Autoionization of water.pdf

Auto ionization of water-self ionization, It is an example of autoprotolysis, and exemplifies the amphoteric nature of water, intermolecular proton transfer that forms a hydronium ion (H3O+) and a hydroxide ion (OH – ):
Dissociation of water-Dissociation constant-K
Molar concentration of water, ion product for water (Kw)
pH, pka-Relating pH and pka with Henderson Hasselbalch equation, Examples for pH and pka problem,
Acids and Bases-Conjugate acids and conjugate base. dissociation constant of water.

1. Theself-ionizationofwater(alsoautoionizationofwater,
andautodissociationofwater)isanionizationreaction
inpurewaterorinanaqueoussolution,inwhichawater
molecule,H
2O,deprotonates(losesthenucleusofoneof
itshydrogenatoms)tobecomeahydroxideion,OH
−
.The
hydrogennucleus,H
+
,immediatelyprotonatesanother
watermoleculetoformhydronium,H
3O
+
.
Itisanexampleofautoprotolysis,andexemplifies
theamphotericnatureofwater.
Shiny C Thomas, Department of Biosciences, ADBU

2. Ionization of Water
The ability of water to ionize is of central importance for
life.
Since water can act both as an acid and as a base, its
ionization may be represented as an intermolecular
proton transfer that forms a hydronium ion (H3O+) and a
hydroxide ion (OH –):

3. •The transferred proton is actually associated with a
cluster of water molecules.
•Since hydronium and hydroxide ions continuously
recombine to form water molecules, an individual
hydrogen or oxygen cannot be stated to be present as
an ion or as part of a water molecule.
•At one instant it is an ion; an instant later it is part of a
water molecule.

4. Individual ions or molecules are therefore not considered.
For dissociation of water,
where the brackets represent molar concentrations (strictly
speaking, molar activities) and K is the dissociation constant.
Since 1 mole (mol) of water weighs 18 g, 1 liter(L) (1000 g)
of water contains 1000 ÷18 = 55.56 mol. Pure water thus is
55.56 molar.

5. How much is a mole of water?
A mole is a unit of measuring the quantity of anything.
It is simple to calculate the weight and volume of a mole
of water.

6. QUICK MOLE REVIEW
A mole is a unit of measuring the quantity of anything.
Asingle moleis set to the number of particles found in
12.000 grams of carbon-12.
This number is 6.022 x 10
23
carbon atoms. The number
6.022 x 10
23
is known as Avogadro's Number.
A mole of carbon-12 atoms has 6.022 x 10
23
carbon-12
atoms. A mole of apples has 6.022 x 10
23
apples.
A moleof water has 6.022 x 10
23
water molecules.

7. MASS OF 1 MOLE OF WATER
How much water is that to most people?
Water (H
2O) is madefrom 2 atomsof hydrogen and 1
atomof oxygen. A mole of water molecules would be 2
moles of hydrogen atoms plus 1mole of oxygenatoms.
From theperiodic tablewe see theatomic
weightofhydrogenis 1.0079 and the atomic weight
ofoxygenis 15.9994.
Atomic massis thenumber of grams per moleof the
element. This means 1 mole of hydrogen weighs 1.0079
grams and 1 mole of oxygen weighs 15.9994 grams.

8. Water would weigh
weight of water = 2(1.0079) g + 15.9994 g
weight of water = 2.0158 g + 15.9994 g
weight of water = 18.0152 g.

9. The molar concentration of water, 55.56 mol/L, is too great
to be significantly affected by dissociation.
It therefore is considered to be essentially constant.
This constant may therefore be incorporated into the
dissociation constant K to provide a useful new constant K w
termed the ion product for water.
The relationship between Kw and K is shown below:

10. Note that the dimensions of K are moles per litre and those
of Kw are moles 2 per liter2. As its name suggests, the ion
product Kw is numerically equal to the product of the molar
concentrations of H+ and OH–:

11. pHisa measure of the concentration of hydrogen ions in an
aqueous solution.
pKa(acid dissociation constant) is related, but more specific,
in that it helps you predict what a molecule will do at a
specific pH.
Essentially, pKatells you what the pH needs to be in order
for a chemical species to donate or accept a proton.

12. pH and pKa
The lower the pH, the higher the concentration of hydrogen
ions, [H
+
]. The lower the pKa, the stronger the acid and the
greater its ability to donate protons.
pH depends on the concentration of the solution. This is
important because it means a weak acid could actually have
a lower pH than a diluted strong acid.
For example, concentrated vinegar (acetic acid, which is a
weak acid) could have a lower pH than a dilute solution of
hydrochloric acid (a strong acid). On the other hand, the pKa
value is a constant for each type of molecule. It is unaffected
by concentration.

13. Even a chemical ordinarily considered a base can have a pKa
value because the terms "acids" and "bases" simply refer to
whether a species will give up protons (acid) or remove
them (base).
For example, if you have a base Y with a pKaof 13, it will
accept protons and form YH, but when the pH exceeds 13,
YH will be deprotonated and become Y. Because Y removes
protons at a pH greater than thepH of neutral water(7), it is
considered a base.

14. RELATING PH AND PKA WITH THE HENDERSON-
HASSELBALCH EQUATION
If you know either pH or pKayou can solve for the other
value using an approximation called theHenderson-
Hasselbalchequation:
pH = pKa+ log ([conjugate base]/[weak acid])
pH = pka+log([A
-
]/[HA])
pH is the sum of the pKavalue and the log of the
concentration of the conjugate base divided by the
concentration of the weak acid.

15. At half the equivalence point:
pH = pKa
It's worth noting sometimes this equation is written for the
K
avalue rather than pKa, so you should know the
relationship:
pKa= -logK
a

16. EXAMPLE PKA AND PH PROBLEM
Find [H
+
] for a solution of 0.225 M NaNO
2and 1.0 M HNO
2.
The K
avalue (from a table) of HNO
2is 5.6 x 10
-4
.
pKa=−logK
a=−log(7.4×10
−4
)=3.14
pH = pka+ log ([A
-
]/[HA])
pH=pKa+log([NO
2
-
]/[HNO
2])
pH=3.14+log(1/0.225)
pH=3.14+0.648=3.788
[H+]=10
−pH
=10
−3.788
=1.6×10
−4

17. ACIDS BASES AND pH
What are acids and bases?
An Acid is a molecule that acts as a proton(hydrogen ion)
donor.
ABaseissimilarlydefinedasaprotonacceptor.
Howreadilyacidsandbasesloseorgainprotonsdepends
onthechemicalnatureofthecompoundsunder
consideration.
Thedegreeofdissociationofacidsinwaterrangesfrom
completedissociationforastrongacidtonodissociation
foraveryweakacid,andanyintermediatevalueis
possible.
HA ⇌H
+
+ A
-
Acid Conjugate base
Ka=[H
+
][A
-
]/[HA]

18. Inthisexpression,thesquarebracketsrefertomolar
concentration–thatis,theconcentrationinmolesperlitre.
Foreachacid,thequantityKahasafixednumericalvalueat
agiventemperature.
Thisvalueislargerformorecompletelydissociatedacids;
thusthegreatertheKa,thestrongertheacid.
Theprecedingacid-basereactionisaproton–transfer
reactioninwhichwateractsasabaseaswellasthesolvent.
Amorecorrectwaytowritetheequationisasfollows:
HA(aq) + H
2O⇌H
3O
+
(aq) + A
-
(aq)
Acid Base Conjugate acid Conjugate base to HA to H
2O

19. Fig: 1
Both the hydrogen ion (H
+)
and the hydroxide ion (OH
-
)
are associated with several water molecules. Fig:2
It is important to have a quantitative estimate of the
degree of dissociation of water. We can start with the
expression
Ka= [H
+
][OH
-
]/[H
2O] * can be considered a constant.

20. (The numerical value is 55.5M, which can be obtained by
dividing the number of grams in 1L, 1000g, by the molecular
weight of water, 18g/mol; 1000/18=55.5M)
Thus,
Ka= [H
+
][OH
-
]/55.5
Kax 55.5 = [H
+
][OH
-
] = Kw ** a new constant Kw, the
ion product constant for water.
Kw= [H
+
][OH
-
= 10
-14
OR (Kw = [H
+
][OH
-
]=(10
-7
) (10
-7
) = 10
-14
This relationship which we have derived for pure water, is
valid for any aqueous solution
,
whether neutral, acidic, or
basic.

21. pH=-log
10[H
+
] is defined as with the logarithm taken to the
base.
When a solution has a pH of 7, it is said to be neutral, like
pure water. Acidic solutions have pH values lower than 7,
and basic solutions have pH values higher than 7.
In biochemistry most of the acids encountered are weak
acids. These have a K
abelow 1.
pK
ahas been defined by analogy with the definition of pH:
pK
a=-log
10K
a
pK
a–numerical measure of acid strength. The smaller its
value, the stronger the acid. This is the reverse of the
situation with K
a, where larger values imply stronger acids.

22. Why do we want to know the pH
An equation connects the Kaof any weak acid with the pH
of a solution containing both that acid and its conjugate
base.This relationship has wide use in biochemistry.
K
a= [H
+
][A
-
]/[HA]
log K
a= log [H
+
] + log [A
-
]/[HA]
-log[H
+
] = -log K
a+ log [A-]/[HA]
We then use the definitions of pHandpK
a:
pH = pK
a+ log [A-]/[HA]
This relationship is known as the Henderson-Hasselbalch
equation. When the concentration of acid, [HA], and the
concentration of the conjugate base [A-], are equal
[HA]=[A-]
The ratio [A-]/[HA] is then equal to 1, and the logarithm of
1 is equal to zero.