Unit 5: Acids and Bases
CH. 14
Acid/Base Definitions
◦Arrhenius Model
◦
◦
Acids produce hydrogen ions in aqueous solutions
Bases produce hydroxide ions in aqueous solutions
◦Bronsted-Lowry Model
◦
◦
Acids are proton donors
Bases are proton acceptors
◦Lewis Acid Model
◦
◦
Acids are electron pair acceptors
Bases are electron pair donors
Acid Dissociation
HA  H+
+ AAcid
Proton Conjugate
base


[ H ][ A ]
Ka 
[ HA]
Alternately, H+ may be written in its hydrated form, H3O+
(hydronium ion)
Dissociation of Strong Acids
◦Strong Acids are assumed to
dissociate completely in
solution.
◦ Does this suggest a large or
small Ka?
◦ Does this suggest a reactant or
product favored equilibrium?
Dissociation Constants: Strong Acids
Strong Acids to know
AP College Board would like you to know the following
strong acids:
◦HCl, HBr, HI, HClO4, H2SO4, HNO3
Dissociation of Weak Acids
Weak acids are assumed to
dissociate only slightly (less
than 5%) in solution.
◦ Does this suggest a large or
small Ka?
◦ Does this suggest a reactant or
product favored equilibrium?
Dissociation Constant: Weak Acids
Auto-Ionization of Water
H2O + H2O

H3O+ + OH-
At 25, [H3O+] = [OH-] = 1 x 10-7M
Kw is a constant at 25 C:
Kw = [H3O+][OH-]
Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14
Amphoterism
The ability to act like both an acid and a base.
Homework
Pg.673-674
◦#16, 21a, 21b, 28-32, 35, 37, 38
Bellwork
Recall with somebody next to you:
◦ What are the 6 strong acids that the AP College Board requires you
to know?
◦ Where is the equilibrium position for strong acids and where is it
for weak acids? (left, right, or center)
The pH Scale
The pH scale is a log based
scale used to measure the
acidity of a solution.
◦ Smaller pH values are more
acidic.
◦ Bigger pH values are more
basic.
◦ pH levels at or very near 7
are considered to be
neutral.
pH and pOH
pH = -log10 (𝐻3 𝑂+ )
−
pOH = -log10 (𝑂𝐻 )
Relationship between pH and pOH
◦pH + pOH = 14
Weak vs Strong and pH
Since strong acids are assumed to completely dissociate, more
hydronium ions will be created.
◦ lower concentration  higher pH
Since weak acids are assumed to dissociate slightly, less hydronium
ions are created.
◦ Higher concentration  lower pH
Since pH depends on both concentration and strength, it takes a
much larger concentration for weak acids to have the pH of a
stronger acid at a smaller concentration.
Finding Concentration from pH or pOH
𝐻3 𝑂+ = 10−𝑝𝐻
𝑂𝐻− = 10−𝑝𝑂𝐻
pH and pOH Calculations
H+
[OH-] = 1 x 10-14
[H+]
OH-
pH
pOH = 14 - pH
pH = 14 - pOH
pOH = -log[OH-]
[OH-] = 10-pOH
pH = -log[H+]
[H+] = 10-pH
[H+] = 1 x 10-14
[OH-]
pOH
Practice
What is the pH of an acid with a [OH-] of 1 x 10-11M? Is this a strong
or weak acid?
pOH = -log(1 x 10-11M) = 11
pH = 14 – pOH = 14 – 11 = 3
Weak acid
Homework
Pg. 674
◦#41-48
Bellwork
Calculate the [H+] for a solution measured to have a pOH of 1.3.
pH = 14 – pOH = 14 – 1.3 = 12.7
[H+] = 10-pH = 10-12.7 = 2.0 x 10-13M
Weak Acid Equilibrium Problem
What is the pH of a 0.50M solution of acetic acid,
HC2H3O2, Ka = 1.8 x 10-5 ?
Step #1: Write the dissociation equation
HC2H3O2  C2H3O2- + H+
Weak Acid Equilibrium Problem cont.
What is the pH of a 0.50M solution of acetic acid, HC2H3O2,
Ka = 1.8 x 10-5 ?
Step #2: ICE it!
HC2H3O2  C2H3O2- + H+
I
C
E
0.50
-X
0.50 - X
0
+X
X
0
+X
X
Weak Acid Equilibrium Problem cont.
What is the pH of a 0.50M solution of acetic acid, HC2H3O2,
Ka = 1.8 x 10-5 ?
Step #3: Set up the law of mass action
HC2H3O2  C2H3O2- + H+
2
( x)( x)
x
1.8 x 10 

(0.50  x) (0.50)
5
Weak Acid Equilibrium Problem cont.
What is the pH of a 0.50M solution of acetic acid, HC2H3O2,
Ka = 1.8 x 10-5 ?
Step #4: Solve for x, which is also [H+]
HC2H3O2  C2H3O2- + H+
2
x
1.8 x10 
(0.50)
5
[H+] = 3.0 x 10-3 M
Weak Acid Equilibrium Problem cont.
What is the pH of a 0.50M solution of acetic acid, HC2H3O2,
Ka = 1.8 x 10-5 ?
Step #5: Convert [H+] to pH
HC2H3O2  C2H3O2- + H+
-35
pH   log(3.0 x 10 )  42.52
.52
Strong Bases
◦Strong bases are metallic hydroxides
◦ Group I hydroxides (NaOH, KOH) are very soluble
◦ Group II hydroxides (Ca, Ba, Mg, Sr) are less soluble
◦pH of strong bases is calculated directly from the
concentration of the base in solution
Weak Bases
Dissociation of Bases
B + H2O  BH+ + OH

[ BH ][OH ]
Kb 
[ B]
Relationship Between Ka and Kb
Kw = [H3

+
O ][OH ]
[OH ][ BH ]
Kb 
[ B ]

[ H 3O ][ A ]
Ka 
[ HA]
Ka x Kb = Kw
Homework
P.674-676
◦#49, 50, 53-59, 61, 63, 65, 66, 71-79, 82, 83, 86
Polyprotic Acids
An acid that can donate more than one proton.
H3PO4 + H2O  H3O+ + H2PO4H2PO4- + H2O  H3O+ + HPO42HPO42- +H2O  H3O+ + PO43With weak acids, all species are minor and so the Hydronium from each
separate reaction affects the pH, but H3PO4 has the dominant equilibrium.
Strong Polyprotic Acids
The initial acid is a strong acid, but the acidic ions produced are not.
H2SO4 + H2O  H3O+ + HSO4HSO4 + H2O  H3O+ + SO42◦ The hydronium produced by the second acid dissociation may not be
large enough to affect the pH.
◦ To find out we can create an ICE box problem and assume x is small. If the
assumption holds true, then the second dissociation doesn’t produce enough
hydronium.
◦ If the assumption proves false, then the second dissociation does affect the pH.
Salts that produce Neutral Soln.
If the salt is made with the cation of a strong base and the anion of a
strong acid, the resulting solution will have no effect on the [H+]
when dissolved in water.
Examples:
◦ KCl
◦ NaNO3
Salts that produce Basic Soln.
If the salt has a cation with a neutral properties (Na+ or K+) and an
anion that is the conjugate base of a weak acid, the resulting
solution is basic.
Example:
◦ NaC2H3O2
Salts that produce Acidic Soln.
Salts that have an anion that is not a base and a cation that is the
conjugate acid of a weak base produce acidic solutions.
Example:
◦ NH4Cl
Salts with a highly charged metal ion also produce an acidic solution.
Example:
◦ AlCl3
Bellwork
Make a list of all of the different legal beverages that you will
normally drink on a weekly basis.
Write down any beverages that your peers legally drink that you
won’t?
Check it out!
Pg. 660
◦ Table 14.6
The above table gives a good break down of what types of salts
produce which type of solution.
Structural Impact
Weaker bonds and increased electronegativity/polarity increases the
strength of the acid.
For acids containing oxygen, the more oxygen atoms attached to the
central atom in the acid, the stronger the acid is.
Oxyacids
H-O-X
Acidic Solution: If X has a high electronegativity, the O-X bond will be
strong and covalent. This will result in a relatively weaker O-H bond
to break releasing a proton.
Basic Solution: If X has a low electronegativity, the O-X bond will be
weak and ionic. This will result in a relatively stronger O-H bond to
stay intact as the O-X bond will break in polar water, releasing the
metal cation and the hydroxide ion
Homework
Pg. 676
◦#93-99, 101, 102, 105, 106, 112-118
Bell Work
Recall: Which acids and bases are strong?
◦ Check with a partner.
Lewis Acids and Bases
Lewis Acid: Electron pair acceptor
Lewis Base: Electron pair donor
Lewis Acids cont.
A stronger Lewis acid is one that is smaller and more
positively charged than another acid.
◦ B3+ is a strong Lewis acid as it has a large positive charge,
making it more attracted to an electron pair.
Practice
Identify the Lewis acid and base for each reaction:
Ni2+(aq) + 6NH3(aq) ↔ Ni(NH3)62+(aq)
H+(aq) + H2O(l) ↔ H3O+(aq)
HCl(aq) + H2O(l)  H3O+(aq) +Cl-(aq)
Solving Acid Base Problems
◦Take a look at each species and identify if it is an acid or a base.
◦Identify the strength
◦ If a strong acid or base is present, assume it goes to completion.
◦ Calculate the concentration of the product
◦ If a weak acid or base is present, draw an ICE box and fill out any given
information.
◦
◦
◦
◦
Assume the acid or base dissociation is small
Define the equilibrium concentrations in terms of “x”
Solve for “x”
Check the assumption
◦ If there the species is polyprotic, choose the reaction that will most affect the
pH.
Homework
Pg. 677
◦ # 119-123
HOCl ↔ OCl- + H+
Hypochlorous acid, HOCl, is a weak acid commonly used as a bleaching
agent. The acid–dissociation constant, K , for the reaction represented above
is 3.2×10–8.
(a) Calculate the [H ] of a 0.14–molar solution of HOCl.
(b) Write the correctly balanced net ionic equation for the reaction that
occurs when NaOCl is dissolved in water and calculate the numerical
value of the equilibrium constant for the reaction.
(e) Household bleach is made by dissolving chlorine gas in water, as
represented below.
Cl + H O → H + Cl + HOCl
Calculate the pH of such a solution if the concentration of HOCl in the
solution is 0.065 molar.
a
+
2(g)
2
+
–
(aq)
FRQ Practice
A comparison of the theories Arrhenius, Brønsted and Lewis shows a
progressive generalization of the acid base concept. Outline the
essential ideas in each of these theories and select three reactions,
one that can be interpreted by all three theories, one that can be
interpreted by two of them, and one that can be interpreted by only
one of the theories. Provide these six interpretations.
Answer
Arrhenius
acid = produce H+ ions in aqueous solution
base = produce OH– ions in aqueous solution
Brønsted–Lowry
acid = proton(H+) donor; base = proton
acceptor
Lewis
acid = e– pair acceptor; base = e– pair donor
Examples:
Interpreted by all three
HCl + H2O → H+(aq) + Cl–(aq)
NaOH + H2O → Na+(aq) + OH–(aq)
Interpreted by two
NH3 + HCl ↔ NH4+ + Cl–
Interpreted by only one
BF3 + NH3 → F3B:NH3
Mo’ Practice… Mo’ Perfect
Predict whether solutions of each of the following salts are
acidic, basic, or neutral. Explain your prediction in each case
(a) Al(NO3)3
(b) K2CO3
(c) NaBr
Answers
(a)
acidic; Al3+ + H2O ↔ AlOH2+ + H+;
hydrolysis of Al3+;
Al(OH2)n3+ as Brønsted acid, etc.
(b)
basic; CO32– + H2O ↔ HCO3– + OH– ; or
hydrolysis of CO32– as conjugate to a weak acid, etc.
(c)
neutral; Na+ from strong base; Br– from strong acid
Because, why not.
Methylamine CH3NH2, is a weak base that ionizes in solution as
shown by the following equation.
CH3NH2 + H2O ↔ CH3NH3+ + OH–
(a) At 25ºC the percentage ionization in a 0.160 molar solution of
CH3NH2 is 4.7%. Calculate [OH–], [CH3NH3+], [CH3NH2], [H3O+], and
the pH of a 0.160 molar solution of CH3NH2 at 25ºC
(b) Calculate the value for Kb, the ionization constant for CH3NH2, at
25ºC.
Answers
(a) CH3NH2; 0.160M x 4.7% = 7.5×10–3 M ionizing
[CH3NH2] = (0.160M – 0.0075M) = 0.0152M @ equilibrium
[CH3NH3+] = [OH–] = 7.5×10–3 M
Kw
-12
+
[H3O ] =
=
1.3

10
M
3
7.5  10
pH = –log [H3O+] = 11.89
+
3 2
[CH 3NH 3 ][OH ] (7.5  10 )

(b) Kb =
=3.710–4
[CH 3NH 2 ]
0.152
And just like the energizer bunny…
Sodium benzoate, C6H5COONa, is the salt of a weak acid, benzoic
acid, C6H5COOH. A 0.10 molar solution of sodium benzoate has a pH
of 8.60 at room temperature.
(a) Calculate the [OH–] in the sodium benzoate solution described
above.
(b) Calculate the value for the equilibrium constant for the reaction:
C6H5COO– + H2O ↔ C6H5COOH + OH–
(c) Calculate the value of Ka, the acid dissociation constant for
benzoic acid.
Answer
(a) pH =8.6, pOH =5.4
[OH–] =10–pOH = 3.9810–6M
(b) [C6H5COOH] = [OH–]
[C6 H 5COOH][OH- ]
(3.98  10 6 )2
–10

Kb =
=
1.58

10
[C6 H 5COO- ]
(0.10  3.98  10 6 )
K w 1.0  10 14
-5

(c) Ka =
=
6.33

10
K a 1.58  10 10
Answer the following questions that relate to the chemistry of halogen oxoacids.
(a) Use the information the table below to answer part (a)(i).
Acid
Ka at 298 K
HOCl
2.9  10-8
HOBr
2.4  10-9
(i) Which of the two acids is stronger, HOCl or HOBr? Justify your answer in terms of Ka.
(ii) Draw a complete Lewis electron-dot diagram for the acid that you identified in part (a)(i).
(iii) Hypoiodous acid has the formula HOI. Predict whether HOI is a stronger acid or a weaker
acid than the acid that you identified in part (a)(i). Justify your prediction in terms of
chemical bonding.
(b) Write the equation for the reaction that occurs between hypochlorous acid and water.
(c) A 1.2 M NaOCl solution is prepared by dissolving solid NaOCl in distilled water at 298 K.
The hydrolysis reaction OCl-(aq) + H2O(l)  HOCl(aq) + OH-(aq) occurs.
(i) Write the equilibrium-constant expression for the hydrolysis reaction that occurs between
OCl-(aq) and H2O(l).
(ii) Calculate the value of the equilibrium constant at 298 K for the hydrolysis reaction.
(iii) Calculate the value of [OH-] in the 1.2 M NaOCl solution at 298 K.
Answer:
(a) (i) HOCl, larger Ka means higher percentage of ionization and, therefore, a stronger acid
gg
gg
gg
gg
(ii) H-O Cl :
(iii) weaker, the H-O bond strength is stronger when the other element bonded to the oxygen
has a lower electronegativity (iodine has a lower EN than chlorine). Thus a H-O-I
molecule holds on to its H more strongly than HOCl and less H+ ionizes
(b) HOCl + H2O  OCl– + H3O+
[HOCl][OH- ]
Kw
1  10 14
(c) (i) & (ii)
= 3.4  10-7
 Kb 

–
8
[OCl ]
K a 2.9  10
(iii) [HOCl] = [OH-] = X
[OCl-] = 1.2 - X
X2
= 3.4  10-7 ; X = 6.4  10-4 M = [OH-]
1.2  X
HOBr(aq) ↔ H+(aq) + OBr–(aq)
Ka = 2.3  10–9
Hypobromous acid, HOBr, is a weak acid that dissociates in water, as represented by the equation
above.
(a) Calculate the value of [H+] in an HOBr solution that has a pH of 4.95.
(b)
Write the equilibrium constant expression for the ionization of HOBr in water, then
calculate the concentration of HOBr(aq) in an HOBr solution that has [H+] equal to 1.8  10–5 M.
(e)
HOBr is a weaker acid than HBrO3. Account for this fact in terms of molecular structure.
Answer:
(a) pH = –log[H+]; 4.95 = –log[H+]
[H+] = 1.12  10–5
[H + ][OBr - ]
(b) Ka =
= 2.310–9
[HOBr]
[H+] = [OBr–] = 1.810–5 M
(1.8  10 5 )2
= 2.310–9
X
X = [HOBr] = 0.14 M
(e) very electronegative oxygen is able to draw electrons away from the bromine and weaken the O–H bond,
making it easier for the hydrogen ion “to leave”.
Give a brief explanation for each of the following.
(a) For the diprotic acid H2S, the first dissociation constant is larger than the second
dissociation constant by about 105 (K1 ~ 105 K2).
(b) In water, NaOH is a base but HOCl is an acid.
Answer:
(a) After the first H+ is lost from H2S, the remaining species, HS–, has a
negative charge. This increases the attraction of the S atom for the bonding
electrons in HS–. Therefore, the bond is stronger, H+ is harder to remove,
and K2 is lower.
(b) Polar H2O can separate ionic NaOH into Na+(aq) and OH–(aq), giving a basic
solution. In HOCl, chlorine has a high attraction for electrons due to its
greater charge density. This draws electrons in the H–O bond towards it
and weakens the bond. H+ can be removed, making an acidic solution.