1. Acid-Base Equilibria

2. Arrhenius Theory
The simplest definition of an acid is a substance capable of producing H+(aq) ions in solution.
For strong acids, this equilibrium lies so far to the right that you can make the mathematical assumption that [HX]E ≈ 0.
These “simple” acids are known as Arrhenius acids. The Arrhenius definition of an acid is restricted to aqueous acid solutions.
HXaq
H+aq + X-aq
An Arrhenius base is a substance that produces OH-(aq) ions in aqueous solution.
MOHaq
M+aq + OH-aq

3. Bronsted Theory
A more inclusive definition of acids and bases is presented by the Bronsted theory
A Bronsted acid is a H+ (proton) donor. A Bronsted base is a proton acceptor.
In this unit, we define the hydronium ion, H3O+(aq). This species forms spontaneously when H+(aq) is produced in solution, which causes the protonation of water.
1 𝐻𝐴 𝑎𝑞 𝐻 + 𝑎𝑞 + 𝐴 − (𝑎𝑞)
2 𝐻 2 𝑂 𝐿 + 𝐻 + 𝑎𝑞 𝐻 3 𝑂 + (𝑎𝑞)
𝑛𝑒𝑡: 𝐻𝐴 𝑎𝑞 + 𝐻 2 𝑂 𝐿 𝐻 3 𝑂 + 𝑎𝑞 + 𝐴 −

4. Bronsted Theory
In the reaction below, water acts as a Bronsted base, which accepts a proton (H+) from the Bronsted acid, HA. As you can see, HA is both an Arrhenius acid and Bronsted acid.
All Arrhenius acids are also Bronsted acids.

5. Bronsted Theory
The Bronsted definition is especially important with regard to bases. Consider ammonia, NH3.
NH3 is a very important base, both industrially and in nature. However, NH3 possesses no –OH groups, so it is not an Arrhenius base. It behaves only as a Bronsted base, accepting protons from water to generate OH- as a product.
Thus, Bronsted bases are not necessarily Arrhenius bases. In the reaction above, we also see that water is behaving as a Bronsted acid.

6. Autoionization of Water
Pure water is equally acidic and basic, and it is therefore considered to be neutral. Because of this, water is capable of reacting with itself:
In water, there are always ions of hydronium and hydroxide present. The equilibrium constant expression for this autoionization reaction is:
At 25oC, the value of Kw is 1 x M2.
In pure water, the concentrations of hydronium and hydroxide are equal.
H2O(l) + H2O(l)
H3O+(aq) + OH−(aq)
𝑲 𝒘 = 𝑯 𝟑 𝑶 + [𝑶 𝑯 − ]

7. Acid Strength
The strength of an acid is measured by its tendency to dissociate (HX  H+ + X-)
The stronger the bond, the poorer the extent of dissociation and the weaker the acid.
There are a class of acids known as strong acids, which are characterized by very weak bonds. These species will completely dissociate in water.

8. Strong Bases
Strong bases are metal salts of hydroxides.

9. Conjugate Acid-Base Pairs
In any acid-base equilibrium, the forward and reverse reactions involve proton transfers.
When an acid, HA, loses its H+(aq), the A-(aq) left behind can behave as a base. Likewise, when a base accepts the H+(aq), it can now act as an acid.
An acid and a base such as HA and A- that differ only in the presence of a proton are called a conjugate acid-base pair. Every acid has a conjugate base, and visa-versa.

10. Relative Strengths of Acids and Bases
The stronger an acid/base, the weaker the conjugate base/acid.
neutral
A strong acid completely transfers its protons into water. Its conjugate base is neutral and has a negligible tendency to be protonated.
𝐇𝐂𝐥 𝐚𝐪 + 𝐇 𝟐 𝐎 𝐋 𝐇 𝟑 𝐎 + 𝐚𝐪 +𝐂 𝐥 − (𝐚𝐪) x
A weak acid partially dissociates, and so it exists as a mixture of reactants and products. The conjugate of a weak acid is a weak base.
neutral

11. Example
Write the conjugate bases of the following acids, and label both species as strong, weak, or neutral
Nitric acid, HNO3 (aq)
Phosphoric acid, H3PO4 (aq)
Formic acid, HCOOH (aq)
Write the conjugate acids of the following bases, and label both conjugates as strong, weak, or neutral
SO32-(aq)
CH3NH2 (aq)
I-(aq)
*NaOH (aq)
HNO3 (strong), NO3- (neutral)
H3PO4 (weak), H2PO4- (weak)
HCOOH (weak), HCOO- (weak)
SO32- (weak), HSO3- (weak)
CH3NH2 (weak), CH3NH3+ (weak)
I- (neutral), HI (strong)
OH- (strong), H2O (neutral)
*Na+ is a spectator

12. pH
The rates of many chemical reactions depend on the concentration of [H3O+]. However, [H3O+] can range vastly.
It is very difficult to plot such a large range of values, so we use a logarithmic scale to measure [H3O+]. We call this the pH scale.
Recall that for pure water, we have the equilibrium:
And in pure water:
So in pure water, 𝑯 𝟑 𝑶 + = 10-7 M, and pH = 7.00
𝐩𝐇=−𝐥𝐨𝐠 [ H3O+ ]
𝐊 𝐰 = 𝐇 𝟑 𝐎 + 𝐎 𝐇 − =𝟏 𝟎 −𝟏𝟒 𝐌 𝟐
𝐇 𝟑 𝐎 + = 𝐎 𝐇 −

13. pOH and the PH scale
In terms of basicity, we can introduce a quantity pOH.
For a basic solution, you can not directly calculate pH. You must calculate pOH first. pH and pOH are related by the following:
𝒑𝑶𝑯=−𝒍𝒐𝒈 [OH−]
𝒑𝑯+𝒑𝑶𝑯=𝟏𝟒
* at 25oC

14. Understanding the pH scale
Since the pH scale is logarithmic, each pH unit corresponds to an order of magnitude
[H3O+] = 10-3 = 0.001M pH = 3
[H3O+] = 10-2 = 0.01M pH = 2
[H3O+] = 10-1 = 0.1M pH = 1
[H3O+] = 100 = 1M pH = 0
[H3O+] = 101 = 10M pH = -1
So, if the pH of a solution is lowered by 3 pH units, the H+ concentration has increased by a factor of 103, or 1000.

15. pH scale and Important Notes
For this class, we will assume the temperature of all aqueous solutions to be exactly 25oC so that the previous expression holds true.
It is important to know the following:
An acidic solution is defined as one in which [H3O+] > [OH-] (pH < 7)
A neutral solution is defined as one in which [H3O+] = [OH-] (pH = 7)
A basic solution is defined as one in which [OH-] > [H3O+] (pH > 7)
𝒓𝒖𝒍𝒆 𝒐𝒇 𝒍𝒐𝒈𝒔: 𝟏 𝟎 𝐥𝐨𝐠⁡(𝒙) =𝒙
this will come in handy

17. Example
Calculate the pH and pOH of a solution having [H3O+] = 9.14 x 10-5M
A solution has a pH of Determine [H3O+] and [OH-]. Is this solution acidic or basic?

18. Example What is the pH of a 0.040M solution of HClO4 at 25oC?
Perchloric acid is a strong acid, so we can assume complete dissociation.
𝐻𝐶𝑙 𝑂 4 𝑎𝑞 + 𝐻 2 𝑂(𝐿)→ 𝐻 3 𝑂 + 𝑎𝑞 +𝐶𝑙 𝑂 4 − 𝑎𝑞
I M
C -(~0.040) (~0.040 M) +(~0.040 M)
E ~ ~0.040M ~ 0.040M
𝑝𝐻=− log =1.40
The dissociation of the acid may also be written as:
𝐻𝐶𝑙 𝑂 4 𝑎𝑞 → 𝐻 + 𝑎𝑞 +𝐶𝑙 𝑂 4 − 𝑎𝑞
0.040M M
It is a matter of preference.

19. Example What is the pH of a 0.028M solution of NaOH at 25oC?
NaOH is a strong base (it is also a salt, and so OH- is directly produced by complete dissociation). To find pH, we must calculate pOH first.
𝑁𝑎𝑂𝐻(𝑎𝑞)→ 𝑂𝐻 − 𝑎𝑞 + 𝑁𝑎 + 𝑎𝑞
E ~0M ~.028M ~.028M
𝑝𝑂𝐻=− log =1.55
𝑝𝐻=14−pOH
𝑝𝐻=12.45

20. Acid Dissociation Constant, Ka
Let’s consider a weak acid. The equilibrium expression of this dissociation is defined by Ka, where the subscript indicates that the species is an acid:
HA (aq)+ H2O(L) H3O+(aq) + A-(aq)
K a = H 3 O + [ A − ] [HA]
The size of Ka describes mathematically how readily an acid dissociates. The stronger the acid, the higher the value of Ka. Any acid that is not a strong acid is assumed to be a weak acid.
We can measure the extent of dissociation of the acid by calculating the % ionization
% 𝑖𝑜𝑛𝑖𝑧𝑎𝑡𝑖𝑜𝑛= [ 𝐻 3 𝑂 + ] 𝐸 [𝐻𝐴 ] 𝐼 𝑥 100%

21. Example
The pH of a 0.40 M formic acid solution (HCOOH) is What is the concentration of H3O+? What is Ka? What % of the acid dissociates?
Formic acid is a weak acid. Weak acids do not fully dissociate
HCOOH (aq)+ H2O(L) H3O+(aq) + HCOO-(aq)
I M
C -x x x
E x x x
2.08=− log 𝑥
𝐾 𝑎 = 𝑥 −𝑥 =1.76 x 1 0 −4
10 −2.08 =[𝑥]
.0083 𝑀=𝑥
𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛= [.0083 𝑀] [0.40 𝑀] 𝑥 100=2.07%

22. Group Example
Calculate the pH of a M acetic acid solution given that Ka = 1.8 x 10-5 M
CH3COOH (aq) + H2O(L) H3O+(aq) + CH3COO-(aq)
1. We set up the ICE table to determine the concentration of H3O+
Concentration
CH3COOH
H3O+
CH3COO-
Initial
0.050 M
Change
- x
+ x
Equilibrium
0.050 – x x
2. Solve for x. We see that Ka is very small because CH3COOH is a weak acid.
𝑥 −𝑥 =1.8 𝑥 1 0 −5 𝑀
𝑥=9.4 𝑥 1 0 −4
3. Calculate pH
𝑝𝐻=− log =3.03

23. Method of Successive Approximations
Weak acids tend to dissociate quite poorly, so the percentage of dissociation is very small. Hence, Ka is very small as well. Let’s revisit the previous example.
Since x is so small:
Therefore:
𝑥 −𝑥 =1.8 𝑥 1 0 −5 𝑀
0.050 −𝑥 ≅0.050
𝑥 =1.8 𝑥 1 0 −5 𝑀
𝑥 2 =9 𝑥 1 0 −7 𝑀 2
Use this method when K < 10-4
𝑥=9.4 𝑥 1 0 −4 𝑀

24. pKa
Ka values range over many orders of magnitude. Therefore, it is more convenient to use the quantity pKa
𝑝 𝐾 𝑎 =− log 𝐾 𝑎
Physically, pKa represents the pH at which 50% of a weak acid is deprotonated.
Lower pKa means stronger acid!

25. Interpreting pKa
By knowing the pKa of an acid, we can predict its form at given pH values.
At pH > pKa +1, the acid is primarily in a deprotonated form.
At pH = pKa, the acid and conjugate base are present in equal ratio
At pH < pKa -1, the acid is primarily protonated
Ex. Predict the form of Acetic acid CH3COOH (pKa = 4.74) at the following pH values:
pH = 3
pH = 4.74
pH = 5.3
pH = 9
𝑝𝑟𝑖𝑚𝑎𝑟𝑖𝑙𝑦 CH 3 COOH
𝐸𝑞𝑢𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡𝑠 𝑜𝑓 CH 3 COOH and CH3CO O −
𝑀𝑜𝑠𝑡𝑙𝑦 CH3CO O − 𝑤𝑖𝑡ℎ 𝑠𝑜𝑚𝑒 CH3COOH
𝑝𝑟𝑖𝑚𝑎𝑟𝑖𝑙𝑦 CH3CO O −

26. Amines
Similar to acids, we can describe the protonation of bases. For example, lets look at the equilibrium involving ammonia in water.
The lone electron pair of electrons on the Nitrogen allows it to accept an H+ ion (proton).
Ammonia falls under a class of bases known as amines, which are composed of a central nitrogen atom with 3 bonding domains.
Amines react with acids to form ammonium salts (positively charged central N atom with 4 bonding domains). The product of the reaction above would be the salt of ammonium hydroxide.

27. Kb and pKb
We can use the base protonation constant Kb, or pKb to describe the degree of the dissociation of a base.
Higher Kb means stronger base
Example: Determine the pH of a 0.75 M hydroxylamine, HONH2(aq) solution given that Kb = 8.7 x 10-9 M
We can write out the equilibrium expression showing the base protonation.
HONH2 (aq) + H2O(L) HONH3+ (aq) + OH-(aq)
𝐾 𝑏 = 𝐻𝑂 𝑁𝐻 [ 𝑂𝐻 − ] [ 𝐻𝑂𝑁𝐻 2 ]

28. Example, contd. Set up ICE table Solve x
Concentration
HONH2
HONH3+
OH-
Initial
0.75 M
Change
- x
+ x
Equilibrium
0.75 – x x
Solve x
𝑥 −𝑥 =8.7 𝑥 1 0 −9 𝑀
𝑥 =8.7 𝑥 1 0 −9 𝑀
𝑥=8.07 𝑥 1 0 −5
THIS IS A BASE. That means that you must calculate pOH FIRST. Then, determine pH.
𝑝𝑂𝐻=− log 𝑥 1 0 −5 =4.09
𝑝𝐻=14−𝑝𝑂𝐻=9.91

29. Acidic and Basic Salts
Suppose we dissolve NaF(s) in water. The result would be an aqueous solution of Na+(aq) and F-(aq). You would expect this solution to be neutral.
However, F-(aq) is the conjugate base of a weak acid, HF
𝐻𝐹 𝑎𝑞 + 𝐻 2 𝑂 𝐿 𝐻 3 𝑂 + 𝑎𝑞 + 𝐹 − (𝑎𝑞)
acid
base
conjugate acid
conjugate base
Therefore, a solution of F-(aq) ions will be weakly basic.
𝐹 − 𝑎𝑞 + 𝐻 2 𝑂 𝐿 𝐻𝐹 𝑎𝑞 + 𝑂𝐻 − (𝑎𝑞)
weak base
negligible acid
weak acid
strong base
Na+(aq) is a cation of a strong base, NaOH. Thus, it has negligible acidity and does not affect the pH. So it is considered a neutral cation.

30. Acidic and Basic Salts
Neutral cations are the cations of strong bases.
Neutral anions are the anions of strong acids.
Basic anions are the anions of weak acids.
Acidic cations are the cations of weak bases

31. Polyprotic Acids
A polyprotic acid is an acid than can donate more than one H+.
For example, sulfuric acid, H2SO4, is a strong, diprotic acid. Thus, it can undergo two deprotonations:
We can assume 100% dissociation of the H2SO4 in step 1. However, the remaining conjugate species, HSO4-, is a much weaker acid. EACH PROTON IS SUBSTANTIALLY HARDER TO REMOVE THAN THE PREVIOUS.
Therefore, the degree of dissociation of the weaker acid is determined from its Ka
1 𝐻 2 𝑆𝑂 4 𝑎𝑞 → 𝐻 + 𝑎𝑞 +𝐻𝑆 𝑂 4 −
2 𝐻𝑆 𝑂 4 − 𝑎𝑞 → 𝐻 + 𝑎𝑞 +𝑆 𝑂 4 2−

32. Polyprotic Acids
Ka values of polyprotic acids are labeled Ka1, Ka2, and so forth to denote each deprotonation.
The table to the left shows pKa values of common polyprotic acids.
Remember, pKa = -log Ka

33. Example Calculate the pH of a 0.10 M H2SO4(aq) solution.
1.) We first have the initial deprotonation. Because sulfuric acid is a strong acid, you can assume that all of it is dissociated at equilibrium.
𝐻 2 𝑆𝑂 4 𝑎𝑞 + 𝐻 2 𝑂(𝐿)→ 𝐻 3 𝑂 + 𝑎𝑞 +𝐻𝑆 𝑂 4 − (𝑎𝑞)
E 𝑀 𝑀 𝑀
2.) Now, we have the second deprotonation. Use the concentrations of H3O+ and HSO4- from above as your initial values. Use pKa values from the table to determine the final concentration of [H3O+]
𝐻 𝑆𝑂 4 − 𝑎𝑞 + 𝐻 2 𝑂 𝐿 𝐻 3 𝑂 + 𝑎𝑞 +𝑆 𝑂 4 2−
𝑝𝐾𝑎 2 =1.99
Concentration
HSO4-
H3O+
SO42-
Initial
(0.10 M)
Change
- x
+ x
Equilibrium
0.10 – x
0.10+ x x
𝐾 𝑎 2 =1 0 −1.99 = .01
𝑥=.0085
𝑝𝐻=− log = .96

34. H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(L) + Na+(aq) + Cl-(aq)
The Reaction of Strong Acids and Strong Bases is A Double-Replacement Reaction Known as a Neutralization Reaction
When acids and bases react, they neutralize each other, and the product is salt and water
HCl (aq) + NaOH(aq) H2O(L) + NaCl(aq)
This is a double replacement reaction. The sodium and chloride ions are spectators (do not undergo any physical change). The net ionic equation is:
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(L) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) H2O(L)

35. Titrations
Knowing that acids and bases neutralize each other, lets imagine that we have an acid or base of unknown concentration.
How can we find the concentration?
Perform a titration

36. Titrations In a titration, an indicator is added to the base solution.
In the example to the right, as long as the pH is above 7 (basic) the indicator will make the solution pink.
An exact volume of an acid solution is added to a buret.
The acid solution is added drop-by-drop until the solution just turns clear (neutralized, pH =7 ).
At this point, you have a stoichiometric equivalence of acid and base.

37. Titrations
Before
titration
After titration
Say we have 100 mL of a basic NaOH solution of an unknown concentration.
We titrate with 5 mL of 1.0 M HCl, and the solution just turns clear.
NaOH(aq) + HCl(aq)  H2O(L) + NaCl(aq)
We know that the acid and base are completely neutralized, and none is left in solution.
Moles of acid added = Stoichiometric equivalent of base
.005 𝐿 𝐻𝐶𝑙 𝑥 1.0 𝑚𝑜𝑙 𝐻𝐶𝑙 𝐿 𝐻𝐶𝑙 𝑥 1 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻 1 𝑚𝑜𝑙 𝐻𝐶𝑙 = .005 𝑚𝑜𝑙 𝑁𝑎𝑂𝐻
.005 𝑚𝑜𝑙 .100 𝐿 =.05 𝑀 𝑁𝑎𝑂𝐻
Concentration of base solution =

38. Examples
It is found that 24.6 mL of 0.30M H2SO4 is required to neutralize 20.0 mL of NaOH (aq). Determine the molarity of NaOH (aq), and the molarity of the salt product.
TRY THIS AT HOME
To test the purity of NaOH tablets, a stock solution is made by dissolving a 0.40g sample into 100 mL of water. This solution is then titrated with a 0.100M HCl (aq) solution. It is found that 25 mL of the NaOH (aq) solution is neutralized by 23.2 g of the HCl (aq) solution. Determine the % of the tablet that is NaOH.