Download analytical chemistry lecture 8

ERT 207
ANALYTICAL CHEMISTRY
ACIDS AND BASES THEORIES
ACID BASE EQUILIBRIA IN WATER
pH SCALE
27 Jan 2011
(MISS NOORULNAJWA DIYANA YAACOB)
1
Brønsted Acids and Bases
• A Brønsted acid is a species that donates a proton.
(a proton donor).
HCl(aq) +
H2O(l)
acid
H3O+(aq) +
hydronium
ion
Cl-(aq)
conjugate
base
A Conjugate base is what remains of the acid after the donation of a proton.
• A Brønsted base is a species that accepts a proton.
(a proton acceptor).
NH3(aq)
base
+ H2O(l)
NH4+(aq)
+
OH-(aq)
conjugate
acid
A Conjugate acid is a newly formed protonated species.
2
Brønsted Acids and Bases
Any reaction,
using the
Brønsted theory,
involves both an
acid and a base.
Water is an
amphoteric
species which can
act as a Brønsted
acid or as a
Brønsted base.
3
Brønsted Acids and Bases
4
Brønsted Acids and Bases
• Exercise:
Identify acids, conjugate acids, bases and
conjugate bases in the reactions below:
(a) H2SO4(aq) + H2O(l)
acid
base
(b) NH4+(aq) + H2O(l)
acid
base
HSO4‒ (aq) +
conjugate base
H3O+(aq)
conjugate acid
+
conjugate acid
• Exercise:
‒
What is the conjugate base of H2PO4 ?
H2PO4‒(aq) + H2O(l)
H3O+(aq)
NH3(aq)
conjugate base
HPO42‒(aq) + H3O+(aq)
conjugate base
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Lewis Acids and Bases
• The Brønsted base is the species that accepts a proton.
• The Brønsted acid is the species that donates a proton.
• A Lewis base is a species that donates a pair of electrons.
• A Lewis acid is a species that accepts a pair of electrons.
H+
+
‒
O
H
O
H
Lewis base
Lewis acid
H
H+
Lewis acid
H
+
N
+
H
H
H
Lewis base
H
N
H
H
6
Lewis Acids and Bases
• Lewis acid-base reactions are more general
than Brønsted acid-base reactions.
• Lewis acid-base reactions result in the
formation of coordinate covalent bonds.
7
ARRHENIUS THEORY-H+ AND OHˉ
• An acid is any substance that ionizes (partially
or completely) in water to give hydrogen ions
• Means associate with solvent to give
hydronium ions
•
HA +H2O ↔ H3O+ + Aˉ
• A base ionizes in water to give hydroxyl ions.
• Weak (partially ionizes) bases ionizes as
follows:
•
B + H2O ⇋ BH+ + OHˉ
For strong bases e.g.: (NaOH)
M(OH)n →M n+ + nOHˉ
• this theory is obviously restricted to water as
a solvent only
Franklin’s theory
• Recognizes the ionization of a solvent to give a
cation and an anion
• E.g:
2H2O ⇋ H3O+ + OHˉ
•
2NH3 ⇋ NH4 + NH2¯
• Acid = a solute that yields the cation of the solvent
• Base = a solute that yields the anion of the solvent
• Eg of strong acid is NH4Cl in liquid ammonia
• Eg of strong base is NaNH2 in ammonia
ACID-BASE EQUILIBRIA IN WATER
• When acid or base is dissolved in water, it will
dissociate
• Amount of ionization depends on the strength
of the acid
• Strong electrolyte completely dissociated
• Weak electrolyte dissociated partially
• Example:
a) Hydrochloric acid (strong acid)
HCl  H 2O  H 3O   Cl 
a) Acetic acid (weak acid)
HOAc  H 2O  H 3O   OAc 
• Equilibrium constant
K 
0
a
aH O   aOAc
3
aHOAc  aH 2O
0
K
• a is the thermodynamic activity constant
• H+ will be used instead of H3O+ for
simplification
• Molar concentration will be replaced by []

H OAc 


Ka

HOAc 
  

K w  H OH

• Ka and Kw are molar equilibrium constants
• [H+]=1.0×10-7 M = [OH-]
The Acid-Base Properties of Water
• Water is a very weak electrolyte. It undergoes
ionization to a very small extent.
H2O (l)
H+ (aq) + OH‒ (aq)
Autoionization of
+
‒
2H2O (l)
H3O (aq) + OH (aq)
water
• The equilibrium expression of water
autoionization is:
Kw = [H+] [OH‒]
or
Kw = [H3O+] [OH‒]
14
Equilibrium Constant in Aqueous
Solutions
• For any aqueous solution:
Kw = [H3O+] [OH‒] = 1.0 × 10-14
at 25C
• Remember that Kw is always constant at a given
temperature.
However, there can be infinite possibilities of
the equilibrium positions for aqueous solution.
– When [H3O+] = [OH‒] ; the solution is neutral.
– When [H3O+] > [OH‒] ; the solution is acidic.
– When [H3O+] < [OH‒] ; the solution is basic.
15
Equilibrium Constant in Aqueous
Solutions
16
Equilibrium Constant in Aqueous
Solutions
• Exercise:
Calculate [OH‒] in a solution in which the
concentration of protons is 0.0012 M at 25C. Is the
solution acidic, basic or neutral?
At 25C, Kw is always equal to 1 × 10-14 .
Kw
= [H3O+] [OH‒]
1 × 10-14 = (0.0012) [OH‒]
[OH‒] = 1 × 10-14 / 0.0012 = 8.3 × 10-12 M
The solution is acidic.
17
The pH Scale
• The pH scale is a convenient way to express the acidity
and basicity of a solution (the concentration of H3O+
ions).
The pH of a solution is defined as:
pH = -log [H3O+] or pH = -log [H+]
• Example:
Two solutions with hydronium ion concentrations of 1.0
× 10-4 M and 5.0 × 10-9 M. Find their pH values.
2 sig. fig.
The first solution;
The second solution;
pH = - log (1.0 × 10-4) = 4.00
pH = - log (5.0 × 10-9) = 8.30
2 decimal
places
The number of sig. fig in the log is equal to
18
the number of decimal places in the answer
pH scale
Example 1
Calculate the pH of a 2.0 x 10-3 M solution
of HCl.
pH scale
Solution
[H+] = 2.0 x 10-3 M
pH = − log10[H + ]
= − log10[2.0 x 10-3 M ]
= 3 - log10 2.0
= 3 – 0.30
= 2.7
The pH Scale
• Exercise:
Find the pH of a neutral solution.
A neutral solution has [H3O+] = [OH‒] .
Since Kw = [H3O+] [OH‒] = 1.0 × 10-14
then [H3O+] = 1.0 × 10-7
pH = - log (1.0 × 10-7) = 7.00
Lower pH values ( < 7.00 )
Neutral
Higher pH values ( > 7.00 )
Higher [H3O+]
Lower [H3O+]
Acidic solution
Basic solution
21
The pOH Scale
• A pOH scale is analogous to the pH scale. It is defined as:
pOH = - log [OH‒]
Also,
[OH‒] = 10-pOH
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• Kw = [H+] [OH-]
• -log Kw = - log [H+][OH-]
= -log [H+] – log [OH-]
pKw = pH + pOH
pOH scale
• PLEASE NOTE THAT AT 25˚C
[H+] x [OH-] = 10-14
pH + pOH = 14
The pOH Scale
• Recall that at 25C:
Kw = 1.0 × 10-14 = [H3O+] [OH‒]
‒ log ([H3O+] [OH‒]) = ‒ log (1.0 × 10-14)
‒ log [H3O+] ‒ log [OH‒] = 14.00
pH
+ pOH = 14.00
pH
13.00
11.00
9.00
7.00
5.00
4.00
1.00
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• Example
Calculate the pOH and the pH of a 5.0 x
10-2 M solution of NaOH
• Solution Method 1
[OH-] = 5.0 x10-2 M
pOH = - log(5.0 x10-2 )
= 2 – log 5.0
= 1.3
pOH scale
• Solution continued
Since
pH + 1.3 = 14.00
pH = 12.7
pH + pOH = 14
pOH scale
Solution Method 2
Since
[H+] x [OH-] = 10-14
[H+] = 1.0 x 10-14 = 2.0 x 10-13
5.0 x 10-12
pH = - log (2.0 x 10-13) = 13 – log 2.0
= 13 – 0.30 = 12.7
The pOH Scale
• Exercise:
What is the pOH of a solution that has a hydroxide ion
concentration of 4.3 x 10-2 M?
pOH  log[OH  ]
pOH  log[4.3 x 102 ]
pOH  1.37
• Exercise:
What is the hydroxide ion concentration of a solution with
pOH 8.35?
[OH ]  10pOH
[OH ]  108.35
[OH  ]  4.5x10 9
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