= 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 ∗100%
0.8 M 0M 0M I C E
0.001= 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡 𝑀
0.8-8x10-4M
8x10-4M
8x10-4M
𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑒𝑑 𝑎𝑚𝑜𝑢𝑛𝑡=8.00𝑥 10 −4 𝑀
𝐾 𝑎 = 𝐻 + 𝐴 − 𝐻𝐴 = 8𝑥 10 −4 8𝑥 10 − −8𝑥 10 −4 = 8.00x 10 −7

16. Polyprotic acids: K=Kw/Ka1 K=Kw/Ka2 What is the Kb of CO32-?
Poly=many protic=protons polyprotic=many proton atoms
Ka1
K=Kw/Ka1
Ka2
K=Kw/Ka2
What is the Kb of CO32-?

17. Molecular Structure and Strength of Acids
Hydrohalic Acids
HX, where X designates a halide (F, Cl, Br, I)
Two competing forces: Enthalpy and Polarity
Enthalpy of HI is much lower than HF, says HI is strongest
Bond polarity would make it seem as if HF should be the strongest
So which wins?
Bond enthalpy
weak
strong
When we are trying to figure out the trend there, there are two things to consider. First is bond enthalpy: if we look at where they are on the periodic table, we can see I is furthest down, making it the largest, which means it will take less energy to break the bond based on size and if we look at the enthalpy number this agrees.
The other thing we need to look at is how the polarity relates. Because F is very electronegative it is most stable as an anion so this would say that HF is most likely to lose the hydrogen and is therefore the strongest acid.
Bond enthalpy ends up winning this battle and this is the proper order.

18. Generalization of Acid Strengths Other Acids
The more you stabilize the anion, the stronger the acid
REMEMBER
Different Central Atoms, Same oxidation number (aka same number of attached groups)
Strength increases with increasing electronegativity of central atom.
Example: HClO3>HBrO3
Same central atom, different number of attached groups
Increases as oxidation number of central atom increases
Example: HClO4 >HClO3>HClO2>HClO
< +1 +3 +5 +7
Oxidation number
Strength of acid

19. Buffers Acid and its conjugate base, or a base and its conjugate acid
Or some combination of components which create this. (next slide)
It works by converting a strong acid into a weak acid, or a strong base into a weak base.
A strong base can’t exist in solution with a weak acid it must react
A strong acid can’t exist in solution with a weak base it must react

20. Can you make a buffer with?
NH3 and HCl?
Yes!
How: when NH3 and HCl react, they form NH4+
NH3 + H+ → NH4+
Gives us a
conjugate
acid base pair
0.20M
0.10M 0M I C F
-0.10M
-0.10M
+0.10M
0.10M 0M
0.10M

21. Buffer Calculation Example
A 100 mL buffer solution is 0.100M Nitrous acid and 0.100M Sodium nitrite. Calculate the pH if
0.005 moles of NaOH is added to the solution
calculate the pH of moles of HCl is added to the solution.
calculate the pH of mols of NaOH is added to the solution. Assume no change in volume.