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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 g

=

𝟏
...
𝟒 𝒙 𝟏𝟎−𝟑

= 7
...
00 – 2
...
15, and then

[OH-] = 10-pOH = 10-11
...
1 x 10-12
06
...
50
...
00 L of this solution to raise the pH to 3
...
5 = 3
...
10= 7
...
00 L conc
...
2 𝑥 10−3 𝑚𝑜𝑙 𝐻3 𝑂+
1 𝐿 𝑐𝑜𝑛𝑐
...
2 x 10-3 mol H3O+

All of the H3O+ in the dilute solution comes from the concentrated solution
...
2 x 10-3 mol H3O+ x

1 𝐿 𝑐𝑜𝑛𝑐
...
9 𝑥 10−4 𝑚𝑜𝑙 𝐻3 𝑂+

= 4
...


Thus, the volume of water to be added is = 3
...

Infinite dilution does not lead to infinitely small hydrogen ion concentrations
...


07
...
0 x 10-3 M HCl solutions
...
Thus, the species are completely ionized and no HCl
will be left in solutions
...
0 x 10-3
Change
-1
...
0

H+(aq)

+

0
...
0 x 10-3
1
...
0
+1
...
0 x 10-3

A positive (+) change represents an increase and a negative () change indicate a
decrease in concentration
...
0 x 10-3 M
pH = -log (1
...
00
08
...
036 M nitrous acid (HNO2) solution (Ka = 4
...
We are given the initial
concentration of a weak acid and asked to calculate the pH of the solution at
equilibrium
...
We ignore water’s contribution to [H+]
...
036
x
0
...
0
+x
x

NO2 (aq)
0
...
5 x 10-4 =

𝑥2
0
...
5 x 104x – 1
...
5 𝑥 10−4 ± √(4
...
62 𝑥 10−5 )
2(1)

= 3
...
8 x 103 M,

pH = -log (3
...
42
09
...
005 M H2SO4 solution is mixed with 50 mL of 0
...
Calculate the pH of the solution before and after mixing
...
valence = 0
...
2 = 0
...
01 = 2
NaOH: [OH] = M
...
05
...
05 M
pOH = 2 – log 5
pH = 14 – (2 – log 5) = 12 + log 5
After mixing

H+ +
Initial
Change
Equilibrium

0
...
5
-

OH  H2O
2
...
5
2

0
...
5

nH+ = [H+]
...
01
...
5 mmol
nOH = [OH]
...
05
...
5 mmol

[OH] =

𝑛𝑂𝐻 −
𝑉

=

𝟐
100

= 0
...
Write down the hydrolysis reactions (if any) of the following salt solutions and
determine whether the solutions are acidic, basic or neutral
...
Na3PO4(aq)

3 Na+(aq) + PO43(aq)

Na+(aq) + H2O(l)

Solutions are base


PO3−
4 (aq) + 3 H2O(l)  H3PO4(aq) + 3OH (aq)

b
...
100 mL of acetic acid (CH3COOH) solution with pH = 3 is reacted with 100 mL of
sodium hydroxide (NaOH) solution with pH = 13
...

Solve
NaOH

CH3COOH
[H+] = √𝐾𝑎
...
𝑀
M = 0
...
V = 0
...
1 = 0
...
1 M
n = M
...
1 x 0
...
01 mol

CH3COOH(aq) + NaOH(aq)
Initial
Change
Equilibrium

[OH] = √

𝑲𝒘
𝑲𝒂

0
...
01
-


...
01
-0
...
1
...
01
0
...
01
0
...
5 M
pOH = 5
...
5 - log√𝟓 ) = 8
...
10−11

0
...
01

12
...
50 M acetic acid (CH3COOH, Ka = 1
...
5 M
sodium acetate (NaC2H3O2)
...

Solve
The major species in the solution are:
CH3COOH,

Na+,

C2H3O2,

and H2O

Weak acid

Neither acid
nor base

Base (conjugate base

Very weak acid
or base

of CH3COOH)

Examination of the solution components leads to the conclusion that the acetic
acid dissociation equilibrium, which involves both CH3COOH and C2H3O2, will
control the pH of the solution:
CH3COOH(aq)  H+(aq) + C2H3O2(aq)
Ka

= 1
...
5
-x
0
...
5
+x
0
...
8 x 10 =

[H+ ][C2 H3 O−
2]
[HC2 H3 O2 ]

=

(𝑥)(0
...
5−𝑥



And x  1
...
8 x 10-5 M and pH = 4
...
5)
0
...
Calculate the change in pH that occurs when 0
...
0 L of the buffered solution described in problem no
...
Compare this pH change
with that which occurs when 0
...
0 L water
...

The solution contains a relatively large amount of the very strong base hydroxide
ion, which has a great affinity for protons
...

1
...
01 mole of CH3COOH has been converted to 0
...

2
...
After the reaction between OH and CH3COOH is
complete, the major species in solution are
CH3COOH, Na+, CH3COO, and H2O
The dominant equilibrium involves the dissociation of acetic acid
...
49
-x
0
...
51
+x
0
...


-5

Ka = 1
...
51+𝑥)
0
...
51)
0
...
7 x 10-5
The approximations are valid (by the 5% rule), so
[H+] = x = 1
...
76
The change in pH produced by the addition of 0
...
76
–
4
...
02
New solution

Original solution

The pH increased by 0
...

Now compare this with what happens when 0
...
0 L water to give 0
...
In this case [OH] = 0
...
0 𝑥 10−14
1
...
0 x 10-12

pH = 12
Thus the pH change is
12
...
00

= + 5
...
00 pH units
...

14
...
0 mL of 0
...
8 x 10-5) by sodium hydroxide after the addition to the acid solution of 10
...
100 M NaOH
...
Therefore, at every stage of the titration
we can calculate the number of moles of base reacting with the acid, and the pH of
the solution is determined by the excess acid or base left over
...

The number of moles of NaOH in 10
...
0 mL x

0
...
00 x 10-3 mole

The number of moles of CH3COOH originally present in 25
...
0 mL x

0
...
50 x 10-3 mole

We work with moles at this point because when two solutions are mixed, the
solution volume increases
...
The changes in number of moles:

CH3COOH(aq) + NaOH(aq)
2
...
00 x 10-3
1
...
00 x 10-3
1
...
00 x 10-3
1
...
pH of the solution:
-5

Ka = 1
...
5 𝑥 10−3 ][1
...
0 𝑥 10−3

= 2
...
7 x 10-5) = 4
...
What is the pH at each of the following points in the titration of 25
...
100M HCl with 0
...
00 mL 0
...
00 mL 0
...

Ionic form:

H3O+(aq) + Cl(aq) + Na+(aq) + OH(aq)

Net ionic form: H3O+(aq) + OH(aq)

Na+(aq) + Cl(aq) + 2 H2O(l)

2 H2O(l)

(a) Before any NaOH is added, we are dealing with 0
...
This solution has
[H3O+] = 0
...
00
(b) The number of millimoles of H3O+ to be titrated is
25
...
100 𝑚𝑚𝑜𝑙𝑒 𝐻3 𝑂+
1 𝑚𝐿

= 2
...
00 mL of 0
...
0 mL x

0
...
40 mmole OH

We can represent the net ionic equation of the neutralization reaction in a
familiar format
...
50

Add
2
...
40
2
...
10
0
The remaining 0
...
00 mL of solution (25
...
00 mL added base)
...
10 𝑚𝑚𝑜𝑙𝑒 𝐻3 𝑂+
49
...
0 x 103 M

pH = log [H3O+] = log (2
...
70
(c) The equivalence point is the point at which the HCl is completely neutralized
and no excess NaOH is present
...

Because Na+ nor Cl hydrolyzes in water, pH = 7

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