1
ACIDโBASE EQUILIBRIA
IN AQUEOUS SOLUTION
2
Arrhenius concept of acids and bases
ุงุฑูููููุณ ูุฃูุญู ุงุถ ูุงูููุงุนุฏูููู ู
The Swedish chemist Svante Arrhenius published his theory of acids and bases
in 1887.
Arrhenius definition of acids: An acid is substance that, when dissolved in water, increases the concentration of H+ ion (or hydronium ion, H3O
+) in aqueous solution.
For example,
Arrhenius definition of bases: A base is substance that, when dissolved in water, increases the concentration of OH- ion (or hydroxide ion, OH-) in aqueous solution.
For example,
ุงูุญุธ ุงู ุุุ
ุงูู ุงุฏุฉ ุงูุชู ุชุชููู ุงู ุชุชุฃูู ูู ุงูู ุงุก ุงูู ุงูููุงุช ุงูููุฏุฑููููู H3Oุงูุชู ุชุชููู ุงู ุชุชุฃูู ูู ุงูู ุงุก ุงูุถุง ุชุนุชุจุฑ ุญู ุถุ ุจููู ุง +
ุชุนุชุจุฑ ูุงุนุฏุฉ. -OHุงูู ุงูููุงุช ุงูููุฏุฑููุณูุฏ
ูู ุงูููุงุช ุงูููุฏุฑููููู ุง ูููุง ุงูู ุงุฏุฉ ุงูุชู ุชุชููู ุงู ุชุชุฃูู H3O .HCl and HNO3)ู )ุชุนุชุจุฑ ุญู ุถุงู ูููุงูุ ู ุซ +
ุงูููุฏุฑููุณูุฏ ุงูู ุงูููุงุช ูููุง ุงูู ุงุฏุฉ ุงูุชู ุชุชููู ุงู ุชุชุฃููOH- ู )ูููุงูุ ู ุซ ุชุนุชุจุฑ ูุงุนุฏุฉ(NaOH and Ba(OH)2.
ุงูู ุงูููุงุช ุงูููุฏุฑููููู ุฌุฒุฆูุง ุงูู ุงุฏุฉ ุงูุชู ุชุชููู ุงู ุชุชุฃูู H3O .CH3COOH)ู )ุ ู ุซุถุนููุงู ุชุนุชุจุฑ ุญู ุถุงู +
ุงูููุฏุฑููุณูุฏ ุงูู ุงูููุงุช ุฌุฒุฆูุง ุงูู ุงุฏุฉ ุงูุชู ุชุชููู ุงู ุชุชุฃููOH- ู )ุ ู ุซุถุนููุฉ ุชุนุชุจุฑ ูุงุนุฏุฉ(NH3.
HCl + H2O H3O+ + Cl- (Hydrochloric acid )
HNO3 + H2O H3O+ + NO-
3 (Nitric acid )
H2SO4 + 2 H2O 2H3O+ + Cl- (Sulfuric acid )
NaOH + H2O Na+ + OH- (Sodium hydroxide)
Ca(OH)2 + H2O Ca++ + 2 OH- (Calcium hydroxide)
NH3 + H2O NH4+ + OH- (Ammonia)
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The BrรธnstedโLowry theory is an acidโbase
reaction theory which was proposed independently
by Johannes Nicolaus Brรธnstedand and Thomas Martin
Lowry in 1923.
Brรธnsted-Lowry concept of acids and bases ูุฃูุญู ุงุถ ูุงูููุงุนุฏ ููุฑู -ุจุฑููุณุชุฏูููู ู
An acid is substance that donates a proton (a hydrogen ion H+) to some other
substance.
A base is substance that accepts a proton (a hydrogen ion H+) from an acid.
For example,
When HCl dissolves in water, as below, HCl acts as a BrรธnstedโLowry acid (it donates a
proton to H2O), and H2O acts as a BrรธnstedโLowry base (it accepts a proton from HCl).
In the reaction between gas phase HCl and NH3, for example, a proton is transferred from the
acid HCl to the base NH3.
Brรธnsted Johannes Nicolaus
Thomas Martin Lowry
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( ู ูููู ุจุฑููุณุชุฏ (Brรธnsted ูููู ุงูุซุฑ ุดู ูุงูู ู ู ู ูููู ุงุฑูููููุณ(Arrhenius) ูุฐูู
ุจููู ุง ู ูููู ุุจุณุจุจ ุงู ู ูููู ุงุฑูููููุณ ูุญุตุฑ ุงูุญู ูุถ ูุงูููุงุนุฏ ูู ุงูู ุญุงููู ุงูู ุงุฆูุฉ ููุท
NH3ู ุน ุบุงุฒ HClุจุฑููุณุชุฏ ูุงูู ุงู ูููุฏูุง ุจุงูู ุญุงููู ุงูู ุงุฆูุฉุ ูู ุซุงู ุฐูู ุงูุชูุงุนู ุงุนุงูู ุจูู ุบุงุฒ
ุงุฌุฉ ุงูู ูุณุท ู ุงุฆู ูุญุฏูุซ ุงูุชูุงุนู.ุฏูู ุงูุญ
ุจุฑููุณุชุฏ ู ูููู ุชุนุชู ุฏ ููุฉ ุงูุญู ุถ ูุงููุงุนุฏู ูู (Brรธnsted): ุนูู
i. ู ุฏู ูุงุจููุฉ ุงูุญู ุถ ูู ูุญ ุงูุจุฑูุชูู ุงูู ุงุฆูH+
.
ii. ู ุฏู ูุงุจููุฉ ุงููุงุนุฏุฉ ูุชูุจู ุงูุจุฑูุชูู ุงูู ุงุฆูH+
.
The stronger acids are those that lose their protons more easily than other acids.
Similarly, the stronger bases are those that hold on to protons more strongly than
other bases.
Conjugate AcidโBase Pairs ุงุฃูุฒูุงุฌ ุงูู ูุชุฑูุฉ ู ู ุงูุญู ุถ ูุงููุงุนุฏุฉ
The term conjugate comes from the Latin word โconjugare,โ meaning โto join together.โ
Reactions between acids and bases always yield their conjugate bases and acids.
The conjugate base is the nitrite ion (NO-2) and the conjugate acid is the
hydronium ion (H3O+).
The conjugate base is the hydroxide ion (OH-) and the conjugate acid is the
ammonium ion (NH+
4).
An acid that has donated its proton becomes a conjugate base.
A base that has accepted a proton becomes a conjugate acid.
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ู ู ุง ุณุจูุุุ
ุญู ุถ ูุงุนุฏุฉ ู ูุชุฑูุฉ ูููู ูุงุนุฏุฉ ุญู ุถ ู ูุชุฑู. ููู
( ูู ูู ุงูุฌุงุฏ ุงูุญู ุถ ุงูู ูุชุฑู ูููุงุนุฏุฉ ุจุฃุถุงูุฉ ุจุฑูุชูู ู ุงุฆูH+
ุ ุจููู ุง ูู ูู ุงูุฌุงุฏ ุงููุงุนุฏุฉ ุงูู ูุชุฑูุฉ (
Hููุญู ุถ ุจูุฒุน ุจุฑูุชูู ู ุงุฆู )+
.)
The strength of acids and bases ููุฉ ุงูุญู ูุถ ูุงูููุงุนุฏ
Strong acids are completely dissociated (ionized) in water. Their conjugate bases
are quite weak. ุจุงููุงู ู ูู ุงูู ุงุก ูุชููู ููุงุนุฏูุง ุงูู ูุชุฑูุฉ ุฌุฏุงู ุถุนููุฉ ูุงุงูุญู ุงุถ ุงููููุฉ ูู ุงูุชู ุชุชูู
Weak acids are only dissociated (ionized) partially in water. Their conjugate
bases are strong bases. ุฌุฒุฆูุงู ูู ุงูู ุงุก ูุชููู ููุงุนุฏูุง ุงูู ูุชุฑูุฉ ูููุฉ ูุงุงูุญู ุงุถ ุงูุถุนููุฉ ูู ุงูุชู ุชุชูู
For example,
1) HCl + H2O H3O+ + Cl- (100 %, Dissociation)
2) HF + H2O H3O+ + F- (3 %, Dissociation)
About 3% of the HF molecules have dissociated, whereas 100% of the HCl molecules have
dissociated in water. Thus, HF is a weaker acid than HCl. F- is strong base. Cl
- is weak base.
ุงูู ูุชุฑูุฉ ุชุตุจุญ ุงุถุนู. ูููู ุง ุฒุงุฏุช ููุฉ ุงูุญู ุถ ูุฅู ูุงุนุฏุช
ุงุถุนู. ูุตุจุญุงูู ูุชุฑูุฉ ูุง ุญู ุถูุฅู ุงููุงุนุฏุฉููู ุง ุฒุงุฏุช ููุฉ
Moreover, the strength of an acid is defined by the equilibrium position of its dissociation
(ionization).
A strong acid is one for which this equilibrium lies far to the right (See the equation of No.
1). This means that almost all the original acid (HA) is dissociated (ionized) at equilibrium.
ุงูุชูุงุฒู ููุฒุงุญ ุงูุชูุงุนู ูุญู ุงููู ููุ ู ู ุง ูุฏู ุนูู ุงู ูู ุงูุญู ุถ ูุฏ ุชููู ุงู ุชุฃูู ) ู ุซุงู ุฐูู ุงูู ุนุงุฏูุฉ ูุญู ุถ ุงูููู ูุนูุฏ ู
(.ุงุณููู aุ ููุฐูู ุงูุฑุณู ุงูุชูุถูุญู ุงุนุงูู 1ุฑูู
(a) A strong acid (b) A weak acid.
Before dissociation After dissociation,
at equilibrium
After dissociation,
at equilibrium
Before dissociation
a b
6
Conversely, a weak acid is one for which the equilibrium lies far to the left. Most of the acid
originally placed in the solution is still present as HA at equilibrium. That is, a weak acid
dissociates only to a very small extent in aqueous solution.
ุจุณุจุจ ุงู ุงูุญู ุถ ุงุงูุตูู ูู ุงูู ุญููู ู ุงุฒุงู ู ุชุจูู ูู ุง ูู ููู ููุฒุงุญ ูุญู ุงููุณุงุฑ ููุญู ุถ ุงูุถุนูู ููุฏ ูุฌุฏ ุงู ุงุงูุชุฒุงู
ุ ููุฐูู -ุงูุญุธ ุงูุณูู -ุงุนุงูู 2) ู ุซุงู ุฐูู ุงูู ุนุงุฏูุฉ ุฑูู ูุชููู )ูุชุฃูู ( ุจุงููุงู ู ุงู ุชููู ุจุดูู ุฌุฒุฆู ุจุณูุท ุงู ุจูู ูุฉ ููููุฉ
(.ุงุนุงูู bุงูุฑุณู ุงูุชูุถูุญู
The two members of each pair are listed opposite each other in the two columns.
ู ุงูุญุธุฉ:
ูุฏ ูุฌุฏ ุงู ุงูููุฉ ุงูุญู ุถูุฉ ุชุฒุฏุงุฏ ู ู ุงููุณุงุฑ ุงูู ุงููู ููุููุฐุฑุฉ ุงูู ุฑูุฒูุฉุ ูู ุงูุฌุฏูู ุงูุฏูุฑู ู ุซูุ
CH4 < NH3< H2O < HF
ุงูุถุง ู ู ุงุงูุนูู ูุงูุณูู ู ุซู:
HF < HCl< HBr < HI
ุฒุฏุงุฏ ุจุฒูุงุฏุฉ ุนุฏุฏ ุงูุฐุฑุงุช ุฐุงุช ุงูุดุญูุฉ ุงูุณุงูุจุฉ ูุงูู ุฑุชุจุทุฉ ุจุงูุฐุฑุฉ ุงูู ุฑูุฒูุฉ ู ุซู:ููุฉ ุงูุญู ุถ ุงูุถุง ุช
H2SO3 < H2SO4
HClO < HClO2 < HClO3 < HClO4
Increasing acid strength
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Because the oxidation number of Y (S or Cl) increases as the number of attached O atoms
increases, this correlation can be stated in an equivalent way: In a series of oxyacids, the
acidity increases as the oxidation number of the central atom increases.
Amphiprotic substances ุงูู ูุงุฏ ุงูู ุชุฑุฏุฏุฉ
An amphiprotic substance is a substance that can donate or accept a proton, H+.
For a substance to be amphiprotic it must :
1- contain a hydrogen atom which is able to be donated to another chemical species.
2- be able to accept a hydrogen ion from another species.
ูู ุงูุชู ุชู ุชูู ุฎูุงุต ุญุงู ุถูุฉ ููุงุนุฏูุฉ ูู ููุณ ููุงู ู ูุงุฏ ุชู ูู ุฎุงุตูุฉ ุงูุชุฑุฏุฏ ูุชุณู ู ุงูู ูุงุฏ ุงูู ุชุฑุฏุฏุฉ ู
:ุงูููุชุ ู ุซู
HCO3-
HSO4-
H2O
Lewis concept of acids and bases ู ูููู ูููุณ ูุงูุญู ุงุถ ูุงูููุงุนุฏ An acid is a substance that accepts a lone pair of electrons.
A base is a substance that donates a lone pair electron.
.ูู ุงูู ุงุฏุฉ ุงูุชู ูุฏููุง ูุงุจููุฉ ูุชูุจู ุฒูุฌ ู ู ุงุงููููุชุฑููุงุช ูุชูููู ุฑุงุจุทุฉ ุชุณุงูู ูุฉ ุงูุญู ุถ
.ูู ุงูู ุงุฏุฉ ุงูุชู ูุฏููุง ูุงุจููุฉ ูู ูุญ ุฒูุฌ ู ู ุงุงููููุชุฑููุงุช ูุชูููู ุงูุฑุงุจุทุฉ ุงูุชุณุงูู ูุฉ ุงููุงุนุฏุฉ
ู ุซุงู:
BF3 + NH3 H3N-BF3
8
Autoionization of water ุชุฃูู ุงูู ุงุก
Water is an amphoteric substance. It can behave either as an acid or as a base
(This is a one of the most important chemical properties of water).
In autoionization reaction for water, one water molecule can donate a proton to
another water molecule:
The autoionization reaction for water can also be written as:
Equation (1) is more accurateโhydrogen ions do not exist in water because they
bond to form hydronium.
ุงูููุฏุฑูุฌูู ุงูุชูุฌุฏ ุจุดูู ุ ูุนูุฏ ุฐูู ุงูู ุงู ุงูููุงุช2ุชููู ุงูุซุฑ ุฏูุฉ ู ู ุงูู ุนุงุฏูุฉ ุฑูู 1ู ุนุงุฏูุฉ ุฑูู ุงู
ูุงู ู ุญุฑู ูู ุงูู ุงุก ูุฐูู ุจุณุจุจ ุงู ูู ูููููุง ู ุฑุชุจุทูู ูุชูููู ุงูููุงุช ุงูููุฏุฑููููู .
Both equation 1 and 2 are referred to as autoionization.
The equilibrium-constant expression for the autoionization of water is:
ู ุงูู ุงุก ุจุงูุชุฑููุฒ ุงูู ูุงูุฑู ูู ุง ููู:ูุนุจุฑ ุนู ุซุงุจุช ุงูุชูุงุฒู ูุชูู
Or
Also,
where Kw, called the ion-product constant or the dissociation constant for
water. ููู ุงุก ุงุฃููููู ูู ุซุงุจุช ุชููู ุงูู ุงุก ุงู ุซุงุจุช ุงูุญุงุตู Kw
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq) (1)
H2O (l) H+ (aq) + OH- (aq) (2)
Kw = [H3O+][OH
-]
Kw = [H+][OH
-]
Kw = [H3O+][OH
-] = [H
+][OH
-]
9
Experiments show that at 25 ยฐC in pure water,
ุ Cยฐ 25ุุนูุฏ ุฏุฑุฌุฉ ุญุฑุงุฑุฉ ุงูุบุฑูุฉูุถุญุช ุงูุชุฌุงุฑุจ ููู ุงุก ุงูููู
[H+] = [OH
-] = 1x 10
-7 M
which means that
Kw = [H+][OH
-] = (1x 10
-7 ) (1x 10
-7 )
Kw = [H+][OH
-] = 1x 10
-14
Problem
Calculate the concentration of H+ (aq) in:
(a) a solution in which [OH-] is 0.010 M,
(b) a solution in which [OH-] is 1.8 x 10
-9 M.
Solution
[H+][OH
-] = 1x 10
-14
[H+] =(1x 10โ14)
[๐๐ปโ]
[H+] =(1x 10โ14)
0.010 = 1x 10โ12 ๐
This solution is basic because [H+] < [OH
-]
ู ุงูุญุธุฉ:
-OHุงู ุง ู ุชุนูู ุงูุชุฑููุฒุงูู ูุงูุฑู [..โฆ] ุนุงูู ุฉ ุงุงูููุงุณ+Hุงู
a)
10
[H+] =(1x 10โ14)
[๐๐ปโ]=
[H+] =(1x 10โ14)
1.8 x 10โ9 = 5.6x 10โ6 ๐
This solution is acidic because [H+] > [OH
-]
The pH Scale ุงุฃูุณ )ุงูุฑูู ( ุงูููุฏุฑูุฌููู
The pH scale provides a convenient way to represent solution acidity.
Hุงุฃูุณ ุงูููุฏุฑูุฌููู ูู ู ููุงุณ ุงุณูู ููุชุนุจูุฑ ุนู ุชุฑููุฒ ุงูููุงุช ุงููุฏุฑูุฌูู +
H3Oุงูุงูููุฏุฑููููู + .
Also, the pH is defined as the negative logarithm of the hydrogen ion
concentration (in mol/L).
Similarly, it can express the concentration of OH- as pOH:
pH: ุงุฃูุณ ุงูููุฏุฑูุฌููู pOH : ุงุฃูุณ ุงูููุฏุฑููุณูุฏู
[H+OH] ุชุฑููุฒ ุงูููุงุช ุงูููุฏุฑููููู ุงู ุงูููุฏุฑูุฌูู :[
- ุชุฑููุฒ ุงูููุงุช ุงูููุฏุฑููุณูุฏ :[
b)
pH = - log [H+]
pOH = - log [OH-]
11
The sum of pH and pOH is always 14 at room temperature.
ุงูููุฏุฑููุณูุฏู ุนูุฏ ุฏุฑุฌุฉ ุงุฃูุณ ููุฏุฑูุฌููู ููู ูู ุงูุญุตูู ุนูู ุชุนุจูุฑ ุงุฎุฑ ู ููุฏุ ููู ู ุฌู ูุน ุงุงูุณ ุง
ูู ุง ููู: 14ุญุฑุงุฑุฉ ุงูุบุฑูุฉ ูููู
ูุงุนุฏุฉ ูุงู ุฉ
Measurement of pH ููุงุณ ุงุฃูุณ ุงูููุฏุฑูุฌููู
ุจุดูู ุนุงู ูู ูู ููุงุณ ุงุงูุณ ุงูููุฏุฑูุฌููู ุจุทุฑููุชูู:
1. Litmus paper ูุฑู ุชุจุงุน ุงูุดู ุณ
โRedโ paper turns blue above ~pH = 8
ูุชุญูู ูุฑู ุชุจุงุน ุงูุดู ุณ ุงุงูุญู ุฑ ุงูู ุงูููู ุงูููู ุงุงูุฒุฑู ูู ุงูู ุญููู ุงููุงุนุฏู
โBlueโ paper turns red below ~pH = 5
ูุฐู ุงูุทุฑููุฉ ุชููู ุงูู ุฏูู ุ ูุชุญูู ูุฑู ุชุจุงุน ุงูุดู ุณ ุงุงูุฒุฑู ุงูู ุงูููู ุงุงูุญู ุฑ ูู ุงูู ุญููู ุงูุญุงู ุถู
ูููุงุณ ุงุงูุณ ุงูููุฏุฑูุฌููู
2. pH meter
ูุนุชู ุฏ ูุฐุง ุงูู ููุงุณ ุนูู ููุงุณ .ุงุณ ุงุงูุณ ุงูููุฏุฑูุฌููู ุงุงููููุชุฑููู ุงูุซุฑ ุฏูุฉูุนุชุจุฑ ุงุณุชุฎุฏุงู ุนุฏุงุฏ ุงู ู ูู
ููุนุฑุถ ุฑูู ุงุฃูุณ ุงูููุฏุฑูุฌููู.ุงูุฌูุฏ ุงูููุฑุจุงุฆู ููู ุญููู ูู ู ุซู
pH 0 7 14
Acidic ุญู ุถู Neutral ู ุชุนุงุฏู Basic ูุงุนุฏู
pH + pOH = 14
12
Litmus paper pH meter
Problem
Calculate pH and pOH for each of the following solution at 25 oC.
a. 1.0 x 10-3
M OH-
b. 1.0 M H+
Solution
[OH-]
13
Problem
Solution
pH + pOH = 14
pOH = 14 โ pH
pOH = 14 โ 7.41
pOH = 6.59
To find [H+] we must go back to the definition of pH:
pH = - log [H+]
Thus
7.41 = -log [H+]
or
log [H+] = - 7.41
[H+] can be calculated by taking the antilog of โpH:
(ุจูุงุณุทู ุงูุญุงุณุจุฉ )ุจุงุฎุฐ ุนูุณ ุงูููุบุงุฑุชู ููููู ุฉ ุงูุณุงูุจุฉ ูุฃูุณ ุงูููุฏุฑูุฌููู
[H+] = antilog (-pH)
[H+] = antilog (โ 7.41)
14
[H+] = 3.90 X 10
-8 M
Also, to calculate [OH-], we must go back to the definition of pOH:
pOH = - log [OH-]
log [OH-]= - pOH
log [OH-]= - 6.59
[OH-]= antilog (- 6.59)
[OH-]= 2.60 x 10
-7 M
Problem
Which of the following aqueous solutions has the highest acidity and
which has the lowest?
1) pH = 7 2) pOH = 2 3) pH = 3 4) pOH = 9
15
Calculating the pH of Strong Acid and Base Solutions
ุญุณุงุจ ุงุฃูุณ ุงูููุฏุฑูุฌููู ูุฃูุญู ุงุถ ูุงูููุงุนุฏ ุงููููุฉ
Strong Acid
Strong acid are strong electrolytes that ionize completely in water (H2O), for
example, HCl, HNO3, HClO4 and H2SO4.
:ุงูู ุงุกุู ุซุงู ุฐูููู ุง ุฐูุฑ ุณุงุจูุงูุ ุงูุญู ุถ ุงูููู ูู ุงูู ุงุฏุฉ ุงูููุชุฑูููุชูุฉ ูููุฉ ูุงูุชู ุชุชุงูู ุจุงููุงู ู ูู
HCl , HNO3, HClO4 , H2SO4
ู ุน ุชุณุงูู ุชุฑููุฒ ุงูุญู ุถ (ุงูููุฏุฑููููู )ุงู ูู ุญุงูุฉ ุงูุญู ูุถ ุงููููุฉ ูุงู ุชุฑููุฒ ุงูููุงุช ุงูููุฏุฑูุฌูู
ู ุงูุญุธุฉ ุนุฏุฏ ุงูููุงุช ุงูููุฏุฑูุฌูู ุงููุงุชุฌู ุจุนุฏ ุงูุชููู.
n :ุนุฏุฏ ุงูููุงุช ุงูููุฏุฑูุฌูู
Ca :ุชุฑููุฒ ุงูุญู ุถ
Cl-(aq) + H
+(aq)
H2O(l) HCl (g)
0 0 0.1 Initial Concentration (M)
ุงูุชุฑููุฒ ุงุงูุจุชุฏุงุฆู
0.1 0.1 0 Final Concentration (M)
ุงูุชุฑููุฒ ุงูููุงุฆู
๐๐๐๐โ(๐๐ช) + 2H
+(aq)
H2O(l) H2SO4 (l)
0 0 0.1 Initial Concentration (M)
ุงูุชุฑููุฒ ุงุงูุจุชุฏุงุฆู
0.1 0.2 0 Final Concentration (M)
ุฑููุฒ ุงูููุงุฆูุงูุช
[H+] = n Ca
16
Strong Base
Strong bases are strong electrolytes that ionize completely in water (H2O), for
example, NaOH, KOH and Ba(OH)2.
OH) ุงูููุงุนุฏ ุงููููุฉ ูุงู ุชุฑููุฒ ุงูููุงุช ุงูููุฏุฑููุณูุฏ ูู ุญุงูุฉ-ู ุน ู ุงูุญุธุฉ ุนุฏุฏ ุงููุงุนุฏุฉุุชุณุงูู ุชุฑููุฒ (
ุงููุงุชุฌู ุจุนุฏ ุงูุชููู. ุงูููุฏุฑููุณูุฏุงูููุงุช
n : ุงูููุฏุฑููุณูุฏุนุฏุฏ ุงูููุงุช
ุชุฑููุฒ ุงููุงุนุฏุฉ : Cb
OH-(aq) + Na
+(aq)
H2O(l) NaOH (s)
0 0 0.1 Initial Concentration (M)
)ูุจู ุงูุชููู( ุงูุชุฑููุฒ ุงุงูุจุชุฏุงุฆู
0.1 0.1 0 Final Concentration (M)
)ุจุนุฏ ุงูุชููู( ุงูุชุฑููุฒ ุงูููุงุฆู
2OH-(aq) + Mg
2+(aq) H2O(l) Mg(OH)2 (s)
0 0 0.1 Initial Concentration (M)
)ูุจู ุงูุชููู( ุงูุชุฑููุฒ ุงุงูุจุชุฏุงุฆู
0.2 0.1 0 Final Concentration (M)
)ุจุนุฏ ุงูุชููู( ุงูุชุฑููุฒ ุงูููุงุฆู
Exercise
Calculate the pH for 5 x 10-4
M of Ba(OH)2?
Solution
2OH-
+ Ba2+
H2O(l) Ba(OH)2
[OH-] = n Cb n= 2, Cb = 5 x 10
-4
[OH-]= 2 x 5 x 10
-4 M
[OH-] = n Cb
17
[OH-]= 1 x 10
-3 M
[H+][OH
-] = 1x 10
-14
[H+] =
1x10โ14
[OHโ]
[H+] =
1x10โ14
1x10โ3
= 1x10-11
M
pH = -log [H+]
pH = -log 1x10-11
pH = 11
Exercise (Home work)
Calculate [H3O+], [OH
-], pH and pOH for the following solutions:
A) 750 mL aqueous solution containing 10 g of HCl (g).
B) 750 mL aqueous solution containing 10 g of H2SO4 (g).
Exercise (Home work)
Calculate [H3O+], [OH
-], pH and pOH for the following solutions:
A) 750 mL aqueous solution containing 10 g of NaOH (s).
B) 750 mL aqueous solution containing 10 g of Mg(OH)2 (g).
18
Calculating the pH of Weak Acid and Base Solutions
ุญุณุงุจ ุงุฃูุณ ุงูููุฏุฑูุฌููู ูุฃูุญู ุงุถ ูุงูููุงุนุฏ ุงูุถุนููุฉ
Weak Acids and acid ionization constant
ุงุงูุญู ุงุถ ุงูุถุนููุฉ ูุซุงุจุช ุชุฃูู )ุชููู( ุงูุญู ุถ ุงูุถุนูู
:ูู ุงูู ุงุกุู ุซุงู ุฐูููุชุฃูู ุฌุฒุฆูุงู ุงูุถุนูู ูู ุงููุฐู ูุชููู ุงูุ ุงูุญู ุถ ูู ุง ุฐูุฑ ุณุงุจูุงู
CH3COOH, HNO2, HF
CH3COO- (aq) + H
+(aq)
H2O(l) CH3COOH (l)
0 0 0.1 Initial Concentration (M)
)ูุจู ุงูุชููู( ุงุงูุจุชุฏุงุฆูุงูุชุฑููุฒ
[CH3COO-] = [H+] [H
+] 0.1-[H
+]
Final Concentration (M)
ุงูุชููู( ุจุนุฏ) ุงูุชุฑููุฒ ุงูููุงุฆู
The equilibrium expression for above reaction would be:
๐๐ =[๐+][๐๐๐๐๐โ]
[๐๐๐๐๐๐๐]
ุบูุฑุฉ ูุจุงูุชุงูู ุชููู ุงุงููููุงุช ุงููุงุชุฌู ูู ูุฉ ุงูุญู ุถ ุงูู ุชูููู ุชููู ุฌุฏุงู ุต ุงูุญู ูุถ ุงูุถุนููุฉ,ูู ุญุงูุฉ
(H+ูู ูุฐู ุงูุถุง ุตุบูุฑุฉุ ููุฐูู ูู ูู ุงุนุชุจุงุฑ ุงู ุชุฑููุฒ ุงูุญู ุถ ุงุงูุจุชุฏุงุฆู ูุงูููุงุฆู ุชูุฑูุจุง ู ุชูุงุฑุจุฉ. (
ุงูุญุงูุฉ ูุชู ุงุณุชุฎุฏุงู ุงููุงููู ุงูุชุงูู:
Ka : ุซุงุจุช ุชููู ุงูุญู ุถ ุงูุถุนูู (Weak acid dissociation, ionization, constant) ููู ุ
ุงู ุชูุงุนู ุงูุชูุงู ุงูุจุฑูุชูู ู ู ุงูุญู ุถ ุงูุถุนูู ุงูู ุงูู ุงุก.ุซุงุจุช ุงุชุฒ
Ca :ุงูุชุฑููุฒ ุงุงูุตูู ููุญู ุถ ุงูุถุนูู original concentration of weak acid .
ู ุงูุญุธุฉ:
ููู ุง ุฒุงุฏุช ููู ุฉKa ุฒุงุฏุช ููู ุฉ[H+ ูู ู ุซู ุฒุงุฏุช ููุฉ ุงูุญู ุถ. [
ู ุงูู ุงุก ููู ูู ุชุชุญุฏุฏ ุจู ูุฏุงุฑ ู ุงูุนุทูุฉ ุงูุญู ุถ ู ู ุงููู ุงูููุฏุฑููููู ุนูุฏ ุชูููู ู ููุฉ ุงูุญู ุถ
. Kaุงุงูุณุชุฏุงูู ุนููุฉ ู ู ููู ุฉ
ูุชุญุฏุฏ ุจู ูุฏุงุฑ ุนุฏุฏ ุงูู ูุงูุช ุงู ูุฒู ุงูุญู ุถ ุงูู ุฐุงุจ ูู ูุชุฑ ู ู ุงูู ุญููู.ุชุฑููุฒ ุงูุญู ุถ
[H+] = โ๐๐ ๐๐
19
ุงูุถุนูู ุงูุณูุจุฉ ุงูู ุฆููุฉ ููุชุฃูู ุงูุญู ุถ
Exercise
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is 2.38. Calculate
Ka for formic acid at this temperature.
Solution
HCOOH + H2O H3O+ + HCOO
-
๐๐ =[๐๐ ๐
+][๐๐๐๐โ]
[๐๐๐๐๐] (๐)
To calculate Ka, we need the equilibrium concentrations of all three things.
We can find [H3O+], which is the same as [HCOO
-], from the pH.
pH = -log [H3O+]
2.38 = -log [H3O+]
-2.38 = log [H3O+]
4.2 10-3
= [H3O+] = [HCOO
-]
From the constant of acid dissociation in (1)
Ka =[4.2 10โ3][4.2 10โ3]
[0.10 ]
Ka = 1.8 10-4
๐๐๐ซ๐๐๐ง๐ญ ๐ข๐จ๐ง๐ข๐ณ๐๐ญ๐ข๐จ๐ง =[๐+]
๐๐ ๐ฑ ๐๐๐
20
Problem
Calculate pH and pOH for 0.1 M for solution of CH3COOH, Ka =
1.8x10-5
.
Solution
[H+] = โ๐๐ ๐๐
[H+] = โ๐. ๐๐๐๐โ๐ ๐ฑ ๐. ๐
[H+] = 1.34 x 10
-3 M
pH= -log [H+]
pH= 2.87
pH + pOH= 14
pOH= 11.13
Exercise (Home work)
0.26 M of weak acid has pH of 2.86. Calculate Ka.
21
Weak Base and Base ionization constant
ุงูููุงุนุฏ ุงูุถุนููุฉ ูุซุงุจุช ุชุฃูู )ุชููู( ุงููุงุนุฏุฉ ุงูุถุนูู
:ู ุซุงู ุฐูู ูู ุงูู ุงุกุุฌุฒุฆูุงู ุชุชุฃูู ุงู ุชุชููู ูู ุงูุชูุงูุถุนูู ุงููุงุนุฏุฉุ ูู ุง ุฐูุฑ ุณุงุจูุงู
NH3, NH2OH, CH3NH2
OH- (aq) + NH4
+(aq)
H2O(l) NH3 (g)
0 0 0.1 Initial Concentration (M)
)ูุจู ุงูุชููู(ุงุงูุจุชุฏุงุฆู ุงูุชุฑููุฒ
[OH -] [NH4
+] = [OH
-] 0.1-[OH
-]
Final Concentration (M)
ุงูุชููู( ุจุนุฏ)ุงูุชุฑููุฒ ุงูููุงุฆู
The equilibrium expression for above reaction would be:
๐๐ =[๐๐๐
+][๐๐โ]
[๐๐๐]
ุงูู ุชูููู ุชููู ุฌุฏุงู ุตุบูุฑุฉ ูุจุงูุชุงูู ุชููู ุงุงููููุงุช ุงููุงุชุฌู ุงููุงุนุฏุฉูู ูุฉ ุงูุถุนููุฉ, ุงูููุงุนุฏูู ุญุงูุฉ
(OH- ูู ูุฐู ุงูุญุงูุฉ ูุชู ุงุณุชุฎุฏุงู ุงููุงููู ุงูุชุงูู: ุงูุถุง ุตุบูุฑุฉุ (
Kb : ุฉุงูุถุนูู ุงููุงุนุฏุฉ ุซุงุจุช ุชููู (Weak base dissociation, ionization, constant) ููู ุ
.ุฉุงูุถุนูู ุงููุงุนุฏุฉุงูู ุงุก ุงูู ู ู ุงูุจุฑูุชููุชูุงุนู ุงูุชูุงู ุซุงุจุช ุงุชุฒุงู
Cb : ุฉุงูุถุนูู ูููุงุนุฏุฉุงูุชุฑููุฒ ุงุงูุตูู Original concentration of weak base .
ู ุงูุญุธุฉ:
ููู ุง ุฒุงุฏุช ููู ุฉKb ุฒุงุฏุช ููู ุฉ[OH- .ุงููุงุนุฏุฉูู ู ุซู ุฒุงุฏุช ููุฉ [
[OH-] = โ๐๐ ๐๐
๐๐๐ซ๐๐๐ง๐ญ ๐ข๐จ๐ง๐ข๐ณ๐๐ญ๐ข๐จ๐ง =[๐๐โ]
๐๐ ๐ฑ ๐๐๐
22
Problem
A) Calculate [H3O+], [OH
-], pH and pOH for an aqueous solution 10
g of NH3 (g) in 750 mL of solution. (Kb,NH3=1.8 x 10-5
).
B) Calculate Kb of a 0.56 M weak base if its pOH =5.75
(N= 14, H=1)
Solution
We need to calculate the original concentration of weak base:
V= 750 mL = 0.75 L
๐ง =๐๐๐ ๐ (๐)
๐. ๐ค๐ก
๐ง =10
17.034= 0.587 ๐๐๐
Then, the original concentration of weak base is:
Molarity = ๐
๐ (๐ฟ)
๐๐ =0.587
0.75= 0.783 ๐
[OH-] = โ๐๐ ๐๐
[OH-] = โ๐. ๐ ๐ ๐๐โ๐ ๐ฑ ๐. ๐๐๐
[OH-] = 1.409 x 10
-5
A)
23
[H+] =
๐๐ฑ๐๐โ๐๐
[๐๐โ]=
๐๐ฑ๐๐โ๐๐
๐.๐๐๐ ๐ ๐๐โ๐ = 7.097 x 10โ10 M
pOH = -log 1.409 x 10-5
= 4.851
pH = 14 - 4.851= 9.149
Kb = 0.56 M, pOH =5.75
pOH = -log [OH-]
[OH-] = antilog (-5.75) =1.78 x 10
-6 M
[OH-] = โ๐๐ ๐๐
1.78 x 10-6
=โKb x 0.56
Kb = 5.65 x10-12
M
B)
24
Ka and Kb
Ka and Kb are related in this way:
Therefore, if you know one of them, you can calculate the other.
The Salt ุงุฃูู ุงูุญ
A salt is defined as the substance that is produced from the reaction
between an acid and a base. There are four different types of salts:
ุชุนุฑู ุงุงูู ุงูุญ ุนูู ุงููุง ุงูู ูุงุฏ ุงูู ุชูููู ู ู ุชูุงุนู ุญู ุถ ู ุน ูุงุนุฏุฉ. ููุฌุฏ ุงุฑุจุน ุงููุงุน ู ุฎุชููุฉ
ู ู ุงุงูู ุงูุญ ูู ุง ููู:
1- Salt from the reaction between a strong acids and a strong base:
HCl + NaOH NaCl + H2O (Neutral, pH=7)
2- Salt from the reaction between a strong acids and a weak base:
HCl + NH3 NH4Cl (Acidic, pH<7)
3- Salt from the reaction between a weak acids and a strong base:
CH3COOH + NaOH CH3COONa + H2O (Basic, pH >7)
Ka Kb = Kw
25
4- Salt from the reaction between a weak acids and a weak base:
CH3COOH + NH3 CH3COONH4
(According to the value of Ka and Kb)
Buffer Solution ุงูู ุญุงููู ุงูู ูุธู ุฉ
A buffered solution is a solution of a weak acid with one of its salts or a weak
base with one of its salts, and it resists drastic changes in pH when small
amounts of strong acid or strong base are added to them.
ุงูุฐู ููุงูู ุฉ ุถุนููุฉ ูู ูุญูุงุ ูุงู ู ู ูุงุนุฏ ุงูู ููู ู ู ุญู ุถ ุถุนูู ูู ูุญุฉ ูู ุงูู ุญููู ุงูู ุญููู ุงูู ูุธู
ุนูุฏ ุงุถุงูุฉ ูู ูุฉ ููููุฉ ู ู ุงูุญู ุถ ุงู ุงููุงุนุฏุฉ ุงููููุฉ. pHุงูุชุบูุฑ ูู ููู ุฉ
ูู ุงุญุชูุงุฆู ุนูู ูุงุฆุถ ู ู ุงูุจุฑูุชููุงุช ุงูู ูุฌุจุฉ ุนูู pHุณุจุจ ุงู ุงูู ุญููู ุงูู ูุธู ููุงูู ุงู ุชุบูุฑ ูู ููู ุฉ
ููุบู ุงุซุฑ ุงู ุญู ุถ ุดูู ุญู ุถ ุถุนูู ููุงุฆุถ ู ู ุงููุงุนุฏุฉ ุนูู ุดูู ุงููููุงุช ุณุงูุจุฉุ ูุจูุฐุง ูุณุชุทูุน ุงู
ุงู ูุงุนุฏุฉ ูุฏ ุชุถุงู ุงููุฉ.
For example, blood is a buffered solution, which can absorb the acids and bases
produced in biologic reactions without changing its pH. A constant pH for blood
is vital because cells can survive only in a very narrow pH range.
How does a buffered solution resist changes in pH when an acid or a
base is added?
1- The addition of a strong acid to a buffer solution.
CH3COO- + H3O
+ CH3COOH + H2O
H3Oู ู ุงูู ุนุงุฏูุฉ ุงุนุงูุฉุ ุนูุฏ ุงุถุงูุฉ ุงูุญู ุถ ุงูููู )+ุงูู ุญููู ุงูู ูุธู ( ูุฃู ุงูููุงุช ุงูุฎุงูุช ู ู ( ุงูู ุงูููุงุช ุงูุฎุงูุช )
ุชูุฑูุจุงู ุซุงุจุชุฉ. pHุชุชูุงุนู ู ุน ุงูููุงุช ุงูููุฏุฑููููู )ุงูุญู ุถ ุงูู ุถุงู( ูุชุฒูู ุงุซุฑูุง ูุชุจูู ููู ุฉ
ุงููู ุงูุฎุงูุช ู ู ุงูู ูุธู ุงูู ุญููู
ุงูุญู ุถ ุงูููู
26
2- The addition of a strong base to a buffer solution.
CH3COOH + OH- CH3COO
- + H2O
OHู ู ุงูู ุนุงุฏูุฉ ุงุนุงูุฉุ ุนูุฏ ุงุถุงูุฉ ุงููุงุนุฏุฉ ุงููููุฉ )- ( ุงูู ุญู ุถ ุงูุฎู )ู ู ุงูู ุญููู ุงูู ูุธู ( ูุฃู ุญู ุถ ุงูุฎู ูุชูุงุนู ู ุน
ุชูุฑูุจุงู ุซุงุจุชุฉ. pHุงูููุงุช ุงูููุฏุฑููุณูุฏ )ุงููุงุนุฏุฉ ุงูู ุถุงูุฉ( ููุฒูู ุงุซุฑูุง ูุชุจูู ููู ุฉ
The equation of Henderson โHasselbalch can be used to calculate
or determine the pH of buffer solution before and/or after addition
small amount of strong acid or strong base.
ูู ุญููู ู ูุธู ุจูุงุณุทุฉ ุชุทุจูู ู ุนุงุฏูุฉ ููุฏุฑุณูู ูุงุณูุจุงูุฎ pHูู ูู ุญุณุงุจ ูุชุญุฏูุฏ
.ุงูุชุงููุฉ
ู ู ุงูุฎู ุญู ุถ ุงูู ูุธู ุงูู ุญููู
ุงููุงุนุฏุฉุฉุงูููู
pH = pKa + log [๐ฌ๐๐ฅ๐ญ]
[๐๐๐ข๐]
pOH = pKb + log [๐ฌ๐๐ฅ๐ญ]
[๐๐๐ฌ๐]
27
Exercise
Calculate pH of a buffer solution that is 0.12 M lactic acid
CH3CH(OH)COOH and 0.1 M sodium lactate CH3CH(OH)COONa.
Calculate the change in its pH after dissolving 0.01 mol of HCl in the
solution. ( Ka CH3CH(OH)COOH = 1.4x10-4
).
Solution
CH3CH(OH)COO- + H3O
+ CH3CH(OH)COOH + H2O
Before: pH = pKa + log [๐ฌ๐๐ฅ๐ญ]
[๐๐๐ข๐]
pH = -log 1.4x10-4 + log
[0.1]
[.12]
pH = 3.775
After: pH = pKa + log [๐ฌ๐๐ฅ๐ญ]
[๐๐๐ข๐]
[Salt] = 0.1- 0.01= 0.09 M
[Acid] = 0.12+ 0.01= 0.13 M
pH = -log 1.4x10-4 + log
[0.09]
[.13]
pH = 3.694
ฮpH= pHafter - pHbefore
ฮpH = 3.694-3.775 = -0.08
28
Homework
Calculate the pH of a buffer solution composed of acetic acid/acetate
in which [CH3COOH] = [CH3COO-] = 1M.
What happens to the pH if we add 0.2 mol of HCl to 1L of this
solution?
What happens to the pH if we add 0.2 mol of NaOH to 1L of this
solution?
.