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Summary (CAIE) Cambridge A Level Chemistry (9701)
Summary & 15’s Acids and Bases Problems with Solutions

Acids and Bases

The Acid – Base Theory
Comparison
Inventor & Years

Arrhenius

Bronsted-Lowry

Lewis

Svante August Arrhenius

Johan Nicolaus Bronsted &
Thomas Martin Lowry
(1923)

Gilbert N
...

The remaining acid ions: Negative ions are
formed from acids after releasing H+ ions
...

The remaining base ions: Positive ions are
formed from bases after releasing OH- ions
...
One of the most glaring is in its treatment of the weak base
ammonia, NH3
...
Where is the OH- in NH3? To get
around this difficulty, chemists began to think of aqueous solutions of NH 3 as containing the compound
ammonium hydroxide, NH4OH, which as a weak base is partially ionized into NH4+ and OH- ions:

NH3(g) + H2O(l)

NH4OH(aq)

NH4OH(aq)  NH4+(g) + OH-(aq)

The Bronsted - Lowry Acid – Base Theory
Acid = proton (H+) donor

Acid  H+ + conjugate base

Base = proton (H+) acceptor

Base + H+  conjugate acid

Example:
Determine the conjugate acid-base pairs in the following reactions
...
This means that water is
acting as a Bronsted-Lowry base
...


Water can also act as an acid
...


CH3COOH(aq) + H2O(l)  CH3COO-(aq) + H3O+(aq)
Acid

Base

Base

Acid

In reaction above, CH3COOH acts as an acid
...
Thus, H2O acts as a base
...

When CH3COOH loses a proton, it is converted into CH3COO-
...
Species that differ by a single proton (H+)
constitute a conjugate acid-base pair
...
Thus, for the reaction above, we can
identify two conjugate acid-base pairs
...

A Lewis acid-base reaction occurs when a base donates a pair of electrons to an acid
...

The boron atom in boron trifluoride, BF3, has only six electrons in its valence shell
...

A decrease of 1 pH unit means that
the acidity of the solution increases
10x
...

A decrease of 1 pOH unit means
that the alkalinity of the solution
increases 10x
...
log 10 = n
pH = - log (a
...
log 10
= n – log a
Hydronium [H3O+] and hydroxide [OH-] ions are present both in pure water and in all
aqueous solutions, and their concentrations are inversely proportional as determined by
the ion product of water (Kw)
...
A
solutions is neutral if it contains equal concentrations of hydronium and hydroxide ions;
acidic if it contains a greater concentration of hydronium ions than hydroxides ions; and
basic if it contains a lesser concentration of hydronium ions than hydroxide ions
...
One such scale that is popular for chemical
concentrations and equilibrium constants is based on the p-function, defined as shown
where “X” is the quantity of interest and “log” is the base-10 logarithm:

pX = -log X
The pH of a solution is therefore defined as shown below, where [H3O+] is the molar
concentration of hydronium ion in the solution:

pH = -log[H3O+]
Rearranging this equation to isolate the hydronium ion molarity yields the equivalent
expression:

[H3O+] = 10-pH
Likewise, the hydroxide ion molarity maybe expressed as a p-function, or pOH:

pOH = -log[OH-]
or

[OH-] = 10-pOH
Finally, the relation between these two ion concentration expressed as p-functions is
easily derived from the Kw expression:

Kw = [H3O+][OH-]
-log Kw = -log [H3O+][OH-] = -log [H3O+] + -log [OH-]
pKw = pH + pOH

At 25°C, the value of Kw is 1
...
0 x 10-7 M at 25°C
...
0 x 10-7) = 7
...
0 x 10-7) = 7
...
In this reaction, one water molecule acts as an acid and the other
acts as a base
...
Therefore, the equilibrium constant for the
autoionization of water is:
Kc = [H3O+][OH-]
Because we use H+(aq) and H3O+(aq) interchangeably to represent the hydrated proton, the
equilibrium constant can also expressed as
Kc = [H+][OH-]
To indicate that the equilibrium constant refers to the autoionization of water, we replace
Kc by Kw

Kw = [H3O+][OH-] = [H+][OH-]

Where Kw is called the ion-product constant, which is the product of the molar
concentrations of H+ and OH- ions at a particular temperature
...
A strong acid yield 100% (or very nearly
so) of H3O+ and A- when the acid ionizes in water
...

The equilibrium constant for an acid is called the acid-ionization constant, Ka:

Ka =

[𝐻3 𝑂+ ][𝐴− ]
[𝐻𝐴]

Where the concentrations are those at equilibrium
...
The larger the
Ka of an acid, the larger of the concentration of H3O+ and A- relative to the concentration of
the nonionized acid, HA
...
The ionization constant increase as the strengths of the acids increase
...
The reaction of a Bronsted-Lowry base with water is given by:

B(aq) + H2O(l)  HB+(aq) + OH-(aq)
Water is the acid that reacts with the base, HB+ is the conjugate acid of the base B, and the
hydroxide ion is the conjugate base of water
...

We can measure the relative strengths of bases by measuring their base-ionization
constant (Kb) in aqueous solutions
...
If A- is a strong base, any protons that are
donated to water molecules are recaptured by A-
...
If A- is a weak base, water binds the proton more
strongly, and the solution contains primarily A- and H3O+ the acid is strong
...


Strong Acid

Weak Acid



Completely ionized ( = 1)



Partially ionized (0 <  < 1)



Example: HCl, HBr, HNO3, H2SO4,




Example: CH3COOH, H3PO4, H2S, etc
...

The ionization reaction goes back and
forth (reversible reaction)
Al(OH)3(s)  Al3+(aq) + 3OH-(aq)
NH3(g) + H2O(l)  NH4+(aq) + OH-(aq)

Sr(OH)2, Ba(OH)2


The ionization reaction proceeds one
way to the right (irreversible reaction)
NaOH(aq)
Na+(aq) + OH-(aq)

Acid and Base Reaction (Neutralization Reaction)
A solution is neutral when it contains equal concentrations of hydronium and hydroxide
ions
...
A strong acid and a strong base, such as HCl(aq) and NaOH(aq) will react to form a
neutral solution since the conjugate partners produced are of negligible strength
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
2
...

3
...
A solution of a weak acid
reacts with a solution of a strong base to form the conjugate base of the weak acid
and the conjugate acid of the strong base
...

However, the conjugate base of the weak acid is a weak base and ionizes slightly in
water
...

4
...

This is the most complex of the four types of reactions
...
Occasionally the
weak acid and the weak base will have the same strength, so their respective
conjugate base and acid will have the same strength, and the solution will be neutral
...


Here is an example of an acid - base neutralization reaction:
a
...
Sulfuric acid solution and potassium hydroxide solution
c
...
Changing the neutralization reaction to: H+(aq) + OH-(aq)
H2O(l)
create an ICE( initial, change, equilibrium) column from the reaction
...
Find moles of H+ and initial moles of OH3
...


Litmus paper

2
...


Artificial indicator

4
...


pH meter

Accuracy
Distinguish between acidic, basic and
neutral solutions
Distinguish between acidic, and basic
solutions
Know the pH range of the solution
Determine the pH to the nearest unit
value
find out the pH value down to its decimal
value (depending on the accuracy of the
tool)

Litmus Paper
Solution
Properties

The Color of Red
Litmus

The Color of Blue
Litmus

Acid
Neutral
Base

Red
Red
Blue

Red
Blue
Blue

Acid-Base Indicators
Certain organic substances change color in dilute solution when the hydronium ion
concentration reaches a particular ion value
...
In more basic solutions where the hydronium ion concentration is less

than 5,0 x 10-9 M (ph > 8,3), it is red or pink
...

Figure below present several indicators, their colors, and their color-change intervals
...
Acid + Base

Salt + Water

2
...
Acid + Base Oxide

Salt + Water

4
...
Ammonia gas + Acid

Salt

Ammonium Salt

Acid Oxides and Base Oxides
Acid Oxides

Base Oxides

Definition

Compounds between non-metallic
elements and oxygen

Examples

CO2, N2O5, SO3, and Cl2O7

Compounds between metallic
elements and oxygen
Na2O, FeO, CuO, and Al2O3

Reaction with water in
solution

In solution reacts with water to form
acids:
SO3 + H2O
H2SO4
N2O5 + H2O
(H2N2O6) 2HNO3
Cl2O7 + H2O
(H2Cl2O8) 2HClO4
P2O3 + 3H2O
(H6P2O6) 2H3PO3
Special oxides of P, As, and Sb are
added 3H2O

Note

1
...
Acid Oxide + Base

K2SO4(aq) + 2H2O(l)
2H2O(l)

Salt + Water

Reaction Equation

SO2(g) + 2NaOH(aq)

Complete Ionic Equation

SO2(g)+ 2Na+(aq) + 2OH-(aq)

Net Ion Equation

SO2(g)+ 2OH-(aq)

Spectator Ions

Na+

SO2 + H2O

2K+(aq) + SO42-(aq) + 2H2O(l)

Na2SO3(aq) + H2O(l)
2Na+(aq) + SO32-(aq) + H2O(l)

SO32-(aq) + H2O(l)

H2SO3 (H+ + SO32-) (Before reacting with NaOH, SO2 reacts with water first)

3
...
Acid Oxide + Base Oxide

CaCl2(aq) + H2O(l)
Ca2+(aq) + 2Cl-(aq) + H2O(l)

Ca2+(aq) + H2O(l)

Salt

Reaction Equation

P2O5(s) + Fe2O3(s)

2Fe2PO4(s)

Complete Ionic Equation

P2O5(s) + Fe2O3(s)

2Fe2PO4(s)

Net Ion Equation

P2O5(s) + Fe2O3(s)

2Fe2PO4(s)

Spectator Ions

-

P2O5 + 3H2O

2H3PO4 ( H+ + PO43- )

(H6P2O8)

5
...
Single Replacement Reaction Group (A + BC
Metal + Dilute Strong Acid

AC + B)

Salt + Hydrogen Gas

 Requirements: The metal must lie to the left of hydrogen (H) in the Volta series
 The Volta series: Li-K-Ba-Ca-Na-Mg-Al-Mn-Zn-Cr-Fe-Ni-Co-Sn-Pb-H-Cu-Hg-Ag-Pt-Au
 For metals that have more than one valence, usually the salt cation is the metal with a
higher valence

Reaction Equation

2Fe(s) + 3H2SO4(aq)

Complete Ionic Equation

2Fe (s) + 6H+(aq) + 3SO42-(aq)

Net Ion Equation

2Fe (s) + 6H+(aq)

Spectator Ions

SO42-

Cu(s) + H2SO4(aq)

Fe2SO4(aq) + 3H2(g)
2Fe3+(aq) + 3SO42-(aq) + 3H2(g)

2Fe3+(aq) + 3H2(g)

It doesn't react because Cu is to the right of hydrogen (H)

7
...
Is an unstable compound (easily decomposed)
2
...
It is a weaker electrolyte than the reactants

AD + CB)

H2O(l) + CO2(g)
H2O(l) + NH3(g)
H2O(l) + SO2(g)

Salt Hydrolysis
Properties of Salt Solutions
Origin of Salt

Salt Properties

Hydrolysis

Strong Acid + Strong Base

Neutral

Not hydrolyzed

Strong Acid + Weak Base

Acid

Partial hydrolysis

Weak Acid + Strong Base

Base

Partial hydrolysis

Weak Acid + Weak Base

Ka > Kb: acid
Ka < Kb: base
Ka = Kb: neutral

Total hydrolysis

Strong Acid

Strong Base

HCl, HBr, HI, HNO3, H2SO4, HClO4

LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2,
Ba(OH)2

Hydrolysis Reaction
Steps to Make a Hydrolysis Reaction
1
...
Make a hydrolysis reaction for each ion (cation and anion)
 Cations / anions derived from strong acids / bases will not be hydrolyzed, while
cations / anions derived from weak acids / bases will be hydrolysed
 Cation hydrolysis produces H3O+ (= H+) ions, while anion hydrolysis produces OH ions

Strong Acid

Anions of strong acids

HCl, HBr, HI, HNO3, H2SO4, HClO4

Cl-, Br-, I-, NO3-, SO4-, ClO4-

Strong Base

Cation of a strong base

LiOH, NaOH, KOH, Ca(OH)2, Sr(OH)2,
Ba(OH)2

Li+, Na+, K+, Ca2+, Sr2+, Ba2+

Example of Salt Hydrolysis Reaction From Strong Acid and Strong Base (NaCl)
NaCl(aq)

Na+(aq) + Cl-(aq)

Na+(aq) + H2O(l)
Cl-(aq) + H2O(l)
So, salts of strong acids and strong bases are not hydrolyzed, so they are neutral
Example of Salt Hydrolysis Reaction From Strong Acid and Weak Base (NH4Cl)
NH4Cl(aq)

NH4+(aq) + Cl-(aq)

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Cl-(aq) + H2O(l)
So, salts of strong acids and weak bases are partially hydrolysed and acidic
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
H+ OH-

NH4+ + OH-

NH4OH (unstable)  NH3

Example of Salt Hydrolysis Reaction From Weak Acid and Strong Base (CH3COONa)
CH3COO-(aq) + Na+(aq)

CH3COONa(aq)

Na+(aq) + H2O(l)
CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH-(aq)
So, salts of weak acids and strong bases are partially hydrolysed and alkaline

Example of Salt Hydrolysis Rea ction From Weak Acid and Weak Base (NH4CH3COO)
CH3COO-(aq) + Na+(aq)

NH4CH3COO(aq)

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
CH3COO-(aq) + H2O(l)  CH3COOH(aq) + OH-(aq)
So, salts of weak acids and weak bases are totally hydrolysed

Calculation of pH of Partially Hydrolyzed Salt Solution
Salts of Strong Acids (HA) and Weak Bases (LOH)
HA(aq) + LOH(aq)
LA(aq)

LA(aq) + H2O(l)

L+(aq) + A-(aq)

L+(aq) + H2O(l)  LOH(aq) + H+(aq) (i)
A-(aq) + H2O(l)
(i) Kc =

[𝑳𝑶𝑯][𝑯+ ]
[𝑳+ ]

Kw = [H+][OH-]
So, Kc =

𝑲𝒘
𝑲𝒃

LOH(aq)  L+(aq) + OH-(aq)
Kb =

x

[𝑂𝐻 − ]
[𝑂𝐻 − ]

[𝐿+ ][𝑂𝐻 − ]

1

[𝐿𝑂𝐻]

𝐾𝑏

=

[𝐿𝑂𝐻]
[𝐿+ ][𝑂𝐻 − ]

When the concentration of LOH equals the concentration of H+ at (i) equation, then

Kc =

[𝑯+ ]𝟐

[H+] = √𝐾𝑐
...
𝑴 (M is molarity of salt and val is valence number of weak ion of salt)

Finding the value of  (degree of hydrolysis)
L+(aq) + H2O(l)  LOH(aq) + H+(aq)
I

M

C M

M

E M(1-)
Kc =

[𝑳𝑶𝑯][𝑯+ ]
[𝑳+ ]

=√

𝐾ℎ
𝑀

=√

=

𝑀
...
𝑴

=√

M

M

M

M

= 2
...
𝑴

Partially Hydrolyzed Salt Formula
Salts of Strong Acids and Weak Bases
𝑲𝒘
Kc =
𝑲𝒃

[H+] = √
=√

𝑲𝒘
𝑲𝒃

𝑲𝒄
𝒗𝒂𝒍
...
𝑴

=√

𝑲𝒘
𝑲𝒃
...
𝑴

𝑲𝒘
𝑲𝒂


...
𝑴

Total Hydrolyzed Salt Solution pH Calculation
Salts of Weak Acids (HA) and Weak Bases (LOH)

[H+] = √

𝑲𝒘
𝑲𝒃

Kc = 𝑲


...


𝑲𝒃

+

L+(aq) + A-(aq) + H2O(l)  LOH(aq) + HA(aq)
Kc =

[𝐿𝑂𝐻][𝐻𝐴] [𝐻 + ][𝑂𝐻 − ]
[𝐿+ ][𝐴− ]


...
[H+][OH-]
]
]

Kc =

1


...
Kw

Kc =

[𝐿𝑂𝐻][𝐻𝐴]

Kc =

[𝐻𝐴]2

[𝐿+ ][𝐴− ]


...
𝑲𝒃

[LOH]  [HA]
[L+]  [A-]

[𝐴− ]2 [𝐻 + ]2

1
𝑲𝒘
= 2
...
𝑲𝒃 𝐾𝑎

[H+] = √

𝑲𝒘
𝑲𝒃


...
Mixing a weak acid with
its salt (conjugate base)
2
...
Mixing a weak base with
its salt (conjugate acid)
2
...
35 – 7
...
8) which results in organ damage
and even death

Component: acid H2PO4- (dihydrogen phosphate ion) and
conjugate base HPO42- (monohydrogen phosphate ion)
Serves to maintain the pH of the intracellular fluid around 7
...
8 and
helps neutralize acids that enter the mouth which can damage
tooth enamel
...
𝒏𝒔

Ka = acid ionization constant
na = acid moles
nbk = conjugate base moles
val = the number of weak ions in the salt
ns = salt moles

[OH-] = Kb

[𝐵 + ][𝑂𝐻 − ]

[𝐵𝑂𝐻]
[𝐵 + ]

[OH-] = Kb

[𝐵𝑂𝐻]

= Kb
𝑛𝑏
𝑛𝑐𝑎

[𝑏𝑎𝑠𝑒]
{𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑎𝑐𝑖𝑑]

= Kb

𝑛𝑏
𝑣𝑎𝑙
...
It is a buffer because it contains both the
weak acid and its salt
...

If we add a base such as sodium hydroxide, the hydroxide ions react with the few
hydronium ions present
...
If we add an acid such as hydrochloric acid, most of the
hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid
molecules:

H3O+ (aq) + CH3COO(aq)

CH3COOH(aq) + H2O(l)

Thus, there is very little increase in the concentration of the hydronium ion, and the pH
remains practically unchanged
...
It is a buffer because it also contains the salt of
the weak base
...

 The purpose of an acid-base titration is to determine the level of an acid or base
solution using certain acid-base indicators
...

 The equivalence point is the point at which the amounts of acid and base are
exactly equivalent
...

The equivalence point  the endpoint of the titration

n H+ = n OHMa
...
vala = Mb
...
valb
Na
...
Vb
N = M
...
Titration involving a weak acid and a weak base are
complicated by the hydrolysis of both the cation and the anion of the salt formed
...
Therefore, these titrations will
not be dealt with here
...
A 0
...
0 ml
of a 0
...
This curve sometimes referred to as a titration curve
...
00
...
Near the
equivalence point the pH begins to rise steeply, and at the equivalence point (that is, the
point at which equimolar amounts of acid and base have reacted) the curve rises almost
vertically
...
Beyond the equivalence point, the pH again increases slowly
with the addition of NaOH
...
A 0
...
0 ml
of a 0
...


At the equivalence point, when we only have sodium acetate present, the pH will be
greater than 7 as a result of the excess OH- ions formed
...
A 0
...
0 ml of a
0
...
As a result of a salt hydrolysis, the pH at the equivalence point
is lower than 7
...
For each of the following, identify the acids and bases in both the forward and
reverse reactions:
a
...
OCl- + H2O  HOCl + OHc
...
HCl + H2PO4-  Cl- + H3PO4
Analyze
Recall that a Bronsted-Lowry acids is one that gives up a proton and a BronstedLowry base is one that takes a proton
...
It gives up a
proton H+, to become ClO2-
...
Now consider H2O
...
Thus, H2O is base, and H3O+ is its conjugate acid
...
OH- produced in this reaction is the conjugate
base of H2O
...
HClO2 + H2O  ClO2- + H3O+
Acid(1)

Base(2)

Base(1)

Acid(2)

b
...
NH3 + H2PO4-  NH4+ + HPO42Base(1)

Acid(2)

Acid(1)

Base(2)

d
...
The
conjugate base of H2PO4- is HPO42- (the deprotonated form of H2PO4-), and the
conjugate acid of H2PO4- is H3PO4 (the protonated form of H2PO4-)
...


02
...
HNO2, PO43-, HCO3-
...

Solve
We know that the formulas of most acids begin with H
...


HNO2(aq) + H2O(l)  NO2-(aq) + H3O+(aq)
HCO3-(aq) + H2O(l)  CO32-(aq) + H3O+(aq)
A negatively charged species will attract a positively charged proton and act as a
base
...
We also know that PO43- must be a
base because it cannot act as an acid – it has no protons to donate – and we know
that all three species have acid – base properties
...
H2O(aq) + OH-(aq)

Notice that HCO3- is the amphiprotic species, acting as both an acid and a base
...
According to the Lewis theory, each of the following is an acid-base reaction
...

Solve
(a) In BF3, the B atom has a vacant orbital and an incomplete octet
...
BF3 is the electron-pair acceptor-the acid
...

(b) We have already identified OH- as a Lewis base, so we might suspect that is the
base and that CO2(aq) is the Lewis acid
...
A rearrangement of an electron pair at one of the double bonds is
also required, as indicated by the smaller red arrow
...
To describe the reaction in (b) in this way requires
to consider the electronic structure of CO2 in terms of molecular orbital theory
...
The vacant orbital in CO2
that accepts the lone pair from OH- is an antibonding -type orbital
...
Identify the Lewis acids and bases in this reactions
...
These two are the electron pair donors, or the
Lewis bases
...
Thus, Al(OH)3 and SnCl4 are the Lewis acids in these
rections
...
Students found that a yogurt sample had a pH of 2
...
What are the [H +] and [OH-]
of the yogurt?
Solve

[H3O+] is readily calculated from pH: [H3O+] = 10-pH = 10-2
...
4 x 10-3 M
...
𝟎 𝒙 𝟏𝟎−𝟏𝟒
𝟏
...
1 x 10-12 or

(2) pH + pOH = 14, pOH = 14 – pH = 14
...
85 = 11
...
15 = 7
...
The pH of solution of HCl in water is found to be 2
...
What volume of water would
you add to 1
...
10?
Solve

[H3O+] is calculated from pH in each case: [H3O+] = 10-pH
[H3O+]conc = 10-2
...
2 x 10-3 M,
[H3O+]dil = 10-3
...
9 x 10-4 M
Amount H3O+ = 1
...
soln x

3
...
𝑠𝑜𝑙𝑛

= 3
...
We
calculate the volume of the dilute solution:

3
...
𝑠𝑜𝑙𝑛
7
...
1 L dilute soln
...
1 L
...
Since
dilution is done with water, the pH of an infinitely dilute solution will approach that
of pure water, namely pH = 7
...
Calculate the pH of a 1
...

Solve
Keep in mind HCl is strong acid
...

The ionization of HCl is:

H+(aq) + Cl(aq)

HCl(aq)

The concentrations of all the species (HCl, H+, Cl-) before and after ionization can
be represented as follows:

HCl(aq)
Initial
1
...
0 x 10-3
Equilibrium
0
...
0
+1
...
0 x 10-3

Cl(aq)
0
...
0 x 10-3
1
...
Thus,

[H+] = 1
...
0 x 10-3) = 3
...
Calculate the pH of a 0
...
5 x 10-4):

HNO2(aq)  H+(aq) + NO2 (aq)
Recall that a weak acid only partially ionizes in water
...

The species that can effect the pH of the solution are HNO2, H+, and the conjugate
base NO2
...

Letting x be the equilibrium concentration of H+, and NO2 ions in mol/L, we
summarize:

HNO2(aq)  H+(aq) +
Initial
Change
Equilibrium

0
...
036x

0
...
0
+x
x

Ka =

[𝐻 + ][𝐻𝑁𝑂2 ]
[𝐻𝑁𝑂2 ]

4
...
036−𝑥

; we must solve the quadratic equation, as follows:

x2 + 4
...
62 x 105 = 0
x=

4
...
5 𝑥 10−4 )2 −4(1)(−1
...
8 x 10-3 M

At equilibrium: [H+] = 3
...
8 x 103) = 2
...
A total of 50 mL of 0
...
05 M NaOH
solution
...

Before mixing

H2SO4: [H+] = M
...
005
...
01 M
pH = -log [H+] = -log 0
...
valence = 0
...
1 = 0
...
5
-0
...
5
-0
...
5
0
...
V = 0
...
50 = 0
...
V = 0
...
50 = 2
...
02 M

pOH = 2 –log 2; pH = 14 – (2 –log 2) = 12 + log 2

10
...

a
...
Ca(NO3)2(aq)

Ca2+(aq) + 2 N𝑂3− (aq)

Ca2+(aq) + H2O(l)
N𝑂3− (aq) + H2O(l)

(neutral)

11
...
If the Ka of acetic acid is 10-5,
determine the final pH of the mixture
...
𝑀

pH = 3

pH =13

10-3 = √10−5
...
1 M
n = M
...
1 x 0
...
01 mol

[OH] = 0
...
V = 0
...
1 = 0
...
01
-0
...
𝑴 = √

pOH = 1

0
...
01
10−14
10−5


...


CH3COONa(aq) + H2O(l)
0
...
01

0
...
2

[OH] = √5 x 105
...
5 - log√5
pH = 14 – (5
...
5 + log√𝟓

= √5
...
01
0
...
A buffer solution contains 0
...
8 x 10-5) and 0
...
Calculate the pH of this solution
...
8 x 10-5 =

[𝐻 + ][𝐶2 𝐻3 𝑂2− ]
[𝐻𝐶2 𝐻3 𝑂2 ]

The concentration are as follows:

CH3COOH(aq)  H+(aq) + C2H3O2(aq)
Initial
Change
Equilibrium

0
+x
x

0
...
5  x

0
...
5 + x

Then
-5

Ka = 1
...
5+𝑥)
0
...
8 x 10-5
The approximations are valid (by the 5% rule), so
[H+] = x = 1
...
74

(𝑥)(0
...
5

13
...
010 mole of solid NaOH is added to
1
...
12
...
010 mole of solid NaOH is added to 1
...

Solve
Since the added solid NaOH will completely dissociate, the major species in
solution before any reaction occurs are CH3COOH, Na+, CH3COO, OH, and H2O
...
The best source of protons is the acetic
acid:
OH(aq) + CH3COOH(aq)
H2O(l) + CH3COO(aq)
Although acetic acid is a week acid, the hydroxide ion is such a strong base that the
reaction above will proceed essentially to completion (until the OH - ions are
consumed)
...
The stoichiometry problem:

Note that 0
...
010 mole of
CH3COO by the added OH
...
The equilibrium problem
...


CH3COOH(aq)  H+(aq) + C2H3O2(aq)
Initial
Change
Equilibrium

0
...
49  x

0
+x
x

0
...
51 + x

Note that the initial concentrations are defined after the reaction with OH  is
complete but before the system adjusts to equilibrium
...
8 x 10 =

[H+ ][C2 H3 O−
2]
[HC2 H3 O2 ]

=

(𝑥)(0
...
49−𝑥



(𝑥)(0
...
49

And x  1
...
7 x 10-5 M and pH = 4
...
01 mole of OH to this buffered
solution is then
4
...
74
= + 0
...
02 pH units
...
01 mole of solid NaOH is added to
1
...
01 M NaOH
...
01 M and

[H+] =

𝐾𝑤
[𝑂𝐻 − ]

=

1
...
0 𝑥 10−2

= 1
...
00
New solution

–

7
...
00

Pure water

The increase is 5
...
Note how well the buffered solution resists a change
in pH as compared with pure water
...
Calculate the pH in the titration of 25
...
100 M acetic acid (CH3COOH,
Ka = 1
...
0
mL 0
...

Solve

CH3COOH(aq) + NaOH(aq)

CH3COONa(aq) + H2O(l)

1 mole CH3COOH  1 mole NaOH(aq)
...
At the equivalence

point, however, the neutralization is complete and the pH of the solution will
depend on the extent of the hydrolysis of the salt formed, which is CH3COONa
...
0 mL is
10
...
100 𝑚𝑜𝑙𝑒 𝑁𝑎𝑂𝐻
1 𝐿 𝑁𝑎𝑂𝐻 𝑠𝑜𝑙𝑛

x

1𝐿
1000 𝑚𝐿

= 1
...
0 mL of solution is
25
...
100 𝑚𝑜𝑙𝑒 𝐶𝐻3 𝐶𝑂𝑂𝐻
1 𝐿 𝐶𝐻3 𝐶𝑂𝑂𝐻 𝑠𝑜𝑙𝑛

x

1𝐿
1000 𝑚𝐿

= 2
...
The molarity will change but the number of moles will
remain the same
...
50 x 10-3
1
...
50 x 10-3

Initial
Change
Equilibrium

1
...
00 x 10-3
0

CH3COONa(aq) + H2O(l)
0
+1
...
00 x 10-3

At this stage we have a buffer system made up of CH3COOH and CH3COO
(from the salt, CH3COONa)
...
8 x 10 =
[H+] =

[H+ ][C2 H3 O−
2]
[HC2 H3 O2 ]

[1
...
8 𝑥 10−5 ]
1
...
7 x 10-5 M

So, pH = log (2
...
57
15
...
00 mL of
0
...
100 M NaOH?
(a) Before the addition of any NaOH ( initial pH)
(b) After the addition of 24
...
100 M NaOH (before the equivalence point)
(c) After the addition of 25
...
100 M NaOH (the equivalence point)

Solve
The titration equation in the ionic and net ionic form
...
100 M HCl
...
100 M and pH = 1
...
0 mL x

0
...
50 mmole H3O+

The number of millimoles of OH present in 24
...
100 M NaOH is
24
...
100 𝑚𝑚𝑜𝑙𝑒
1 𝑚𝐿

= 2
...

H3O+ + OH2 H 2O
Initially present
2
...
40
Changes
2
...
40
After reaction
0
...
10 mmole of H3O+ is present in 49
...
00 mL
original + 24
...

+

[H3O ] =

0
...
00 𝑚𝐿

= 2
...
0 x 103) = 2
...
As seen in the ionic form of the equation for the
neutralization reaction, the solution at the equivalence point is simply NaCl (aq)
...
00



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