Acids bases : Acids bases
introduction : introduction The word “acid”has been derived from a latin word “acidus” meaning sour.
Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with metals.
Similarly, bases are known to turn red litmus paper blue, taste bitter and feel soapy.
When acids and bases are mixed in the right proportion they react with each other to give salts.
Arrhenius Concept of Acids andBases : Arrhenius Concept of Acids andBases According to Arrhenius theory, acids are:
substances that dissociates in water to give hydrogen ions H+(aq)
and bases are substances that produce hydroxyl ions
OH–(aq).
HX (aq) → H+(aq) + X– (aq)
or
HX(aq) + H2O(l) → H3O+(aq) + X –(aq)
Slide 4 : A bare proton, H+ is very reactive and cannot exist freely in aqueous solutions. Thus, it bonds to the oxygen atom of a solvent water molecule to give trigonal pyramidal hydronium ion, H3O+.
Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation:
MOH(aq) → M+(aq) + OH–(aq)
Limitations : Limitations Arrhenius concept of acid and base, however, suffers
from the limitation of being applicable only to aqueous solutions and also, does not account for the basicity of substances like, ammonia which do not possess a hydroxyl group.
The Brönsted-Lowry Acids and Bases : The Brönsted-Lowry Acids and Bases According to Brönsted-Lowry theory, acid is
a substance that is capable of donating a hydrogen ion H+
HCl H+ + Cl-
H2SO4 H+ + HSO4-
CH3COOH H+ + CH3COO-
A base is a substance which can behave as a proton acceptor. Any species which has a lone pair of electrons can thus behave as a base:
Slide 7 : NH3 + H+ NH4+
OH- + H+ H2O
CO32- + H+ HCO3-
In short, acids are proton donors and bases are
proton acceptor.
Consider the example of dissolution of NH3
in H2O represented by the following equation:
Slide 8 : In this reaction, water molecule acts as proton donor and
ammonia molecule acts as proton acceptor and are thus, called Lowry-Brönsted acid and base, respectively. In the reverse reaction, H+is transferred from NH4+ to OH–. In this case,NH4+ acts as a Bronsted acid while OH– acted as a Brönsted base
The acid-base pair that differs only by one proton is called a conjugate acid-base pair.
Therefore, OH– is called the conjugate base of an acid H2O and NH4+ is called conjugate acid of the base NH3.
If Brönsted acid is a strong acid then its conjugate base is a weak base and vice versa.
conjugate acid has one extra proton and each conjugate base has one less proton.
Lewis Acids and Bases : Lewis Acids and Bases Lewis defined an acid as a species which accepts electron pair and base which donates an electron pair.
A typical example is reaction of electron deficient speciesBF3 with NH3.
BF3 does not have a proton but still acts as an acid and reacts with NH3 by accepting its lone pair of electrons.
The reaction can be represented by,
BF3 + :NH3 → BF3:NH3
Electron deficient species like AlCl3, Co3+,Mg2+, etc. can act as Lewis acids while species like H2O, NH3, OH– etc. which can donate a pair of electrons, can act as Lewis bases.
Slide 10 : strong acids have very weak conjugate bases.
Strong acids likeperchloric acid (HClO4), hydrochloric acid(HCl), hydrobromic acid (HBr), hydroiodic acid
(HI), nitric acid (HNO3) and sulphuric acid
(H2SO4) will give conjugate base ions ClO4–, Cl,Br–, I–, NO3– and HSO4– , which are much weaker
bases than H2O.
Slide 11 : Weak acids have very strong conjugate bases.
Forexample, NH2–, O2– and H– are very good proton
acceptors and thus, much stronger bases than H2O.
Similarly a very strong base would give a very weak conjugate acid
Slide 12 : Classify the following species into Lewis acids and Lewis bases and show how these act as such:
(a) HO– (b)F – (c) H+ (d) BCl3
Solution
(a) Hydroxyl ion is a Lewis base as it can donate an electron lone pair (:OH– ).
(b) Flouride ion acts as a Lewis base as it can donate any one of its four electron lone pairs.
(c) A proton is a Lewis acid as it can accept a lone pair of electrons from bases like hydroxyl ion and fluoride ion.
(d) BCl3 acts as a Lewis acid as it can accept a lone pair of electrons from species like ammonia or amine molecules.
Slide 13 : 2.In each of the following reactants, one of the reactants acts as an acid. Identify it:
H2O(l) + NH3(aq) NH4+(aq) + OH-(aq)
CH3COOH(aq) + HClO4(aq) CH3COOH2+(aq) + ClO4-(aq)
HCO3-(aq) + HSO4-(aq) H2O(l) + CO2(g) + SO42-(aq)
H3O+(aq) + OH-(aq) 2H2O(l)
Slide 14 : 3.Identify the acid-base conjugate pairs in the following reactions:
HCO3-(aq) + H2O(l) CO32-(aq) + H3O+(aq)
HCO3-(aq) + H3O+(aq) CO2(g) + H2O(l) + H2O(l)
H2SO4(aq) + HNO3(aq) HSO4-(aq) + NO2+(aq) + H2O(l)
HSO4-(aq) + OH-(aq) SO42-(aq) + H2O(l)
Slide 15 : What are the conjugate base of the each of the following:
1.HS-;2.H3O+;3.H2PO4-;4.CH3COOH.
Give the conjugate acids of the following:
1.OH-;2.HCO3-;3.CO32-;4.HCOO-
The Ionization Constant of Waterand its Ionic Product : The Ionization Constant of Waterand its Ionic Product 2H2O(l) H3O+(aq) + OH-(aq)
This is known as the auto-ionisation of water.
The dissociation constant is represented by,
K = [H3O+] [OH–] / [H2O]
The concentration of water under these conditions is not changed significantly, since the proportion of water which dissociates into its ions is small.
The water concentration [H2O]; can thus be assumed to be constant and it can be incorporated into Kc to give a new constant, Kw, :
Slide 17 : Kc[H2O] = Kw = [H3O+][OH-]
Kw = [H3O+][OH-]
The expression [H3O+][OH-] is known as the ionic product of water and has a value of 1.0 x 10-14 mol2dm-6 at 25oC.
This value is a constant at a given temperature.
In pure water, the [H3O+] and [OH-] concentrations are equal.
It follows that [H3O+][OH-] = [H3O+]2 = Kw.
So [H3O+] = [OH-] = √Kw = 1.0 x 10-7 moldm-3.
Slide 18 : We can distinguish acidic, neutral and basic aqueous solutions by the relative values of the H3O+ and OH– concentrations:
Acidic: [H3O+] > [OH– ]
Neutral: [H3O+] = [OH– ]
Basic : [H3O+] < [OH–]
Defining pH : Defining pH Acidic solution has pH < 7
Basic solution has pH > 7
Neutral solution has pH = 7
Slide 20 : Now again, consider the equation at 298 K
Kw = [H3O+] [OH–] = 10–14
Taking negative logarithm on both sides of equation, we obtain
–log Kw = – log {[H3O+] [OH–]}
= – log [H3O+] – log [OH–]
= – log 1 x 10 –14
pKw = pH + pOH = 14
Slide 21 :