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(ii) Aprotic Solvents :
Such solvents do not have replaceable hydrogen in them. These can be classified into three categories
(a) Non polar or very weakly polar, nondissociated liquids, which do not solvate strongly.
Ex . CCl , hydrocarbons, C H , C H etc.
4 6 6 6 12
(b) Non-ionised but strongly solvating, generally polar solvents.
E x . Acetonitrile CH3CN, DMF, DMSO (dimethyl sulfoxide), THF and SO2.
(c) Highly polar, autoionising solvents.
E x . Inter halogen compounds (BrF , IF and trichloro phosphine)
35
(a) 2 BrF3 BrF2+ + BrF4–
(b) 2 IF IF + + IF –
5 46
(c) 2Cl PO Cl PO+ + Cl PO–
3 24
Levelling Solvents :
(i) The Bronsted - Lowery theory can be extended to acid - base reactions in non-aqueous solvents. It can be
used in differentiating the acid strength of a particular acid and in titration of weak bases.
(ii) In water solvent, mineral acids appear to be equally strong because of their complete ionisation, water is
called here a levelling solvent because it levels all the acids to the same strength.
(iii) If instead of water solvent, we take mineral acids in pure acetic acid solvent (which is poor proton acceptor
as compared to water) it is found acids become weak and can be differentiated.
E x . HCl + CH COOH Cl– + CH COOH +
3 32
Acid Base Base Acid
In above example acetic acid and Cl– ions both compete for protons and the former being a poor proton
acceptor does it much less effectively than water. Thus HCl in acetic acid solvent appears to be a much weaker
acid than that in water.
(iv) Mineral acids in acetic acid solvent follow the following order of their strengths.
HNO < HCl < H SO < HBr < HClO
3 24 4
(v) A weak base like acetamide or acetanilide in aqueous medium can not be titrated with acids. If how ever, the
weak base is taken in glacial acetic acid solvent, the former behaves as a strong base and can be titrated.
This is because acetic acid (which acts as a better proton donor) exerts a levelling effect on the base.
Lux - Flood Concept (1939 & 1947)* :
(i) The proton plays an important role in explaining the acid-base behaviour in the Bronsted-Lowery concept.
Lux observed that acid - base reactions are also feasible in oxide systems without the aid of protons.
(ii) Above approach was extended by Flood and applied to non-protonic systems, which were not covered by
the Bronsted Lowery concept.
(iii) According to this concept a base (like CaO, BaO or Na O) is an oxide ion (O2–) donor and an acid (like SiO ,
22
CO2 or P4O10) is an oxide ion (O2–) acceptor.
E x . Base Acid
(a) CaO + SiO CaSiO
(b) 23
(iv)
Ex. 6Na O + PO 4Na PO
Note : 2 4 10 34
(i)
Substances are termed amphoteric if they show a tendency of losing as well as accepting an oxide ion.
(ii)
ZnO, Al O
23
The concept is particularly applicable to reactions which take place at high temperature i.e. in
metallurgical operations or during the manufacture of ceramics and glass.
The approach can be extended to include other negative ion systems (like halides, sulphide or carbanion).
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Example base on : Bronsted - Lowery Concept
E x . Select the aprotic solvent from the following is :
(A) H O (B) C H (C) HF (D) NH
2 66 3
S o l . (B) according to the classification of solvents. (D) NO –
2
E x . The strongest conjugate base is -
(D) ClO –
(A) Cl– (B) CH COO– (C) SO 2– 4
3 4
(D) Bronsted base
S o l . (B) CH COOH is weakest acid among HCl, CH COOH, HSO – and HNO .
3 3 42 (D) OH–
E x . Which is the strongest Bronsted base in the following anions :
(A) ClO– (B) ClO – (C) ClO –
2 3
S o l . (A), HClO is weakest acid among HClO, HClO , HClO and HClO .
23 4
Ex . Give appropriate equation and label acid and base.
Sol. NH CONH + NH NH + + NH CONH–
22 3 4 2
Acid base Acid base
In liquid NH solution urea can show weak acidic nature.
3
E x . Ammonium ion is-
(A) A Lewis acid (B) Lewis base (C) Bronsted acid Ans. (C)
Ans. (A)
S o l . Correct answer is (C)
Self Practice Problem :
1 . (a) Write conjugate acids of
SO42–, RNH2, NH2–, F –
(b) Write conjugate base of
HNO2, OH–, H2CO3, HClO4.
(c) Write conjugate acids and conjugate base of amphoteric species.
HS–, NH3, H2O, HSO4–
2 . Which of the following is the strongest base-
(A) NH2– (B) CH3COO– (C) C2H5O–
Lewis concept (electronic concept) :
An acid is a molecule/ion which can accept an electron pair with the formation of a coordinate bond.
Acid e– pair acceptor
E x . Electron deficient molecules : BF3, AlCl3
Cations : H+, Fe2+, Na+
Molecules with vacant orbitals : SF4, PF3
A base is any molecule/ion which has a lone pair of electrons which can be donated.
Base (One electron pair donate)
E x . Molecules with lone pairs : NH3, PH3, H2O, CH3OH
Solved Examples
E x . In which of the following reaction does NH act as an acid ?
3
(A) NH3 + H+ NH4+ 1
(B) NH3 + Na NaNH2 + 2 H2
(C) NH3 + HCl NH4Cl (D) none of these
(B)
Sol.
In the following reaction, NH3 changes of NaNH2 which contains NH2 ion. This means that NH3 has donated
a proton to Na and hence acts as an acid.
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E x . Sulphanilic acid is a/an (C) Neither (A) or (B) (D) Both (A) and (B)
(A) Arrhenius acid (B) Lewis base
SO3H
S o l . (D) Sulphanilic acid is .
NH2
The SO3H group is capable of donating H+ and hence it acts as arrhenius acid while nitrogen in the NH2
group contains lone pair of electrons and hence can act as lewis base by donating it.
1 . Hard Acids :
(i) It includes ions of alkali metals, alkaline earth metals, lighter transition metals in higher oxidation states.
E x . Ti4+, Cr3+, Fe3+, Co3+ and H+.
(ii) Small size, high polarising power, high electronegativity and high positive oxidation state are the characteristic
features of hard acids.
(iii) Their outer electrons or orbitals are not easily distorted.
Note :
(i) It should be remembered that we are considering ions like Li+, Na+ etc. These ions are highly electronegative
because their subsequent ionisation energies are very high.
2 . Hard Bases :
(i) The ligands which preferably combine with hard acids are called hard bases.
E x . NH , R N, H O and F – etc.
33 2
(ii) Hard bases (ligands) are also small, not very polarisable and having high electronegativity.
3 . Soft Acids :
(i) It includes ions of the heavier transition metals and those in lower oxidation states.
E x . Cu+, Ag+, Hg 2+, Pd2+, Pt2+ and Hg2+.
2
(ii) They are large sized and their outer electrons or orbitals are easily distorted. They have low polarising power
and low electronegativity.
4 . Soft Bases :
(i) The ligands which preferably combine with soft acids are called soft bases.
E x . R P (phospines), R S (thioethers), CO, CN–, H– etc.
32
(ii) Soft bases (ligands) are large sized, more polarisable and having low electronegativity.
5 . Pearson's Principle :
(i) It is a simple rule to predict qualitatively the relative stability of acid - base reactions.
(ii) "Hard acids prefer to bind to hard bases and soft acids prefer to bind to soft bases". Thus the complex A : B
is the most stable when both A and B are either hard or soft.
(iii) However, when one of the reactants is very hard and the other very soft, the complex will not be much
stable.
(iv) Hard-hard interactions involve ionic bonding and soft-soft interactions result primarily from covalent bonding.
(v) Soft-soft interaction can be explained on the basis of -bonding. Soft acids have low oxidation states and
possess large number of d-electrons. Thus they can act as -bond donors and soft bases are generally
-bond acceptors. The presence of d-orbitals on the ligands (except on CO) helps to strengthen the
-bonding.
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6. Classification of Acids :
Table - 2
Hard Border line Soft
H+, I(A) Group ions, Fe2+, Co2+, Ni2+, Co(CN) 3–, Pd2+, Pt2+,
5
II(A) Group ions, Be(CH ) , Cu2+, Zn2+, B(CH ) ,
32 33 Pt4+, Cu+, Ag+, Au+,
Sc3+, Ce4+, Ti4+, VO2+, GaH , R C+, C H +, Cd2+, Hg 2+, Hg2+, CH Hg+,
33 65 23
Cr3+, Cr6+, MoO3+, WO4+, Sn2+, Pb2+, NO+, Sb3+, BH , Ga(CH ) , GaCl ,
3 33 3
Mn2+, Mn7+, Fe3+, Co3+, Bi3+, SO2 GaBr3, GaI3, Tl+, Tl(CH3)3,
BF , BCl , B(OR) , Al3+, CH , Carbenes, -acceptors
33 3 2
Al(CH ) , AlCl , AlH , Ga3+, (TNB, chloranil, quinones, tetracyanoethylene)
33 3 3
CO , Si4+, Sn4+, N3+, As3+, HO+, RO+, RS+, Te4+, Br ,
2 2
SO3, RSO2+, Cl3+, Cl7+, I5+, Br+, I2, I+, ICN, O, Cl, Br, I, N,
RO–, M° (metal atoms and bulk metals)
I7+, HX (H-bonding molecules)
7. Classification of Bases : Table - 1 Soft
Hard Border line
NH , RNH , N H , H O, Ph-NH , C H N, N –, H–, R–, C H , C H , CN–, RNC
3 2 24 2 2 55 3 24 66
OH–, O2–, ROH, RO–, R O, N , NO –, SO 2–, Br– CO, SCN–, R P, (RO) P, R As,
2 22 3 3 33
CH , CO 2–, NO –, PO 3–, R S, RSH, RS–, S O 2–
33 34 2 23
SO42– ClO4–, COO–, F–, Cl–
PROPERTIES OF WATER :
Amphoteric (amphiprotic) Acid/Base nature :
Water - an acid as well as base according to Bronsted - Lowry theory but according to Lewis concept it can only
be taken as base only.
In pure water [H+] = [OH–] so it is neutral.
Molar concentration / Molarity of water :
1000 g / litre
Molarity = No. of moles/litre = 18 g / mole = 55.55 mole/litre = 55.55 M (density = 1 g/cc)
Ionic product of water :
According to arrhenius concept
H2O H+ + OH– so, ionic product of water, kw = [H+][OH–] = 10–14 at 25° (exp.)
Dissociation of water is endothermic, so on increasing temperature Keq increases.
Kw increases with increase in temperature.
Now pH = –log[H+] = 7 and pOH = –log[OH–] = 7 for water at 25°C (experimental)
pH = 7 = pOH neutral at 25 °C
pH < 7 or pOH > 7 acidic
pH > 7 or pOH < 7 Basic
Ionic product of water is always a constant whatever has been dissolved in water since its an equilibrium
constant so will be dependent only on temperature.
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Degree of dissociation of water :
H2O H+ + OH– = no. of moles dissociated
Total no. of moles initially taken
10 7 = 18 × 10–10 or 1.8 × 10–7 %
=
(a) 55.55
Absolute dissociation constant of water :
Ex.
Sol. H O H+ + OH– K = K = [H ][OH ] = 107 107 = 1.8 × 10–16
2 ab [H2O] 55.55
So, pKa = pKb = – log(1.8 × 10–16) = 16 – log1.8 = 15.74
Acidity and pH scale :
Acidic strength means the tendency of an acid to give H3O+ or H+ ions in water.
So greater then tendency to give H+, more will be the acidic strength of the substance.
Basic strength means the tendency of a base to give OH– ions in water.
So greater the tendency to give OH– ions, more will be basic strength of the substance.
The concentration of H+ ions is written in a simplified form introduced by Sorenson known as pH scale.
pH is defined as negative logarithm of activity of H+ ions.
pH = – log aH (where aH is the activity of H+ ions)
Activity of H+ ions is the concentration of free H+ ions or H3O+ ions in a solution.
The pH scale was marked from 0 to 14 with central point at 7 at 25 °C taking water as solvent.
If the temperature and the solvent are changed, the pH range of the scale will also change. For example
0 - 14 at 25 °C (K = 10–14) Neutral point, pH = 7
w Neutral point, pH = 6.5
0 - 13 at 80 °C (Kw = 10–13)
pH can also be negative or > 14
pH Calculation of different Types of solutions :
Strong acid solution :
(i) If concentration is greater than 10–6 M.
In this case H+ ions coming from water can be neglected,
so [H+] = normality of strong acid solution
(ii) If concentration is less than 10–6 M
In this case H+ ions coming from water cannot be neglected.
So [H+] = normality of strong acid + H+ ions coming from water in presence of this strong acid
Solved Examples :
Calculate pH of 10–8 M HCl solution.
H2O H+ + OH–
10–8+x x
kw = [H+][OH–]
10–14 = x(x + 10–8)
x2 + x × 10–8 – 10–14 = 0
108 1016 4 1014 108 107 4 1 = ( 401 1)108 = 0.95 × 10–7
x= = 100 2
22
[H+] = 10.5 × 10–8 = 1.05 × 10–7
pH = –log [H+]
pH = 7 – log 1.05 6.98
10–9 M HCl pH 7
10–16 M HCl pH 7
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(b) Strong base solution : Calculate the [OH–] which will be equal to normality of the strong base solution and
then use Kw = [H+] × [OH–] = 10–14, to calculate [H+]
Ex. Solved Examples :
Sol.
Calculate pH of 10–7 M of NaOH solution
[OH–] from NaOH = 10–7
[OH–] from water = x < 10–7 M (due to common ion effect)
H2O OH– + H+
– (x + 10–7) x
Kw = [H+] [OH–] = 10–14 = x (x + 10–7)
x2 + 10–7x – 10–14 = 0
x = 5 1 × 10–7 = 0.618 × 10–7 ( 5 = 2.236)
2
[OH–] = 10–7 + 0.618 × 10–7 = 1.618 × 10–7
pOH = 7 – log(1.618) = 6.79
pH = 14 – 6.79 = 7.21
Self Practice Problem :
1. Calculate pH of a KOH solution having
Ans. (a) 5.6 g of KOH mixed in 50mL water
(c)
(b) if it is further diluted to make 100 mL.
(a) 14.3 (b) 14
pH of mixture of two strong acids : If V1 volume of a strong acid solution of normality N1 is mixed with
V2 volume of another strong acid solution of normality N2, then
Number of H+ ions from I-solution = N1V1
Number of H+ ions from II-solution = N2V2
If final normality is N and final volume is V, then
NV = N1V1 + N2V2
[dissociation equilibrium of none of these acids will be disturbed as both are strong acid]
[H+] = N = N1V1 N 2 V 2
V1 V2
(d) pH of mixture of two strong bases :
similar to above calculation
Ex. [OH–] = N = N1V1 N 2 V 2 [H+] = 10 14
Sol. V1 V2 [OH ]
Solved Examples :
Calculate pH of mixture of (400 mL, 11 M HCl) + (200 mL of water)
M H SO ) + (400 mL,
200 24 100
N1V1 = 1 × 400 = 4 4 , H+ ions from water will be neglected
100 1000 1000 , N2V2 = 1000
N1V1 + N2V2 = 8 × 10–3 [H+] = 8 10 3 = 8 × 10–3
pH = 3 – log8 = 2.1 1
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Ex. 500 mL of 10–5 M NaOH is mixed with 500 mL of 2.5 × 10–5 M of Ba(OH) . To the resulting solution 99 L
Sol. 2
water is added. Calculate pH.
[OH–] = 500 10 –5 500 2 2.5 105
1000
= 3 × 10–5 M
M1 = 3 × 10–5 M
V2 + V1 = 1 L
V1 = 100 L
no. of moles of [OH–] initially = no. of moles of [OH–]
3 × 10–5 = M2 × 100
M2 = 3 × 10–7 < 10–6
H2O H+ + OH–
x (x + 3 × 10–7)
Kw = x (x + 3 × 10–7) = 10–14
13 3 × 10–7
x = 2
x = 0.302 × 10–7
13 3 3 13
[OH–]Net = 3 × 10–7 = × 10–7 = 3.302 × 10–7
2 2
Self Practice Problem :
1. Calculate the pH for-
Ans. (a) 50 mL of 0.1 M HCl, 25 mL of 0.1M H2SO4, 25 mL of 0.2 M HNO3 + 100 mL of H2O
(e)
(b) 50 mL of 0.2 M NaOH + 100 mL of 0.1 M RbOH the resulting solution is diluted by 350 mL H O.
2
(a) 1.123 (b) 12.6
pH of mixture of a strong acid and a strong base :
Acid Base neutralisation reaction will take place.
The solution will be acidic or basic depending on which component has been taken in excess.
If V1 volume of a strong acid solution of normality N1 is mixed with V2 volume of a strong base solution of
normality N2, then
Number of H+ ions from I-solution = N1V1
Number of OH– ions from II-solution = N2V2
If N V > N V If N V > N V
11 22 22 11
[H+] = N = N1 V1 N2 V2 [OH–] = N = N1 V1 N2 V2
V1 V2 V1 V2
Solution will Solution will
be acidic in nature be basic in nature
[H+] = 10 14
[OH ]
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Solved Examples :
Ex. 11
Calculate pH of mixture of (400 mL, 200 M Ba(OH)2) + (400 mL, 50 M HCl) + (200 mL of water)
Sol.
Ex. 400 1 400 1 2
Sol. [H+] = 50 200 = 4 × 10–3, so pH = 3 – 2 log2 = 2.4
1000
What will be the resultant pH when 150 mL of an aqueous solution of HCl (pH = 2.0) is mixed with 350 mL
of an aqueous solution of NaOH (pH = 12.0) ?
pH of HCl = 2 pOH = 2 [NaOH] = 10–2 M
[HCl] = 10–2 M
pH of NaOH = 12,
HCl + NaOH NaCl + HO
150×10–2 2
Meq. initial 350×10–2 0 0
Meq. final
= 1.5 = 3.5
0 0.2 1.5 1.5
[OH–] from NaOH = 2 = 4 × 10–3 M
500
pOH = – log[OH–] = –log (4 × 10–3)
pOH = 2.3979
pH = 14 – pOH = 14 – 2.3979 = 11.6021
Self Practice Problem :
1 . Calculate pH of mixture 200 mL of 0.2 M H2SO4 + 300 mL of 0.2 M NaOH + 200 mL of 0.1 M KOH.
2 . Calculate the pH when 200 mL of 0.25 M H2SO4 is mixed with 200 mL of 0.2 M Ba(OH)2
Ans. 1. 7 2. 1.30
( f ) pH of a weak acid (monoprotic) Solution :
Weak acid does not dissociated 100 % therefore we have to calculate the percentage dissociation using
Ka dissociation constant of the acid.
We have to use Ostwald's Dilution law (as have been derived earlier)
HA H+ + A–
t=0 C 00
[H ][A ] C2
teq C(1 – ) CC Ka = [HA ] =
1
If << 1 (1 – ) 1 Ka C2 = K a (is valid if < 0.1 or 10%)
C
[H+] = C = C Ka = 1
C K a C So pH = 2 (pKa – logC)
on increasing the dilution C = and [H+] pH
Solved Example :
Ex.
Calculate pH of (a) 10–1 M CH3COOH (b) 10–3 M CH3COOH (c) 10–6 M CH3COOH
Sol. Take Ka = 2 × 10–5
( a ) CH3COOH CH3COO– + H+
C 00
C(1 – ) C C
C 2 Ka = 2 105 = 2 104 (<< 0.1)
Ka = 1 = C 101
So, [H+] = 10–1 × 2 × 10–2 pH = 3 – 1
log 2 = 2.85
2
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(b) = K a = Ka = 2 105 2 102 ( > 0.1)
C C 103 =
So we have to do the exact calculations
K= C 2 × 10–5 = 103 2 = 13.14 %
a 1 2 1 pH = 4 – log(1.314) 3.8
[H+] = 10–3 × 0.1314 = 1.314 × 10–4
( c ) If approximation is used the, = 2 105 = 20 > 1,
1 0 6
so we have to do the exact calculation, 2 × 10–5 = 10–6 2 0.95 or 95%
1
[H+] = 0.95 × 10–6 = 9.5 × 10–7 pH = 7 – log(9.5) = 6.022
At very low concentration (at infinite dilution) weak electrolyte will be almost 100%
dissociate, so behave as strong electrolyte
(pH) of 10–6 M HCl pH of 10–6 M CH3COOH 6)
( g ) pH of a mixture of weak acid (monoprotic) and a strong acid solution :
Weak acid and Strong acid both will contribute H+ ion.
For the first approximation we can neglect the H+ ions coming from the weak acid solution and calculate
the pH of the solution from the concentration of the strong acid only.
To calculate exact pH, we have to take the effect of presence of strong acid on the dissociation equilibrium
of the weak acid.
If [SA] = C1 and [WA] = C2, then [H+] from SA = C1
the weak acid will dissociate as follows.
HA H+ + A–
C2 0 0
C (1 – ) C2+C1 C2 K= (C2 C 1 )C2 ( <<< 1)
2 a
C2 (1 )
(The weak acids dissociation will be further suppressed because of presence of strong acid, common ion
effect)
Ka = (C2 + C1)
Total H+ ion concentration = C1 + C2
If the total [H+] from the acid is more than 10–6 M, then contribution from the water can be neglected, if
not then we have to take [H+] from the water also.
RELATIVE STRENGTH OF WEAK ACIDS AND BASES :
The relative strength of weak acids and bases are generally determined by their dissociation constants K and K
ab
respectively. For weak acid, i.e. CH COOH
3
CH COOH CH COO– + H+
3 3
C 00
C(1 – ) C C
K = C.C = C2 K = C2 (if << 1)
a C(1 ) (1 ) a
Similarly, for weak base, i.e. NH OH
4
NH OH NH + + OH–
4 4
C 00
C(1 – ) C C
K = C2
b
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K and K are just the equilibrium constants and hence depends only on temperature. Greater the value of
ab
dissociation constant of the acid (K ), more is the strength of the acid and similarly greater the value of dissociation
a
constant of the base, more is the strength of the base. For two acids of equimolar concentrations.
Strength of acid (I) K a1
=
Strength of acid (II) K a2
Similarly for bases, Strength of base (I) K b1
=
Strength of base (II) K b2
The modern method is to convert K as a power of 10 and express acid strength by power of 10 with sign
a
changed and call this new unit pK . Thus, if K for acid is equal to 10–4, pK = 4. So higher pK value means
aa aa
lower acid strength,
that is, pK = – log K
aa
Also, pK = – log K
bb
E x . 1 K for acid HA is 2.5 × 10–8 calculate for its decimolar solution at 25°C.
a
(i) % dissociation
(ii) pH
(iii) OH– ion concentration
S o l . HA H+ + A–
C 00
C(1–)
K [H ][A ] K = C.C = C2 C2
a = a C(1 ) (1 )
[HA ]
(i) = K a = 2.5 108 (C = 1/10 M)
C 1 /10
= 5 × 10–4 = 0.05%
(ii) [H+] = C = 1 5 10 4 = 5 × 10–5 mol/L So pH = 5 – log 5 = 4.30
10
(iii) [H+] [OH–] = 1 × 10–14
[OH–] 10 14 = 2 × 10–10 mol/L
= 5 10 5
E x . 2 Determine the degree of dissociation of 0.05 M NH at 25°C in a solution of pH = 10.
3
S o l . NH OH NH + + OH–
44
C 00
Given, pH = 10
[H+] = 10–10
[H+] [OH–] = 1 × 10–14
[OH–] = 1 10 14 = 10–4 = C
1 0 10
= [OH ] = 10 4 = 2 × 10–3 or 0.2 %
C 0.05
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Ex .3 Two weak monobasic organic acids HA and HB have dissociation constants as 1.6 × 10–5 and 0.4 × 10–5
respectively at 25° C. If 500 mL of 1 M solutions of each of these two acids are mixed to produce 1 litre of
mixed solution, what is the pH of the resulting solution ?
S o l . In such cases, we have to consider H+ from both HA and HB simultaneously. The concentration of HA and HB
in the mixture = 0.5 M [equal volumes are mixed] = say 'c'
HA H+ + A–
HB H+ + B–
Let, x = [H+] from HA and y = [H+] from HB
[H+] = x + y
final
K HA (x y)x and (x y)y
c K HB c
[H ]HA x k HA C
[H ]HB y kHB C
x 1.6 105 2
y 0.4 105
x = 2y x
y =
2
x x2 xy
Substitute for y = in K HA
2 c
1.6 × 10–5 = 2x2 3x2
2 0.5
3x2 = 1.6 × 10–5 x2 = 5.33 × 10–6
x = 2.30 × 10–3 M, y = 1.15 × 10–3 M
[H+] = x + y = 2.30 × 10–3 + 1.15 × 10–3 = 3.45 × 10–3 M
Final
pH = –log (3.45 × 10–3)
10
pH = 2.462
E x . 4 Saccharin (Ka = 2 × 10–12) is a weak acid represented by formula HSaC. A 8 × 10–4 mole amount of saccharin
is dissolved in 400 cm3 water of pH = 3. Assuming no change in volume, calculate the concentration of
SaC– ions in the resulting solution at equilibrium.
Sol. [HSaC] = mole 8 104 = 2 × 10–3 M
=
litre 400 / 1000
The dissociation of HSaC takes place in presence of [H+] = 10–3
HSaC H+ + SaC–
Conc. Before dissociation 2×10–3 10–3 0
In presence of H+, the dissociation of HSaC is almost negligible because of common ion effect. Thus, at
equilibrium
[HSaC] = 2 × 10–3 ; [H+] = 10–3 M
K = [H ][SaC ]
a [HSaC]
2 × 10–12 = [103 ][SaC 1 ]
2 103
[SaC–] = 4 × 10–12 M
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E x . 5 A solution contains 0.08 M HCl, 0.08 M CHCl COOH and 0.1 M CH COOH. The pH of this solution is 1. If K
23 a
for acetic acid is 10–5, calculate K for CHCl COOH.
a2
S o l . pH will be decided by [H+] furnished by HCl and CHCl COOH.
2
CHCl COOH CHCl COO– + H+
22
Initial conc. 0.08 0 0.08 (from HCl)
Final conc. (0.08 – x) x (0.08 + x)
[H+] = 0.08 + x ;
but pH = 1,
[H+] = 10–1 = 0.1
0.08 + x = 0.1 M
x = 0.02
K for CHCl COOH can be given as
a2
K = [H ][CHCl2COO ] 0.10.02 = 3.33 × 10–2
a [CHCl2COOH] = (0.08 0.02)
ISOHYDRIC SOLUTIONS :
(i) Solutions of electrolytes are said to be isohydric if the concentration of the common ion present in them is the
same and on mixing such solutions, there occurs no change in the degree of dissociation of either of the
electrolyte.
(ii) Let the isohydric solution is made by HA and HA acids, then [H+] of both acids should be equal i.e.
12
K a1C1 = K a2 C2 or K a1 C2
K a2 C1
EXAMPLE BASED ON : Dissociation of Acid and base and pH Calculation
Ex.1
The degree of dissociation of pure water at 25°C is found to be 1.8 × 10–9. Find K and K at 25°C.
Wd
(A) 3.24 × 10–18 ; 5.83 × 10–20 (B) 1 × 10–14 ; 1.8 × 10–16
(C) 1.8 × 10–16 ; 1 × 10–14 (D) 1 × 10–14 ; 1 × 10–14
Sol. (B)
Since = 1.8 × 10–9
and for water c = 1000 = 55.56
18
[H+] = [OH–] = c = 55.56 × 1.8 × 10–9 = 1 × 10–7 M
K = [H+] × [OH–] = (1 × 10–7)2 = 10–14
W
and K [H ][OH ] KW 10 14 10–16
== = = 1.8 ×
d [H2O] [H2O] 55.56
Ex.2 The concentration of [H+] and [OH–] of the 10–1 M aqueous solution of 2% ionised weak acid is :
Sol.
(A) 2 × 10–3 M and 5 × 10–12 M (B) 1 × 10–3 M and 3 × 10–11 M
Ex.3
Sol. (C) 2 × 10–4 M and 5 × 10–11 M (D) 3 × 10–2 M and 4 × 10–13 M
(A) = 2 × 10–3 M
[H+] = C
10 14
or [OH–] = [H ] = 5 × 10–12 M
When a 0.1 N solution of an acid at 25°C has a degree of ionisation of 4%, the concentration of OH– present is :
(A) 2.5 × 10–3 (B) 2.5 × 10–11 (C) 2.5 × 10–12 (D) 2.5 × 10–13
(C)
[H+] = C = 0.1 × 4 × 10–2 = 4 × 10–3 M
10 14
or [OH–] = [H ] = 2.5 × 10–12 N
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Ex.4 Calculate the molar concentration of a solution of acetic acid (HOAc) that has a pH of 4.00.
Sol.
(K = 1.8 × 10–5) : (B) 1.0 × 10–6 (C) 0.057 × 10–2 (D) 0.010
Ex.5 a
Sol.
Ex.6 (A) 1.0 × 10–3
(C)
K= [H ][OAc ]
a
[HOAc]
(10 4 )2
or 1.8 × 10–5 = [HOAc]
or [HOAc] = 5.56 × 10–4 M
Select the correct option from the following ?
(A) pK increases with increase of temperature (B) pK decreases with increase of temperature
W W
(C) pK = 14 at all temperatures (D) pK = pH at all temperatures
W W
(B)
For a 0.218 N solution of CH COOH at 25°C the value of v = 16.4 mho equiv–1 and = 391 mho equiv–1.
3
Calculate ionisation constant of acetic acid :
Sol. = v = 16.4 = 4.19 × 10–2 and C = 0.218 N
391
Ex.7
Sol. Ka = C2 (since < 0.05)
= 0.218 (4.19 × 10–2)2
K = 3.83 × 10–4
a
How much water must be added to 200 mL of 0.2 M solution of CH COOH for the degree of dissociation of
3
the acid to double ? K for the acetic acid = 1.8 × 10–5:
a
C112 = C222 or C = C 1 2 = C1
2 1 2 4
0.2
so M = 0.2 M ; M = M
1 24
V = 200 mL, V = ?
12
MV = MV
11 22
or V = M1V1 = 0.2 200 4 = 800 mL
2 M2 0.2
Ex.8 so 800 – 200 = 600 mL water should be added.
Sol. The degree of dissociation of acetic acid in a 0.1 M solution is 1.32 × 10–2. Calculate dissociation constant of
acid and its pK value :
a
CH COOH CH COO– + H+
3 3 0
Initially 0.1 0
at equilibrium 0.1(1–0.0132) 0.1× 0.0132 0.1× 0.0132
K = [CH 3COO ][H ] = 0.1 0.0132 0.1 0.0132 = 1.76 × 10–5
a [CH3COOH] 0.1 (1 0.0132)
Ex.9 pK = – log K = –log (1.76 × 10–5) = 4.75
Sol. aa
A solution having pH = 13, calculate the no. of H+ ions present in 1 mL of this solution :
pH = 13 so [H+] = 10–13 M
moles of H+ in one mL = 10 13 = 10–16 mol.
No. of H+ ions 103
= 10–16 × 6.022 × 1023
= 6.022 × 107
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(h) pH of a mixture of two weak acid (both monoprotic) solution :
Both acids will dissociate partially.
Ex.
Sol. Let the acid are HA & HA and their final concentrations are C & C respectively, then
12 12
HA1 H+ + A1– HA2 H+ + A2–
t=0 C1 00 C2 00
At eq. C1(1 – ) C11 + C22 C11 C2(1 – ) C22 + C11 C22
K a1 = C11 (C11 C 22 ) Ka2 = (C22 C11 )C22
C1 (1 1 ) C2 (1 2 )
(Since 1, 2 both are small in comparision to unity)
K a1 = (C11 + C22)1 ; K a2 = (C11 + C22)2 K a1 = 1
K a2 2
[H+] = C11 + C22 = C1K a1 + C2K a2 [H+] = C1K a1 C2K a2
C1K a1 C2K a2 C1K a1 C2K a2
If the dissociation constant of one of the acid is very much greater than that of the second acid then
contribution from the second acid can be neglected.
So, [H+] = C11 + C22 C11
Solved Example :
Calculate pH of solution obtained by mixing equal vol. of 0.02 M HOCl & 0.2 M CH COOH solution given
3
that Ka1 (HOCl) = 2 × 10–4
Ka2 (CH COOH) = 2 × 10–5
3
also calculate OH–, OCl–, CH3COO–
Final solution volume become double
C1 = 0.01, C2 = 0.1
[H+] = K a1 C1 K a2 C2
= 2 104 0.01 2 105 0.1 = 2 106 2 106 = 2 × 10–3
pH = 3 – log2 = 3 – 0.3010 = 2.69
2 104 = 10–1 2 105 = 10–2
1 = 2 103 2 = 2 103
HOCl H+ + OCl– CH3COOH H+ + CH3COO–
C11 C22
C (1 – 1) C11 + C22 C (1 – 2) C11 + C22
1 2
[OCl–] = C11 [CH COO–] = C22
3
= 0.01 × 10–1 = 0.1 × 10–2
= 1 × 10–3 = 1 × 10–3
[OH–] = K w = 1014 = 0.5 × 10–11
[H ] 2 103
= 5 × 10–12 M
[HOCl] = 10–2 (1 – 0.1) = 9 × 10–3 M
[CH3COOH] = 10–1(1 – 0.01) 10–1
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(i) pH of a solution of a polyprotic weak acid :
Diprotic acid is the one, which is capable of giving 2 protons per molecule in water. Let us take a weak
diprotic acid (H2A) in water whose concentration is c M.
In an aqueous solution, following equilbria exist.
If
1 = degree of ionization of H2A in presence of HA– Ka1 = first ionisation constant of H2A
2 = degree of ionisation of HA– in presence of H2A K a2 = second ionisation constant of H2A
I step II step
H2A + H2O HA– + H3O+ HA– + H2O A2– + H3O+
at eq. c(1 – 1) c1(1 – 2) (c1 + c12) at eq. c1 (1 – 2) c12 (c1 + c12)
(Keq)1 [H2O] = [H3O ][HA ] = K a1 (Keq)2 [H2O] = [H3O ][A2 ] = K a2
[H2 A ] [HA ]
K a1 = (c1 c12 ) [c1 (1 2 )] Ka2 = (c1 c12 ) [c12 )]
c(1 1 ) c1 (1 2 )
= [c1 (1 2 )][1 (1 2 )] ....... (i) [c1 (1 2 )]2
1 1 = 1 2 ...... (ii)
Knowing the values of K a 1 , K a 2 and c, the values of 1 and 2 can be calculated using equations (i) and (ii)
After getting the values of 1 and 2, [H3O+] can be calculated as
[H3O+]T = c1 + c12
Finally, for calculation of pH
If the total [H3O+] < 10–6 M, the contribution of H3O+ from water should be added.
If the total [H3O+] > 10–6 M, then [H3O+] contribution from water can be ignored.
Using this [H3O+], pH of the solution can be calculated.
Approximation :
For diprotic acids, K a 2 << K a 1 and 2 would be even smaller than 1
1 –2 1 and 1 + 2 1
Thus, equation (i) can be reduced to K a 1 = C 1 1
1 1
This is expression similar to the expression for a weak monoprotic acid.
Hence, for a diprotic acid (or a polyprotic acid) the [H3O+] can be calculated from its first equilibrium
constant expression alone provided K a2 << Ka1
Solved Examples :
Ex.
Calculate pH of [HS–], [S2–], [Cl–] in a solution which is 0.1 M HCl & 0.1 M H2S given that K a 1(H2S) = 10–7,
Sol. Ka2 (H2S) = 10–14 also calculate 1 & 2.
HCl + H2S
0.1 0.1
C1 = C2 = 0.1 (most of [H+] comes from HCl]
pH = 1 H+ + HS–
H2S 10–1 C1 = 0.1 1
0.1(1 – 1)
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Ka1 = C1 101 1 0 7 = 1
C(1 1 ) = + H+
1 = 10–6
1 0 1
HS– S2–
C1(1 – 2) C12 0.1
10–14 = 0.1 × 2
2 = 10–13
[S2–] = C12
= 10–6 × 10–1 × 10–13 = 10–20 M
( j ) pH of a mixture of a polyprotic weak acid and a strong acid :
pH can be calculated by taking the concentration of strong acid only (for first approximation)
For precise calculation we should take only the first dissociation constant of the weak polyprotic acid.
(As can be predicted from the equations we have presented so far for different cases.)
All these steps can be followed for the calculation of pOH for weak base and their mixtures
(we just have to replace Ka with Kb)
( k ) pH of a mixture of a weak acid/weak base with weak/strong base/acid respectively.
For this type of mixtures there can be two cases in general,
(i) if the acids and bases are mixed in equal amounts (equivalents)
(ii) if the acids and bases are mixed in different amounts (equivalents)
First case will lead to phenomenon of Salt hydrolysis and second case will lead to formation of Buffer
Solution.
SALTS :
(i) Salts are the ionic compounds formed when its positive part (Cation) come from a base and its negative part
(Anion) come from an acid.
(ii) Salts may taste salty, bitter, astringent or sweet or tasteless.
(iii) Solution of salts may be acidic, basic or neutral.
(iv) Fused salts and their aqueous solutions conduct electricity and undergo electrolysis.
(v) The salts are generally crystalline solids.
1 . Classification of salts :
The salts may be classified into four categories.
1. 1 Simple salts :
The salts formed by the neutralisation process between acid and base. These are of three types.
(i) Normal salt :
(i) The salt formed by the loss of all possible protons (replaceable H+ ions)
Ex. NaCl, NaNO3, K2SO4, Ca3(PO4)2, Na3BO3, Na2HPO3, NaH2PO2 etc.
(ii) Acid salts :
(i) Salts formed by incomplete neutralisation of polybasic acids. Such salts contain one or more replaceable H
atom.
Ex. NaHCO , NaHSO , NaH PO , Na HPO etc.
3 4 24 2 4
(ii) Above salts when neutralized by base form normal salts.
(iii) Basic salts :
(i) Salts formed by in complete neutralisation of poly acidic bases are called basic salts. These salt contain one
or more hydroxyl groups.
Ex. Zn(OH)Cl, Mg(OH)Cl, Fe(OH) Cl, Bi(OH) Cl etc.
22
(ii) Above salts when neutralised by acids form normal salts.
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1.2 Double salts :
(i) The addition compounds formed by the combination of two simple salts are termed as double salts.
Ex . FeSO (NH ) SO . 6H O (Ferrous ammonium sulphate), K SO Al (SO ) .24H O (Alum) and other alums.
4 42 4 2 242 43 2
(ii) Above salts are stable in solid state only.
(iii) When dissolved in water, it furnishes all the ions present in the simple salt from which it has been constituted.
(iv) The solution of double salt shows the properties of the simple salts from which it has been constituted.
1. 3 Complex salts :
(i) These are formed by combination of simple salts or molecular compounds.
Ex . K Fe(CN) , Co(NH ) SO etc.
46 36 4
(ii) FeSO4 6 KCN K 4Fe(CN)6 K2SO4
simple salt complex salt
(iii) CoSO + 6NH Co(NH ) SO
4 3 36 4
Simple
Molecular complex
salt compound salt
(iv) These are stable in solid states as well as in solutions.
(v) On dissolving in water, if furnishes a complex ion.
K Fe(CN) H2O 4K+ + [Fe(CN) ]4–
46 6
complex ion
(vi) The properties of the solution are different from the properties of the substance from which it has been
constituted.
1 . 4 Mixed salts :
(i) The salt which furnishes more than one cation or more than one anion when dissolved in water is called
mixed salt.
Ex. CaOCl , NaKSO , NaNH HPO etc.
2 4 44
OCl Na Na
Ca , SO4 , NH4 PO4
Cl K H
Hydrolysis of Salt :
Salt Hydrolysis :
Salt hydrolysis is defined as the process in which water reacts with salt to form acid & base.
Water + Salt Acid + Base
H = +ve
It is always an endothermic process because it is reverse of acid – base neutralization reaction which is always
exothermic.
Hydrolysis constant Acid Base
Kh = Salt
Here H2O is a solvent (in excess) so active mass of H2O is 1.
Note : Equilibrium constant K is equal to 1
Kh
i.e. K = 1
Kh
because salt hydrolysis is reverse of acid base equilibrium reaction.
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TYPES OF SALT HYDROLYSIS
( 1 ) Hydrolysis of strong acid - weak base [SA - WB] type salt –
E x . CaSO4, NH4Cl, (NH4)2SO4, Ca(NO3)2, ZnCl2, CuCl2, CaCl2
NH4Cl + H2O NH4OH + HCl
NH4++Cl– + H2O NH4OH + H+ + Cl–
NH4+ + H2O NH4OH + H+
(1) In this type of salt hydrolysis, cation reacts with H2O, therefore called as cationic hydrolysis.
(2) Solution is acidic in nature (SA WB) as [H+] is increased.
(3) pH of the solution is less than 7.
( a ) Relation between Kh , KW & Kb
NH + + H O NH OH + H+
42 4
Hydrolysis constant K
h
N H 4 O H H
Kh ..... (1)
[N H 4 ]
For weak Base NH OH NH + + OH–
44
N H O H
4
Kb ..... (2)
NH4 OH ..... (3)
H2O H+ + OH–
For water
K = [OH–] [H+]
w
Now multiplying Eq. (1) & (2) = Eq. (3)
N H O H H NH O H
4
4 × = [H+] [OH–]
NH N H 4 OH
4
i.e. K × K = K
hb w
Kh Kw
Kb
( b ) Degree of hydrolysis – Represented by h (initial concentration at equilibrium)
NH4+ + H2O NH4OH + H+
C 00
C–x xx
nx = a
1x = Ch
x = Ch
C – Ch Ch C h
NH 4 O H H Ch Ch
C Ch
Kh = =
NH
4
C2h2 Ch2
= C 1 h = 1 h
Since h <<<< 1
then (1 – h) 1
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Kh = Ch2 h h
h2 = Kh C
C
Kh Kw Kw h Kw
Kb h K b Kb C
C
( c ) pH of the solution :
pH = – log [H+]
[H+] = Ch = C Kw [H+] = Kw C
Kb C Kb
On taking – log on both sides
Kw C 1 2
Kb
– log [H+] = – log pH = – log K w C
K b
1
pH = [ log Kw + logC – log Kb ]
2
1 11
pH = logKw logC (–log Kb)
2 2 2
pH = 1 pK w 1 log C 1 pK b
2 2 2
11
pH = 7 2 pK b 2 log C
SUMMARY :
(1) Kh Kw (2) h K h K w
Kb C Kb C
(3) H Ch Kw C (4) pH = – log [H+]
Kb
pH = 7 1 pK b 1 log C
2 2
Ex.1 Find out the K of centi normal [10–2 N]solution of NH Cl (SA - WB) if dissociation constant of NH OH is
Sol. h4 4
10–6 and K = 10–14. Find out degree of hydrolysis and also find [H+] and pH of solution ?
w
Given : Kw = 10–14 ; Kb = 10–6
(1) Kh Kw = 10 14 = 10–8
Kb 10 6
(2) h K h = 108 = 106 = 10–3
C 1 0 2
(3) [H+] = Ch
= 10–2 × 10–3
= 10–5
(4) pH = – log [H+]
= – log [10–5]
= + 5 log10
=+5×1
=5
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Ex.2 Find out the Kh at 363 K (90°C) of a salt of [Strong Acid – Weak Base] if the value of Kb is 10–5
[At 90°C Kw = 10–12]
Sol. Kh Kw = 10 12 = 10–7
Kb 1 0 5
E x . 3 How many grams of NH Cl should be dissolved per litre of solution to have a pH of 5.13 ? K for NH is
4 b3
1.8 × 10–5.
S o l . NH Cl is a salt of strong acid and weak base for solutions of such salts.
4
1 [pKW – log C – pKb]
pH =
2
10.26 = 14 – log C – 4.74
log C = 9.26 – 10.26 = –1.0
C = 10–1 M
[NH Cl] = 10–1 M
4
WNH4NO3 = 10–1 × 53.5 gL–1 = 5.35 gL–1
Ex.4 What is the pH of 0.4 M aqueous NaCN solution? (Given pK of CN – = 4.70)
Sol. b
pK for HCN = 14 – 4.7 = 9.30
a
NaCN + H O NaOH + HCN
2
C 00
C(1–h) Ch Ch
[OH–] = Ch = C K h = KhC KwC
C Ka
11
pOH = (pK – pK – log C) = [14 – 9.30 – log (0.4)] = 2.548
2w a 2
pH = 14 – 2.5 = 11.45
Ex.5 The acid ionization constant for
Sol.
Zn2+ + HO Zn(OH)+ + H+
2
is 1.0 × 10–9. Calculate the pH of 0.10 M solution of ZnCl . Also calculate basic dissociation constant of
2
Zn(OH)+.
Zn2+ + HO Zn(OH)+ + H+
2
[H+] = C.h = C Kh KhC
C
KwC w h e r e K is ba s ic dis socia tio n c o n sta nt o f Zn ( O H )
Kb
b
= i.e. Zn(OH ) Zn2 OH
We know Zn2+ and Zn(OH)+ are conjugate acid and base.
Ka × Kb = 10–14
or K= 10 14 = 10–5
b
10 9
Now, [H+] = 1014 0.1 = 106 = 10–3
10 9
pH = 3
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( 2 ) Hydrolysis of [WA – SB] type salt –
Ex. KCN, NaCN, K2CO3, BaCO3, K3PO4
NaCN +H2O NaOH + HCN
Na+ + CN– + H2O Na+ + OH– + HCN
CN– + H2O HCN + OH–
(1) In this type of salt hydrolysis anion reacts with water therefore called as anionic hydrolysis.
(2) Solution is basic in nature as [OH–] increases.
(3) pH of the solution is greater than 7.
( a ) Relation between K , K , K
hwa
CN– + H2O HCN + OH–
H C N O H
Kh
CN ..... (1)
For weak acid HCN CN– + H+
C N H
Ka ..... (2)
HCN ..... (3)
For water H2O H+ + OH–
Kw = [OH–] [H+]
eq. (1) × eq. (2) = eq. (3)
HCN [OH – ] [CN – ][H ] = [H+][OH–]
[CN ] [HCN]
Kh × Ka = Kw
Kh Kw
Ka
( b ) Degree of hydrolysis :
CN– + H2O HCN + OH–
C0 0
Initial concentration at equlibrium
x
C–x x
nx = a
1x = Ch
x = Ch
C – Ch Ch Ch
H C N O H C2h2
Kh = Ch Ch = C 1 h
CN C Ch
Kh Ch2
1 h
Since h <<<< 1
therefore (1 – h) 1
K = Ch2
h
h2 Kh h h
C C
h Kw
Ka C
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( c ) pH of the solution
[OH–] = Ch
[OH–] = C × Kw
Ka C
OH Kw C
Ka
Kw = [OH–] [H+]
H Kw
O H
H Kw
Kw C
Ka
H Kw Ka
C
On taking – log on both sides
– log [H+] = – log K w K a
C
Kw Ka 1 2
C
pH = – log
1
pH = [ log K + logK – logC ]
2 wa
11 1
pH = logKw logKa logC
2 2 2
11 1
pH = pK + pK + logC
2 w2 a 2
11
pH = 7 + 2 pKa + 2 logC
SUMMARY :
(1) Kh Kw (2) h K h = K w
Ka C Ka C
(3) [OH–] = Ch = K w C (4) [H+] = K w K a
Ka C
(5) pH = – log [H+]
11
pH = 7 + 2 pKa + 2 log C
Ex1. Find out pH, h and [OH–] of milli molar solution of KCN 10–3 M, if the dissociation constant of HCN
Sol. is 10–7.
(1) 1 pKa 1 log C = 7 1 7 1 log103
pH = 7 + + 22
2 2
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= 7 7 3 log10 14 7 3 = 21 3 = 18 9
=
22 2 22
(2) h = Kh = Kw = 1 0 1 4 1014 1010 = 104 = 10–2
C Ka C =
107 103
(3) [OH–] = Kw C = 1014 103 = 1017 107 = 10 10 = 10–5
Ka 1 0 7
Ex.2 Calculate the pH and degree of hydrolysis of 0.01 M solution of NaCN, K for HCN is 6.2 × 10–12.
Sol. a
NaCN is a salt of strong base NaOH and weak acid HCN. K+ does not react with water whereas CN– reacts
with water as here under
CN– + H O HCN + OH–
2
K= [HCN][OH ] = Kw = 10 14 = 1.6 × 10–3
h [CN ] Ka 6.2 10 –12
Let, x moles of salt undergo hydrolysis then concentrations of various species would be
[CN–] = (0.01 – x) 0.01, [HCN] = x
[OH–] = x
K = x.x = 1.6 × 10–3
h 0.01
x2 = 1.6 × 10–5
x = 4 × 10–3
[OH–] = x = 4 × 10–3 M
[H O+] = KW ] = 10 14 = 0.25 × 10–11
3 [OH – 4 10 3
pH = – log(0.25 × 10–11) = 11.6020
Degree of hydrolysis = x 4 10 3 = 4 × 10–11
=
0.01 0.01
E x . 3 Calculate for 0.01 N solution of sodium acetate
(i) Hydrolysis constant (ii) Degree of hydrolysis (iii) pH
Given K of CH COOH = 1.9 × 10–5.
a3
Sol. For CH COONa + H O CH COOH + NaOH
32 3
Initial C 00
After C(1–h) Ch Ch
(i) K = Kw 10 14 = 5.26 × 10–10
h Ka = 1.9 10 5
(ii) h = K h = 5.26 10 10 = 2.29 × 10–6 M
C 0.01
(iii) [OH–] from NaOH, a strong base = Ch = 0.01 × 2.29 × 10–4 = 2.29 × 10–6 M
pOH = 5.64
pH = 14 – 5.64 = 8.36
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3 . Hydrolysis of (WA - WB) type salt :
E x . NH CN, CaCO , (NH ) CO , ZnHPO
4 3 42 3 3
NH4CN + H2O NH4OH + HCN
NH4+ + CN– + H2O NH4OH + HCN
(1) Solution is almost neutral but it may be acidic or basic depending upon the nature of acid & base & pH of the
solution is near to 7.
( a ) Relation between Kh, Kw, Ka & Kb
NH4+ + CN— + H2O NH4OH + HCN
Kh NH4OH HCN
N H 4 CN ..... (1)
For weak base NH4OH NH4+ + OH–
N H O H
4
Kb
N ..... (2)
H O H
4
For weak acid HCN H+ + CN–
H C N
Ka ..... (3)
HCN ..... (4)
For water H2O H++ OH–
Kw = [OH–] [H+]
Multiply Eq. (1) × Eq. (2) × Eq. (3) = Eq. (4)
NH4OH HCN N H O H H C N
4
× × = [H+] [OH–]
NH CN NH HCN
4 OH
4
Kh × Kb × Ka = Kw
Kh Kw
Ka Kb
(2) Degree of Hydrolysis –
NH4+ + CN– + H2O NH4OH + HCN
CC 00 Initial concentration
at equilibrium
nx = a C–x C–x xx
x = Ch Ch Ch
C – Ch C – Ch
Kh NH4OH HCN = Ch Ch = C2h2 = h2
N H 4 CN C ChC Ch C 1 h C 1 h 1 h 2
Since h <<<< 1
Then (1 – h) 1
or Kh h2
h2 KW ..... (5)
Ka Kb
h KW
Ka Kb
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( b ) pH of the solution
From eq. (3)
H C N
Ka
HCN
H K a HCN
C N
H Ka Ch = Ka h
C Ch 1h
Since h <<<< 1
(1 – h) 1
[H+] = K × h [Now put the value of h from eq. (5)]
a
= Ka Kw
Ka Kb
[H] K w K a
Kb
On taking – log on both sides
1 2
– log [H+] = – log K w K a
Kb
1
pH = – [ log (K × K ) – log K ]
2 wa b
1
pH = – 2 [ log Kw + log Ka – log Kb ]
1 11
pH = – [ log K ] – [log K ] – [ – log K ]
2 w2 a2 b
1 11
pH = + 2 pKw + 2 pKa – 2 pKb
11
pH = 7 + pK – pK
2 a2 b
SUMMARY :
(1) Kh Kw Kw
Ka Kb (2) h K h = K a K b
(3) [H+] = Kw Ka = Ka.h (4) pH = – log [H+]
Kb
11
pH = 7 + 2 pKa – 2 pKb
Note : Degree of hydrolysis of [WA – WB] type salt does not depend on the concentration of salt.
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E x . 1 Salt of weak acid and weak base
(i) Calculate pH of the mixture (25 mL of 0.1 M NH OH + 25 mL of 0.1 M CH COOH).
43
Given that K : 1.8 × 10–5, and K = 1.8 × 10–5
ab
Sol. NH OH + CH COOH CH COONH + H O
4 3 34 2
Initial milli moles 25 × 0.1 25× 0.1 0 0
= 2.5 = 2.5 –
Final milli moles 0 0 2.5 2.5
As salt is formed (salt of weak acid and weak base) and pH will be decided by salt hydrolysis
pH = pKw pKa pK b 1 (–log 10–14 – log 1.8 × 10–5 + log 1.8 × 10–5) = 7
=
22
E x . 2 In the following which one has highest / maximum degree of hydrolysis.
(1) 0.01 M NH Cl
(2) 0.1 M 4
(3) 0.001 M
(4) Same NH Cl
4
NH Cl
4
Sol. Kh
Ex.3 [3] h C if C decreases, h increases
Sol. In the following which one has lowest value of degree of hydrolysis.
Ex.4
Sol. (1) 0.01 M CH3COONH4
(2) 0.1 M CH3COONH4
(3) 0.001 M CH3COONH4
(4) Same
[4]
Find out the concentration of [H+] in 0.1M CH3COONa solution (Ka = 10–5)
Salt is [WA – SB] type
[H+] = Kw Ka = 1014 105 = 1019 10 1 = 10 18 = 10–9
C 1 0 1
Ex.5 Calculate the degree of hydrolysis of a mixture containing 0.1N NH4OH and 0.1N HCN
Sol. Ka = 10–5 & Kb = 10–5
Salt is [WA – WB]
h= Kw = 1 0 1 4 1014 1010 = 104 = 10–2
Ka Kb =
105 105
( 4 ) Hydrolysis of [SA – SB] type salt –
E x . NaCl, BaCl2, Na2SO4, KClO4 etc.
NaCl + H O NaOH + HCl
2
Na+ + Cl– + H2O Na+ + OH– + H+ + Cl–
H2O H+ + OH– (It is not salt hydrolysis)
(1) Hydrolysis of salt of [SA – SB] is not possible (2) Solution is neutral in nature (pH = pOH = 7)
(3) pH of the solution is 7
Hydrolysis of polyvalent anions or cations :
The hydrolysis of these species will take place in steps (just like dissociation of weak acids).
Out of different steps generally first step hydrolysis dominants mainly because of two reasons
The hydrolysis constant of second and further steps is generally negligible in comparison to first step
hydrolysis constant.
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The second and further step hydrolysis will be suppressed in presence of ions produced due to first step
hydrolysis.
Ex.
Sol. For a polyprotic acid (H2S, H3PO4, H2CO3, H2C2O4) we already know that the disscociation always takes
place in steps, so for example for H3PO4
[H ] [H 2 P O ]
4
H3PO4 H+ + H2PO4– K a1 = ..... (1)
[H3PO4 ]
HPO –2 [H ] [H PO 2 ]
4 4
H PO – H+ + K a2 = ..... (2)
24
[H P O 4 ]
2
HPO42– H+ + PO4–3 [H ] [P O 3 ]
4
Ka3 = ..... (3)
2
[H P O 4 ]
For all acids we always have K a1 >> K a2 >> K a3
pH of the solution can be calculated from Ist step only because [H+] from IInd & IIIrd step can be neglected as
(a) Ka1 >> K a2 >> K a3
(b) [H+] from Ist dissociation will suppress the dissociation of IInd & IIIrd step.
Now for the hydrolysis of polyvalent ions of salts (like K3PO4, Na2C2O4, ZnSO4, FeCl3, (NH4)2C2O4 or
ions like PO43–, C2O42–, Zn2+, Fe3+ etc).
Consider the hydrolysis in step
PO43– + H2O HPO42– + OH–
C 00
[O H ] [H P O 2 ]
4
C(1 – h) Ch Ch K h1 = ..... (4)
[P O 3 ] ..... (5)
4 ..... (6)
..... (7)
HPO42– + H2O H2PO4– + OH– [O H ] [H P O ]
4
K h2 = 2
[H PO 2 ]
4
H PO – + HO H PO + OH– K h3 = [OH][H3PO4 ]
24 2 34
[H P O ]
2 4
H2O H+ + OH– Kw = [H+] [OH–]
From above equations we get.
K a1 × K h3 = Kw
K a2 × K h2 = Kw
K a3 × K h1 = Kw
Genarally pH is calculated only using the first step hydrolysis
K h1 = ChCh = Ch2 Ch2
C(1 h) 1h
h= K h1 [OH–] = Ch = Kh1 C [H+] = Kw = Kw Ka3 = kw Ka3
C [OH ] KwC C
So pH = 1 pK a3 + logC]
[pK +
2w
Solved Example
What is the pH of 1.0 M Na3PO4 in aqueous solution ?
PO43– + H2O HPO42– + OH– ; Kb = 2.4 × 10–2
K (HPO 2–) = 1014 = 4.17 × 10–13
a 4 2.4 10 2
pKa = – logKa = 12.38
11 11
or pH = 7 + pKa + logC = 7+ (12.38) + log(1) = 13.19
2 2 2 2
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Hydrolysis of Amphiprotic Anion.
(Cation is not hydrolysed)
NaHCO , NaHS, etc., can undergo ionisation to from H+ ion and can undergo hydrolysis to from OH– (Na+
3
ion is not hydrolysed)
(a) (i) HCO3– + H2O ionisation CO32– + H3O+ (acid)
(ii) HCO – + H O hydrolysis H CO + OH– (base)
32 23
pH(HCO3–) = pK a1 pK a2
2
(b) Similarly for H2PO4– and HPO42– amphiprotic anions.
p K a1 pKa2 p K a2 pKa3
2 2
p H ( H PO ) = and p H ( H P O 2 ) =
4 4
2
Cation is also hydrolysed :
(i) Salts like NH HCO , NH HS in which HCO – and HS– are amphiprotic respectively but NH + will also
434 3 4
hydrolysed.
(ii) Equilibria in such solutions will be :
(Hydrolysis of anion)
HCO – + H O H CO + OH–
32 23
(Hydrolysis of cation)
NH + + HO NH OH + H+
4 2 4
so, [H+] = K a1 Kw K a2
Kb
EXAMPLE BASED ON : Salt Hydrolysis
E x . 1 Select the compound whose 0.1 M solution is basic :
(A) ammonium chloride (B) ammonium acetate (C) ammonium sulphate (D) sodium acetate
S o l . (D), since sodium acetate is salt of (WA + SB) so its pH > 7.
E x . 2 If one equivalent of a strong acid is added to one equivalent of a weak base, the resulting solution will be.
(A) neutral (B) acidic (C) alkaline (D) none of these
S o l . (B), since after neutralisation salt of (SA + WB) will form and its pH < 7.
E x . 3 Which is the correct option for hydrolysis constant of NH CN ?
4
(A) Kw Kw (C) Kb Ka
Ka (B) K a K b Kc (D) K b
S o l . (B), Since NH CN is a salt of (WA + WB).
4
Equilibrium constant of hydrolysis of WA + WB is K w
Ka Kb
E x . 4 Increasing order of pH of 0.1 M solution of the following salts is :
(A) NaCl < NH Cl < NaCN (B) NH Cl < NaCl < NaCN
4 4
(C) NaCN < NH Cl < NaCl (D) NaCl < NaCN < NH Cl
4 4
Sol. (B), Since NH Cl is the salt of (WB + SA) so pH < 7, NaCl is salt of (SA + SB) so pH = 7 and NaCN is salt of
4
(WA + SB) so pH < 7.
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Ex.5 When a solution of 0.01 M CH COOH is titrated with a solution of 0.01 M NaOH. Calculate the pH at the
Sol. 3
equivalence point. (pK of CH COOH is 4.74)
a3
CH COOH + NaOH CH COONa + H O
3 32
Let acid be = V mL
V mL of 0.01 M CH COOH will require V mL of 0.01 M NaOH. But CH COONa formed will make solution
33
alkaline due to hydrolysis
CH COONa + H O CH COOH + NaOH
32 3
[CH COONa] = 0.01 = 0.005 M
32
for pH of salt of weak acid and strong base
pH = 7 + pK a + log C
22
=7+ 4.74 log 0.005
+ = 8.22
22
E x . 6 Calculate the pH of 0.5 M Na PO in aqueous solution ?
34
PO 3– + H O HPO 2– + OH– ; K (PO –3) = 2.4 × 10–2
42 4 b4
Sol. HPO 2– and PO –3 are conjugate acid and base so K × K = 10–14
44 ab
K (HPO 2–) = 10 14 = 4.17 × 10–13
a4 2.4 102
Ex.7 pK = – log K = 12.38
Sol. aa
11
or pH = 7 + pK + log C
2 a2
pH = 13.04
What is degree of hydrolysis, K and pH of 1 M urea hydrochloride solution in water. K (urea) = 1.5 × 10–14.
hb
NH CONHCl is a salt of (SA + WB)
2
so h = K w = 10 14
K b.C 1.5 1014 1
or h = 81.65 %
Kh = Kw 1 0 14 = 6.667 × 10–1
Kb = 1.5 1014
11
pH = 7 pK – log C
2 b2
= 71 (13.82) 1 log(1) or pH = 0.09
–
22
BUFFER SOLUTIONS :
A solution that resists change in pH value upon addition of small amount of strong acid or base (less than 1 %)
or when solution is diluted is called buffer solution.
The capacity of a solution to resist alteration in its pH value is known as buffer capacity and the mechanism
of buffer solution is called buffer action.
Types of buffer solutions
(A) Simple buffer solution
(B) Mixed buffer solution
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(A) SIMPLE BUFFER SOLUTION :
A salt of weak acid and weak base in water e.g. CH3COONH4, HCOONH4, AgCN, NH4CN.
Buffer action of simple buffer solution
Consider a simple buffer solution of CH3COONH4 ,since it is a salt will dissociated completely.
CH3COONH4 CH3COO + NH4+
If a strong acid such as HCl is added then
HCl H+ + Cl
The H+ ions from the added acid (HCl) combine with CH3COO ions to form CH3COOH, which is a weak
acid so will not further ionized.
Thus there is no rise in H+ ion concentration and the pH remains constant.
CH3COO + H+ CH3COOH (Weak acid)
If a strong base is added as NaOH
NaOH Na+ + OH
NH4+ + OH NH4(OH) (Weak base)
Thus change in OH ion concentration is resisted by NH+ ions by forming NH4OH which is a weak base. So
4
it will not further ionized and pH remains constant.
pH of a simple buffer solution :-
11
pH 7 2 pka 2 pkb
( B ) MIXED BUFFER SOLUTIONS :
( a ) Acidic buffer solution :
An acidic buffer solution consists of solution of a weak acid and its salt with strong base. The best known
example is a mixture of solution of acetic acid and its salt with strong base (CH COONa). Other example :
3
HCN + KCN, (H CO + NaHCO ) blood
23 3
CH COOH CH COO + H+ (Weakly ionised)
3 3
CH COONa CH COO + Na+ (Highly ionised)
3 3
When a few drops of an acid (HCl) are added to it, the H+ ions from the added acid (HCl) combine with the
CH3COO ions to form CH3COOH. Thus there is no rise in H+ ion concentration and the pH of solution
remains constant. On the other hand, when a few drops of base(NaOH) are added, the OH of the added
base reacts with acetic acid to form unionise water and acetate ions.
CH3COOH + OH H2O + CH3COO. Thus there is no increase in OH ion concentration and hence
the pH of the solution remains constant.
pH of a acidic buffer solution (Handerson equation) :-
Consider a buffer mixture (acidic buffer)
HA + NaA (CH3COOH + CH3COONa) where A = CH3COO, A– = CH3COO–
HA H+ + A
NaA Na+ + A
Applying law of mass action to dissociation equilibrium of HA
[H ][A ] so [H+] = Ka [HA ]
Ka = [HA ] ; [A ]
taking log
log [H+] = log Ka [H A ]
+ log [A ]
–log [H+] = – log Ka [H A ]
– log [A ]
= pKa [A ]
pH + log
[H A ]
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[A] = Initial concentration of salt as it is mainly comes from salt.
[HA] = Initial concentration of the acid.
pH = pKa [Salt]
+ log (it is known as Handerson-Hasselbalch equation.)
[Acid]
Note : A solution can act as buffer only if ratio of concentration of salt to acid is between 0.1 to 10.
CH3COOH : CH3COONa
1 10 pH = pK a 1
10 1 pH = pK a 1
Thus pH range of an acidic buffer solution is pK a 1 to pK a 1
pH range = pK a 1
Maximum buffer action will be only when ratio of concentration of acid and salt is 1. So for maximum buffer action.
pH = pKa + log1/1 pH = pKa
E x . 1 Calculate the pH after the addition of 80 mL and 100 mL respectively of 0.1 N NaOH to 100 mL, 0.1 N
CH COOH. (Given pK for CH COOH = 4.74)
3 a3
Sol. If 80 mL of 0.1 N NaOH is added to 100 mL of 0.1 N CH COOH, acidic buffer will form as
3
H CCOOH + NaOH H CCOONa + H O
3 32
Initial 0.01 eq. 0.008 eq. 00
Final 0.002 eq. 0 0.008 eq.
pH = pKa + log [CH 3 CO O ] = 4.74 + log 8 = 4.74 + 0.6020 = 5.342
[CH3COOH] 2
If 100 mL of 0.1 N NaOH is added is added to 100 mL of 0.1 N CH COOH, complete neutralization takes
3
0.1
place and the concentration of H3CCOONa = 2 M = 0.05 M
11 log C
Now, pH = 7 + pKa + 2
2
pH = 8.72
Ex.2 How much volume of 0.2 M solution of acetic acid should be added to 100 mL of 0.2 M solution of sodium
Sol.
acetate to prepare a buffer solution of pH = 6.00? (pKa for acetic acid is 4.74)
pH = pK + log [Salt]
a
[Acid]
[Salt] pH pK = [Salt] 18.2
log = – a 6.00 – 4.74 = 1.26 =
[Acid] [Acid]
Moles of CH COONa in solution 100 0.2 = 0.02
3 1000
Let, volume of 0.2 acetic acid added = V mL
Moles of acetic acid = V 0.2
1000
0.02 18.2
V 0.2
1000
V = 5.49 mL
Ex.3 Calculate the pH of a solution when 0.20 moles of HCl is added to one litre solution containing ?
Sol.
(a) 1 M each of acetic acid and acetate ion ? (b) 0.1 M each of acetic acid and acetate ion ?
Given K for acetic acid is 1.8 × 10–5.
a
(a) Initially [Acetic acid] =1 M
[Acetate] = 1 M
Now 0.2 moles of HCl are added to it.
HCl + CH COO– CH COOH + Cl–
3 3
Mole before reaction 0.2 1 10
Mole after reaction 0 0.8 1.2 0.2
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New [CH COOH] = 1.2 ; [CH COO–] = 0.8
3 3
pH = pk + log [conjugate]
a
[acid]
pH = – log 1.8 × 10–5 + log 0.8 = 4.5686
1.2
(b) In II case initially [Acetic acid] = 0.1 M
[Acetate] = 0.1 M
Now 0.2 mole of HCl are added to it
HCl + CH COO– CH COOH + Cl–
3 3
Mole before reaction 0.2 0.1 0.1 0
Mole after reaction 0.1 0 0.2 0.1
[H+] from free HCl = 0.1 = 10–1M
pH = 1
Note : CH COOH no doubt gives H+ but being weak acid as well as in presence of HCl does not dissociate appreciably
3
and thus, H+ from CH COOH may be neglected.
3
E x . 4 Calculate the amount of NH and NH Cl required to prepare a buffer solution of pH = 9 when total concentration
34
of buffering reagents is 0.3 mol L–1. (pK for NH = 4.7, log 2 = 0.30)
b3
[salt]
S o l . pOH = – log K + log
b [Base]
5 = 4.7 + log a a 2
bb
a = 2b
Given a + b = 0.3
2b + b = 0.3
3b = 0.3
or b = 0.1 mole/L
Amount of base = 0.1 × 17 = 1.7 g/L
a = 0.2 mole/L
Amount of salt = 0.2 × 53.5 = 10.7 g/L
Thus, [Salt] = 0.2 M and [Base] = 0.1 M
Ex.5 Calculate the ratio of pH of a solution containing 1 mole of CH COONa + 1 mole of HCl per litre and of
Sol. 3
other solution containing 1 mole CH COONa + 1 mole of acetic acid per litre.
3
Case I : pH when 1 mole CH COONa and 1 mole HCl are present.
3
CH COONa + HCl CH COOH + NaCl
3 3
Before reaction 1 1 00
After reaction 0 0 11
[CH COO] = 1 M
3
[H+] = C. = C Ka = (K a .C ) = (Ka ) C=1
C
1
pH1 = – 2 log K a
Case II : pH when 1 mole CH COONa and 1 mole of CH COOH ; a buffer solution
33
[salt] [Salt] = 1M
pH = – log K + log [Acid] = 1M
2 a [acid]
pH = – log K
2a
pH1 1
=
pH2 2
( b ) Basic buffer solution :
A basic buffer solution consists of a mixture of a weak base and its salt with strong acid. The best known
example is a mixture of NH4OH and NH4Cl.
NH4OH NH4+ + OH (Weakly ionised)
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NH4Cl NH4+ + Cl (Highly ionised)
When a few drops of a base (NaOH) are added, the OH ions from NaOH combine with NH4+ ions to form feebly
ionised NH4OH thus there is no rise in the concentration of OHions and hence the pH value remains constant.
NH4+ + OH NH4OH
If a few drops of a acid (HCl) are added the H+ from acid combine with NH4OH to form H2O and NH4+ ions.
NH4OH + H+ NH4+ + H2O
Thus the addition of acid does not increase the H+ ion concentration and hence pH remains unchanged.
pH of basic buffer solution :
NH OH NH + + OH
4 4
NH4Cl NH4+ + Cl
N H O H
4
Kb
NH4 OH
O Kb N H 4 OH K b Base
S alt
H
N
H 4
( N H mainly comes from salt)
4
taking –log on both side
–log OH– = –log K b Base pOH = –log Kb – log B a se
S a lt S a lt
pOH = pK + log Salt pH = 14 – pOH
b Base
pOH range :
A solution can act as buffer solution only if ratio of concentration of salt to base is from 0.1 to 10.
NH4OH : NH4Cl
1 10 pOH = pKb + 1
10 1 pOH = pKb – 1
S o pOH range is pKb ± 1
Condition for maximum buffer action :
[NH4OH] : [NH4Cl]
1 1
pOH = pK b log 1
1
pOH = pKb and pH = 14 – pKb
Maximum buffer action because pH remains constant.
Ex. A solution of weak base LiOH was titrated with 0.1 N HCl. The pH of the solution was found to be 10.04 and
Sol.
9.14 after the addition of 5 mL and 20 mL of the acid respectively. Find the dissociation constant of the base.
Case I : LiOH + HCl LiCl + H O
0.1× 5=0.5 2
Millimole before reaction a
00
Millimole after reaction (a–0.5) 0 0.5 0.5
pOH = – log K + log [LiCl] .........(i)
b .........(ii)
[LiOH]
pH = 10.04 so pOH = 3.96
0.5
3.96 = – log Kb + log (a 0.5)
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Case II : LiOH + HCl LiCl + H O
2
Millimole before reaction a 0.1× 20=2 0 0
Millimole after reaction (a–2) 0 22
pOH = – log K + log [LiCl] .........(iii)
b
[LiOH]
pH = 9.14 pOH = 4.86
4.86 = – log K + log 2 .........(iv)
b
(a 2)
Ex. Solving Eqs. (ii) and (iv), K = 1.81 × 10–5
Sol. b
An organic base B has K value equal to 1 × 10–8. In what amounts should 0.01 M HCl and 0.01 M solution
b
of B be mixed to prepare 1 L of a buffer solution having pH = 7.0 ?
B + HO BH+ + OH–
2
[XH ][OH ]
K= [B] = 1 × 10–8
b
pOH = pK + log [BH ]
b
[B]
[BH ] [BH ]
7 = – log (10–8) + log [B] 7 = 8 + log [B]
[BH ]
log [B] = –1
[BH ]
[B] = 10–1 = 0.1
Let, volume of HCl taken = xL
Volume of base taken = (1 – x) L
After the reaction, millimole of BH+ formed = 0.01 × (x)
Millimoles of base left = 0.01 (1 – 2x)
[BH ] x
[B] = [1 2x] = 0.1
x = 0.083 L = Volume of HCl
Volume of base = 0.0917 L
Note : In Handerson-Hasselbalch equation concentration should be place in terms of N × V, if volume is given, If
volume is not given then concentration should be placed in terms of normality.
BUFFER CAPACITY :
It is defined as the number of moles of acid (or base) require by one litre of a buffer solution for changing its
pH by one unit.
No.of moles of acid or bases added per litre
Buffer capacity =
change in pH
E x . When 2 moles of HCl is added to 1 lit. of an acidic buffer solution, its pH changes from 3.4 to 3.9. Find its
buffer capacity.
S o l . B.C. = 2 =4
0.5
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EXAMPLE BASED ON : Buffer solutions
E x . Which of the following buffers containing NH OH and NH Cl show the lowest pH value?
44
conc. of conc. of
NH OH (mol L–1) NH Cl (mol L–1)
4 4
0.50
(A) 0.50
(B) 0.10 0.50
(C) 0.50 1.50
(D) 0.50 0.10
Sol. ( B ) pOH = pk + log [salt] for NH Cl = 0.5 and NH OH = 0.1
Ex.
b [base] 4 4
pOH will be maximum and so pH will be minimum.
pH of a mixture containing 0.2 M X– (base) and 0.4 M HX with pK (X–) = 4 is :
b
(A) 4 + log 2 (B) 4 – log 2 (C) 10 + log 2 (D) 10 – log 2
Sol. (D). HX H+ + X– , K 10 14 = 10–10
Ex. or =
a Kb
[H+] = K H X = 1 0 10 0.4
a [X ] 0.2
pH = 10 – log 2
pH of a mixture of 1 M benzoic acid (pK = 4.2) and 1 M sodium benzoate is 4.5, in 150 mL buffer, benzoic
a
acid is :
(A) 200 mL (B) 150 mL (C) 100 mL (D) 50 mL
S o l . ( D ) pH = pK + log [salt]
a [Acid]
4.5 = 4.2 + log (150 x) where x is the volume of benzoic acid
x
(150 x) (150 x)
0.3 = log x or 2.0 = x
or x = 50 mL
Ex. Buffering action of a mixture of CH COOH and CH COONa is maximum when the ratio of salt to acid is
Sol. 33
Ex.
equal to :
Sol.
(A) 1.0 (B) 100.0 (C) 10.0 (D) 0.1
(A). The buffer action of a buffer mixture is effective in the pH range pK ± 1. It is maximum when pH = pK .
aa
What amount of sodium propanoate should be added to one litre of an aqueous solution containing 0.02 mole
of propanoic acid to obtain a buffer solution of pH 4.74 ? What will be the pH of 0.01 mol of HCl is dissolved
in the above buffer solution ? Compare the last pH value with the pH of 0.01 molar HCl solution. Dissociation
constant of propanoic acid at 25°C is 1.34 × 10–5.
Using Henderson's expression
pH = pK + log [salt]
a [acid]
We get 4.74 = – log (1.34 × 10–5) + log [Salt]
0.02
Which gives 4.74 = 4.87 + log [Salt] or [Salt] = 0.74 or [Salt] = 1.48 × 10–2 M
0.02 0.02
Hence, amount of sodium propanoate to be added = 1.48 × 10–2 × 96 g = 1.42 g
The addition of 0.01 mol of HCl converts the equivalent amount of sodium propanoate into propanoic acid.
Hence, we will have
(0.01482 0.01) mol L1
pH = 4.87 + log (0.02 0.01) mol L1
pH = 4.87 + log (0.160) = 4.87 – 0.79 = 4.08
(The pH of 0.01 molar HCl solution would be pH = – log (0.01) = 2)
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Ex. A solution of a weak acid was titrated with NaOH the equivalence point was reached when 25.06 mL of
Sol. 0.1 N NaOH have been added. Now 12.53 mL of 0.1 N HCl were added to the titrated solution, the pH was
found to be 4.92. What is K of the acid.
a
For complete neutralisation, meq. of acid = meq. of NaOH
= 25.06 × 2.506 = 3.612 meq.
HA + NaOH NaA + HO
2
Initial 2.506 2.506 0 0
At equivalence point 0 0 2.506 2.506
Now 1.253 meq. of HCl are added so,
NaA + HCl NaCl + HA
2.506 1.253 0 0 (Before)
1.253 0 1.253 1.253 (After)
Now mixture contains HA and NaA so it will be Acidic buffer,
pH = pK + log [Salt] = pK + log 1.253
a [Acid] a 1.253
or pH = pK
a
or K = antilog (–4.92)
a
or K = 1.2 × 10–5
a
SOLUBILITY (s) AND SOLUBILITY PRODUCT (Ksp) :
This is generally used for sparingly soluble salts. We will be dealing with the solubilities in the following type
of solution
1 . Simple solution in H2O
2 . Effect of common ions on solubility
3 . Simultaneous solubility
4 . Condition for precipitation
5 . Solubility in a buffer solution
6 . Solubility due to complex formation
Solubility product (ksp) is a type of equilibrium constant, so will be dependent only on temperature
for a particular salt.
Following examples will illustrate the different type of solubilities and the effects of different factors or situation
on solubility of a salt.
Simple solubility
Let the salt is Ax By, in solution in water, let the solubility in H2O = 's' M, then
AxBy xAy+ + yB–x ksp = (xs)x (ys)y = xx, yy.(s)x+y
– xs ys
Solved Examples
Ex. Calculate ksp of Li3Na3[AlF6]2
Sol. ksp = 33.33.22.(s)8 = 36.4 (s)8 = 2916 s8
Self Practice Problem
Ex. Calculate ksp of Mg3(PO4)2
Sol. 108 s5
Effect of common ions on solubility :
Because of the presence of common ion the solubility of the salt decreases
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Solved Example
Ex.
Calculate solubility of silver oxalate in 10–2M Potassium oxalate solution given that ksp of silver oxalate
Sol. = 10–10
[oxalate] = 10–2 + x, where x is the solubility of silver oxalate, this can be neglected in comparison to
10–2. so
ksp = 10–10 = 10–2 × (2x)2 1 0 8 = x2 x = 5 × 10–5
22
Self Practice Problem
Ex. Calculate the solubility of BaCl2 in presence of 'c' mol/litre NaCl in terms of Ksp(BaCl2).
Sol. Ksp/c2
Calculation of simultaneous Solubility
When two sparingly soluble salts are added in water simultaneously, there will be two simultaneous
equilibria in the solution.
Self Practice Problem
Ex. Calculate solubility of BaSO4 when BaSO4 and CaSO4 are dissolved in water simultaneously KspCaSO4 = p,
Ksp BaSO4 = q and solubility of CaSO4 is 'b' mol/litre.
Sol. bq/p
Ex.1 Solved Example
Sol. Calculate simultaneous solubility of silverthiocyanate and sliver bromide in water given that ksp of silver
thiocyanate = 10–12 and ksp of silver bromide = 5 × 10–13 respectively.
Let the solubility of AgSCN be x and that of AgBr is y, then
AgSCN Ag+ + SCN– AgBr Ag+ + Br–
x+y x x+y y
10–12 = x (x + y)
5 × 10–13 = y(x + y) ...... (i)
...... (ii)
On solving we get, x = 2y
So y = 4.08 × 10–7 and x = 8.16 × 10–7
Ex.2 50 mL of 0.02 M solution of Ca(NO ) is added to 150 mL of 0.08 M solution of (NH ) SO . Predict whether
Sol. 32 42 4
CaSO will be precipitated or not, K (CaSO ) = 4 × 10–5.
4 sp 4
Calculation of Ca2+ concentration, M1V1 = M2V2
0.02 × 50 = M × 200
2
[Ca(NO ) ] after mixing = 5 × 10–3 mol L–1
32
Since Ca(NO ) is completely ionized, [Ca2+] = [Ca(NO ) ] = 5 × 10–3 mol L–1
32 32
Calculation of SO 2– ion concentration
4
Applying M ' V ' = M ' V '
11 22
0.08 × 150 = M' × 200
2
M' = 0.08 150 = 6 × 10–2 M
2
200
[(NH4)2SO4] is completely ionized, [SO42–] = [(NH4)2SO4) = 6 × 10–2 mol/L
Ionic product = [Ca2+] [SO 2–] = [5 × 10–3] [6 × 10–2] = 3 × 10–4
4
Since Ionic product (3 × 10–4) is greater than solubility product of CaSO (4 × 10–5), hence precipitate of
4
CaSO will be formed.
4
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Ex.3 What is the maximum volume of water required to dissolve 1 g of calcium sulphate at 25°C. For calcium
Sol.
sulphate, K = 9.0 × 10–6.
sp
CaSO (aq) Ca2+ (aq) + SO 2– (aq)
4 4
If S is the solubility of CaSO in moles L–1
4
K = [Ca2+] × [SO 2–] = S2
sp 4
S = K sp = 9.0 106
= 3 × 10–3 mol L–1
= 3 × 10–3 × 136 g L–1 = 0.408 gL–1
For dissolving 0.408 g of CaSO water required = 1 L
4
1
For dissolving 1g CaSO water required =L = 2.45 L
4 0.408
Ex.4 A weak acid HA after treatment with 12mL of 0.1M strong base BOH has a pH of 3. At the end point, the
Sol. volume of same base required is 26.6 mL. Calculate K of acid.
Ex.5 a
Sol.
For neutralization :
Total Meq. of acid = Meq. of base = 26.6 × 0.1 = 2.66
Now for partial neutralization of acid
Meq. before reaction HA + BOH BA + H2O
2.66 1.2 0 0
Meq. after reaction 1.46 0 1.2 1.2
The resultant mixture acts as a buffer and [HA] and [BA] may be placed in terms of Meq. since volume of
mixture is constant.
pH = – log K + log [Salt] [1.2]
a or 3 = – log K + log
[Acid]
a [1.46]
K = 8.219 × 10–4
a
Calculate simultaneous solubility of AgCNS and AgBr in a solution of water. K of AgBr = 25 × 10–13 and
sp
K of AgCNS = 5 × 10–12.
sp
Let solubility of AgCNS and AgBr in a solution be a and b mol litre–1 respectively.
AgCNS (s) Ag+ + CNS–
AgBr (s) a a
Ag+ + Br–
bb
For AgCNS K spAgCNS [Ag ][CNS ]
5 × 10–12 = (a + b) (a)
..............(1)
For AgBr K spAgBr [Ag ][Br ] ..............(2)
By Eqs. (1) and (2), 25 × 10–13 = (a + b) (b)
a 5 1012 2 or a = 2b
b 25 1013
By Eq. (1), (2b + b) (2b) = 1 × 10–12
6b2 = 5 × 10–12
b = 0.912 × 10–6 mol litre–1 = 9.12 × 10–5 M
By Eq. (1), (a + a/2) (a) = 5 × 10–12
a = 1.82 × 10–6 mol litre–1
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Ex.6 Equal volumes of 0.04 M CaCl and 0.0008 M Na SO are mixed. Will a precipitate form?
Sol. 2 24
K for CaSO = 2.4 × 10–5
sp 4
CaCl + Na SO CaSO + 2NaCl
2 24 4
Millimole added 0.04 V 0.0008× V 0 0
Suppose V mL of both are mixed
0.04 V
[Ca2+] = 2V
[SO2–] = 0.0008 V
4
2V
[Ca2+] [SO2–] = 0.04 V 0.0008 V = 8 × 10–6
4 ×
2V 2V
Thus, [Ca2+] [SO2–] in solution < K
4 sp
8 × 10–6 < 2.4 × 10–5
CaSO will not precipitate.
4
Ex.7 Calculate the [OH–] of a solution after 50 mL of 0.2 M MgCl is added to 500 mL of 0.4 M NaOH. K of
Sol. 2 sp
Mg(OH) is 1.2 × 10–11.
2
MgCl + 2NaOH Mg(OH) + 2NaCl
2 2
m mole before reaction 10 20 0 0
00 10 20
Thus, 10 m mole of Mg(OH)2 are formed. The product of [Mg2+] [OH–]2 is therefore 10 × 20 2 = 4 × 10–3
100
100
which is more than Ksp of Mg(OH)2. Now solubility (S) of Mg(OH)2 can be derived by
K = 4S3
sp
S = 3 K sp 3 1.2 10 11 = 1.4 × 10–4 M
[OH–] = 2S = 2.8 × 10–4 M
Ex.8 Will a precipitate of Mg(OH) be formed in a 0.002 M solution of Mg(NO ) , if the pH of solution is adjusted
Sol. 2 32
to 9 ? Ksp of Mg(OH)2 = 8.9 × 10–12.
pH = 9
[H+] = 10–9 M
or [OH–] = 10–5 M
Now if Mg(NO ) is present in a solution of [OH–] = 10–5 M, then,
32
Product of ionic conc. = [Mg+2] [OH–]2 = [0.002] [[10–5]2
= 2 × 10–13 lesser than K of Mg(OH) i.e, 8.9 × 10–12
sp 2
Mg(OH)2 will not precipitate.
Condition of precipitation
Ex.9 For precipitation ionic product [IP] should be greater than solubility product ksp
You are given 10–5 M NaCl solution and 10–8 M AgNO3 solution, they are mixed in 1 : 1 volume ratio, predict
whether AgCl will be precipitated or not, if solubility product of AgCl in 10–2 M AgNO3 is = 10–10 mole per
litre.
Sol. Ionic product = 105 × 108 = 25 × 10–15 < ksp
2 2
Hence no precipitation will take place.
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Solubility in appropriate buffer solutions.
Appropriate buffer means that the components of buffer should not interfere with the salt or only H+ or OH–
ions should be interacting with the ions of the salt.
Selective Precipitation: When the k values differ then one of the salt can be selectively precipitated.
sp
E x . 1 0 What [H+] must be maintained in saturated H2S(0.1 M) to precipitate CdS but not ZnS, if [Cd2+] = [Zn2+] = 0.1
initially ?
Sol. Ksp = (CdS) = 8 × 10–27
Ksp = (ZnS) = 1 × 10–21
Ka = (H2S) = 1.1 × 10–21
In order to prevent precipitation of ZnS
[Zn2+] [S2–] < Ksp(ZnS) = 1 × 10–21
(ionic product)
or (0.1) [S2–] < 1 × 10–21
or [S2–] < 1 × 10–20
This is the maximum value of [S2–] before ZnS will precipitate. Let [H+] to maintain this [S2–] be x. Thus for
H2S 2H+ + S2–
Ka = [H ]2 [S2 ] = x2 (1 1020 )
[H2S] 0.1
= 1.1 × 10–21
or x = [H+] = 0.1 M
No ZnS will precipitate at a concentration of H+ greater than 0.1 M
Effect on solubility because of complex formation
E x . 1 1 What must be the concentration of aq. NH3(eq.) which must be added to a solution containing 4 × 10–3 M Ag+
and 0.001 M NaCl, to prevent the precipitation of AgCl. Given that Ksp(AgCl) = 1.8 × 10–10 and the formation
constant of [Ag(NH3)2]+ is K format ion = 108 .
6
S o l . Calculate silver ion concentration which can be allowed to remain in the solution,
1.8 × 10–10 = [Ag+][Cl–]
[Ag+] = 1.8 1010 = 1.8 × 10–7 M,
0.001
This quantity is so small that almost all the Ag+ ion will be consumed.
Ag+ + 2NH3 [Ag(NH3)2]+ 108
4 × 10–3 b 0 K=
6
1.8 × 10–7 (b – 8 × 10–3) 4 × 10–3 K = 108 4 103 b = 0.0445
6 = 1.8 107 (b 8 103 )2
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E x . 1 2 The solubility of Mg(OH) is increased by addition of NH + ion. Find
24
(i) K for the reaction
c
Mg(OH)2 + 2NH4+ 2NH3 + 2H2O + Mg2+
(ii) Calculate solubility of Mg(OH) in a solution containing 0.5 M NH Cl
S o l . (i) 24
( K =sp [Mg(OH )2 ] 1.0 × 10–11, K =b (NH3 ) 1.8 × 10–5)
For the reaction
Mg(OH) + 2NH + 2NH + 2H O + Mg2+
24 32
K = [NH 3 ]2 [Mg2 ] ..........(1)
c
[N H ]2
4
Also for the reaction
NH3 + H2O NH4+ + OH–
K b(NH3 ) [ N H ][O H ]
4
= ..........(2)
[NH3 ]
[As water is in large excess in both the cases]
[NH 3 ]2 [Mg2 ] [ N H ]2 [O H ]2
4
Now, K × K2 = ×
c b [N H ]2 [NH 3 ]2
4
= [Mg+2] [OH–]2 = K sp[Mg (OH )]2
Kc = K sp 10 11 = 3.08 × 10–2
K 2 = (1.8 10 5 )2
b
(ii) Now, let us, assume that 'a' moles of Mg(OH) be dissolved in presence of 0.5 M NH Cl.
24
Initial
Mg(OH) + 2NH + 2NH + 2H O + Mg2+
2 4 3 2
— 0.5 0— 0
Eqm. — (0.5–2a) 2a — a
K= a (2a)2 4a3 = 3.08 × 10–2
c (0.5 2a)2
0.25
or a = 0.124 M
E x . 1 3 0.10 mol sample of AgNO is dissolved in one litre of 2.00 M NH . Is it possible AgCl(s) form the solution by
33
adding 0.010 mol of NaCl ? ( K sp(AgCl) = 1.8 × 10–10, K f[Ag(NH3 )2 ] = 1.6 × 107)
Sol. Ag+ + 2NH [Ag(NH ) +]
3 32
0.10 M 2.00 0
0.10–0.10 (2–0.20 M) 0.10 M
= 0 = 1.80 M
It is assumed that all Ag+ ions have been complexed and only x amount is left
Kf = [Ag(NH 3 )2 ] 1.6 × 107 = 0.10
[Ag ] [NH 3 ]2 x (1 .8 0 )2
x = 1.93 × 10–9 M = [Ag+] undisolved
[Cl–] = 1.0 × 10–2 M
[Ag+] [Cl–] = 1.93 × 10–9 × 1.0 × 10–2 = 1.93 × 10–11 < 1.8 × 10–10 [K ]
sp(AgCl)
Hence, AgCl (s) will not precipitate.
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Ex.14 How many grams of CaC O will dissolve in distilled water to make one litre of saturated solution
24
(K = 6.25 × 10–10 and its molecular mass is 128) :
sp
(A) 0.0064 g (B) 0.0128 g
(C) 0.0032 g (D) 0.0640 g
Sol. (C)
[Ca2+] [C O 2–] = 6.25 × 10–10 so solubility = 2.5 × 10–5 mol/L = 2.5 × 128 × 10–5 = 3.2 × 10–3 g/L
24
E x . 1 5 Select the sulphides which has maximum solubility in water ?
(A) CdS (Ksp = 36 × 10–30) (B) FeS (Ksp = 11 × 10–20)
(C) HgS (K = 32 × 10–54) (D) ZnS (K = 11 × 10–22)
sp sp
Sol. (B)
All salt are AB type so solubility will be K sp . Higher the value of Ksp, the maximum will be solubility.
E x . 1 6 If equal volumes of the following solutions are added, precipitation of AgCl
(K = 1.8 × 10–10) will occur only with. (B) 10–5 M (Ag+) and 10–5 M (Cl–)
sp
(A) 10–4 M (Ag+) and 10–4 M (Cl–)
(C) 10–6 M (Ag+) and 10–6 M (Cl–) (D) 10–10 M (Ag+) and 10–10 M (Cl–)
Sol. (A)
One can calculate ionic product from given data and for precipitation Ionic product > K .
sp
E x . 1 7 If hydrolysis of any one of the ions will occur, after the dissolution of a sparingly soluble salt, then -
(A) solubility of salt decreases. (B) solubility of salt increases
(C) there will be no effect on solubility (D) question is absurd
Sol. (B)
Dissolution equilibria shift towards right side due to hydrolysis of cation or anion.
E x . 1 8 What is the concentration of Ag+ ions in 0.01 M AgNO that is also 1.0 M NH ? Will AgCl precipitate from
33
a solution that is 0.01 M AgNO , 0.01 M NaCl and 1 M NH ?
33
K (Ag[NH ] +) = 5.88 × 10–8 ;
d 32
K (AgCl) = 1.8 × 10–10.
sp
S o l . Let us first assume that 0.01 M AgNO shall combine with 0.02 NH to form 0.01 M Ag(NH ) + and the
3 3 32
consider its dissociation.
AgNO + 2NH Ag(NH ) + ...Initial conc.
3 3 32
0.01 M 1M 0
0 (1–0.02)=0.98M 0.01 M ...at eq. conc.
Ag(NH ) + Ag+ + 2NH
32 3
(0.01 – x) x (0.98 + 2x)
= 0.01 M 0.98 M
....Equib. conc.
Since x <<< 1
K= [Ag ][NH 3 ]2 = 5.88 × 10–8
d [Ag(NH 3 )2 ]
[Ag+] = 5.88 108 0.01 = 6.12 × 10–10 M
(0.98)2
Further, ionic product of AgCl = [Ag+] [Cl–] = (6.12 × 10–10) (0.01) = 6.12 × 10–12
Because the ionic product is smaller than K = 1.8 × 10–10, no precipitate should form.
sp
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INDICATORS :
An indicator is a substance which changes its colour at the end point or neutral point of the acid-base titration
i.e. the substance which is used to indicate neutral point of acid-base titration are called indicators. At end
point N1V1 = N2V2
Indicators are of two types
(i) Acidic (ii) Basic
( i ) Phenolphthalein (HPh) :- HPh is acid indicator. It ionises in water to give colourless H+ ions and pink
coloured anions.
HPh H+ + Ph
Colourless Pink
If [Hph] > [Ph–] Colourless
[Hph] < [Ph–] Pink
In acidic medium the dissociation of HPh is almost nil so it gives no colour because acid suppress the
ionisation of HPh due to the presence of common ion H+ and the solution remains colourless.
HPh H+ + Ph–
HCl H+ + Cl–
In alkaline medium, the OH ions combine with H+ ions of the indicator to form water.
HPh H+ + Ph
NaOH Na+ + OH–
Thus Ph ions gives pink colour in alkaline medium.
( i i ) Methyl orange (MeOH) :- It is a weak base and dissociates as :-
MeOH Me+ + OH
Yellow Red Colourless
If [MeOH] > [Me+] Yellow
[MeOH] < [Me+] Red
MeOH is not dissociated in alkaline medium due to the presence of common ions OH and the solution
remains yellow.
MeOH Me+ + OH–
NaOH Na+ + OH–
In acidic medium OH combine with H+ thus increase the ionisation of MeOH. Hence yellow colour of
solution change to red colour.
MeOH Me+ + OH–
HCl H+ + Cl–
MODERN QUINONOID THEORY :
According to this theory,
(i) An acid-base indicator is a dynamic equilibrium mixture of two alternative tautometric forms ; ordinarily one
form is benzenoid while the other is quinonoid.
(ii) The two forms have different colours.
(iii) Out of these one form exist in acidic solution, while the other in alkaline solution.
(iv) The change in pH cause the transition of benzenoid form to quinonoid form and vice-versa and consequently
change in colour.
Ex.
(a) For methyl orange
Na+ +–O3S N= N N(CH3)2 (AHcid( A)lkHali) Na+ +–O3S N–N= N(CH3)2
H
Yellow benzenoid form (in bases) Red quinonoid form (in acids)
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(b) For Phenolphthalein ABYSS Learning
OH
O
HO CO H HO C
C= O H COO–
Colourless benzenoid form (in Acid) Red quinonoid form (in Alkali)
THEORIES OF INDICATORS :
( a ) Ostwald theory :- According to this theory :
(i) Indicators are organic, aromatic weak acids or weak bases.
(ii) The colour change is due to ionisation of the acid - base indicator. The unionised form has different
colour than the ionised form.
(iii) Every indicator shows colour changes in opposite medium due to the conversion of unionized part into
ionized part.
For example phenolpthalein shows pink colour in basic medium and methyl orange shows red colour in
acidic medium.
Note : For oxidation reduction (Redox) reactions indicators are not used because these reactions are very
fast. Indicators are not used in coloured solution also.
TITR ATION OF STRONG ACID AGAINST STRONG ALK ALI :
The graph (A) shows how pH changes during the titration of 50 cm3 of 0.1 M HCl with 0.1 M NaOH.
NaOH (aq) + HCl (aq) NaCl (aq) + HO ()
2
The pH of 0.1 M solution of HCl in the beginning would be 1. As alkali is added, the pH changes slowly in the
beginning. However, at the equivalence point pH changes rapidly from about 3.5 to 10. It can be shown by
simple calculations that pH of the solution is 3.7 when 49.8 cm3 of 0.1 M NaOH solution have been added.
The pH suddenly changes to 10 after addition of 50.1 cm3 of the NaOH solution. Thus, any indicator having
pH range between 3.5 to 10 will identify the equivalence point. This means that any one of phenolphthalein,
methyl orange or bromothylmol blue could be used as an indicator.
14 14
10 Phenolphthalein 10 Phenolphthalein
pH 7 pH 7
(A) 4 Methyl orange (B) 4 Methyl orange
0 50 100 0 50 100
Volume of base/cm3 Volume of base/cm3
14
10 Phenolphthalein 14
10 Phenolphthalein
pH 7 7
(C) 4 Methyl orange
pH
(D) 4 Methyl orange
0 50 100 0 50 100
Volume of base/cm3 Volume of base/cm3
Titration curves : (A) strong base with strong acid ; (B) weak base with strong acid ; (C) strong base with weak acid ;
(D) weak base with weak acid.
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TITR ATION OF STRONG ACID AGAINST WEAK ALK ALI :
The graph (B) shows how pH changes during titration of 50 cm3 of 0.1 M HCl with 0.1 M NH .
3
HCl (aq) + NH OH (aq) NH Cl (aq) + HO ()
4 4 2
In this case, the pH changes rapidly from 3.5 to 7.0 at the equivalence point. Methyl orange, methyl red and
bromocresol green are suitable indicators for this type of titration. Phenolphthalein is unsuitable because its pH
range lies outside the vertical portion of the curve.
TITR ATION OF WEAK ACID AGAINST STRONG BASE :
The graph (C) shows how pH changes during titration of 50 cm3 of 0.1 M CH COOH with 0.1 M NaOH.
3
CH COOH (aq) + NaOH (aq) CH COONa (aq) + H O ()
3 32
The vertical portion of this titration curve lies between pH range 7 to 10.6. Phenolphthalein is suitable indicator
for this titration. Methyl orange is not suitable for this titration because its pH range lies on the flat portion of
the curve.
TITR ATION OF WEAK ACID AGAINST WEAK BASE :
The graph (D) represents the titration curve obtained for titration of 50 cm3 of 0.1 M CH COOH with 0.1 M NH .
33
CH3COOH (aq) + NH4OH (aq) CH3COONH4 (aq) + H2O ()
For this type of titration there is no sharp increase in pH at the equivalence point. No indicator is suitable for
this type of titration.
Indicator pH range Colour change pK
a
Methyl orange 3.2 – 4.5 Pink to yellow
Methyl red 4.4 – 6.5 Red to yellow Neutral colour pH
Litmus 5.5 – 7.5 Red to blue 3.7
Phenol red 6.8 – 8.4 Yellow to red 5.1
Phenolpthalein 8.3 – 10.5 Colourless to pink 7.0
7.8
9.6
Ex. Bromophenol blue is an indicator with a value of K = 6.84 × 10–6. At what pH it will work as an indicator?
Sol. a
Also report the % of this indicator in its basic form at a pH of 5.84.
HBPh H+ + BPh–
Ka [H ][BPh ] when BPh– = HBPh, indicator will work. Thus
=,
[HBPh]
[H+] = 6.84 × 10–6
pH = 5.165
Also if pH = 5.84
or [H+] = 1.44 × 10–6, then
K= [H ][BPh ] or 6.84 × 10–6 = 1.44 106.C or = 0.83 or 83 %
a
[HBPh] C(1 )
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MEMORY TIPS
1 . A strong electrolyte is defined as a substance which dissociates almost completely into ions in aqueous solution
and hence is a very good conductor of electricity e.g., NaOH, KOH, HCl, H SO , NaCl, KNO etc.
24 3
2 . A weak electrolyte is defined as a substance which dissociates to a small extent in aqueous solution and hence
conducts electricity also to a small extent e.g. NH OH, CH COOH etc.
43
3 . Degree of dissociation :- The fraction of the total amount of an electrolyte which dissociates into ions is called
the degree of dissociation (),
Number of moles dissociated
i.e. = Number of moles taken
4 . According to Arrhenius concept of acids and bases, an acid is a substance which gives H+ ions in the aqueous
solution whereas a base is a substance which gives OH– ions in the aqueous solution.
5 . According to Bronsted-Lowry concept of acids and bases, an acid is a substance which can give a proton and
a base is a substance which accepts a proton.
6 . According to Lewis concept of acids and bases, an acid is a substance which can accept a lone pair of electrons
whereas a base is a substance which can donate a lone pair of electrons.
Types of Lewis bases
(i) Neutral molecules containing a lone pair of electrons on the central atom like : NH , R O H , H2O : etc. (ii) All
3
negative ions like F–, Cl–, Br–, l–, OH– etc.
Types of Lewis acids
(i) Molecules having central atom with incomplete octet e.g. BF , AlCl etc.
33
(ii) Simple cations e.g. Ag+ , Cu2+, Fe3+ etc.
(iii) Molecules having central atom with empty d-orbitals e.g. SnCl , SiF , PF etc.
4 45
(iv) Molecules containing multiple bonds between different atoms e.g. O = C = O.
7. According to Ostwald's dilution law, for the solution of a weak electrolyte with concentration
C, mol L–1 and as the degree of dissociation,
K C2 C2 or = Ka / C = KaV
=
a 1
8 . Relative strength of two weak acids is given by Strength of acid HA1 = K a1
Strength of acid HA 2 K a2
9 . Ionic product of water, Kw = [H+] [OH–] or [H3O+] [OH–]. Its value at 25°C = 10–14
10. pH = –log [H O+], pOH = –log [OH–], pK = –log K , pK = – log K
3 a ab b
11. As K = [H+] [OH–] = 10–14 therefore pK = pH + pOH = 14.
ww
12. Solubility product of a sparingly soluble salt Ax By is given by
K = [Ay+]x × [B x–]y
sp
e.g. for AgCl, K = [Ag+] [Cl–], for Ca (PO ) , K = [Ca2+]3 [PO 3–]2 etc.
sp 3 4 2 sp 4
13. If two solutions are mixed in which ions can combine to form a precipitate, concentration of ions in the solution
after mixing are calculated. Then ionic product is calculated using the same expression as for K . If ionic
sp
product > solubility product, a precipitate is formed.
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ABYSS Learning
14. To calculate the solubility of a salt like AgCl in the pre sence of a strong electroly te like NaCl, total
[Cl–] is calculated (Cl– ions from AgCl being negligible). Knowing K , [Ag+] can be calculated.
sp
15. pH of an acidic buffer is given by Henderson equation viz
[Salt]
pH = pK + log
a [Acid]
16. pH of a basic buffer is given by
pOH = pK + log [Salt] and then pH = 14 – pOH
b [Base]
17. Buffer capacity = No.of moles of the acid or base added to1 litre of buffer = n
Change in pH pH
18. pH of boiling water is 6.5625. It does not mean that boiling water is not neutral. It is due to greater dissociation
of H O into H+ and OH–.
2
19. pH can be zero in 1 N HCl solution or it can be negative for more concentrated solution like 2N, 3N, 10 N etc.
20. The buffer system present in blood is H2CO3 + NaHCO3.
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ABYSS Learning
SOLVED EXA MPLES
Ex.1 Stomach acid is a solution of HCl with concentration of 2.2 × 10–3 M. what is the pH of stomach acid :
Sol.
(A) 3.92 (B) 2.65 (C) 4.92 (D) 1.92
Ex.2
Sol. (B)
Ex.3
Sol. HCl is 100 % ionised so
[H O+] = 2.2 × 10–3 M
3
pH = – log (2.2 × 10–3 M) or pH = 2.65
Calculate the [H O+] of blood, the pH of which is 7.2 (slightly basic).
3
(A) 5 × 10–8 M (B) 6.3 × 10–8 M (C) 5 × 10–9 M (D) 4 × 10–7 M
(B)
As pH = 7.2 so [H O+] = antilog (–7.2) = 6.3 × 10–8 M
3
The pH of an aqueous solution at 25°C made up to 0.3 M, with respect to NaOH and 0.5 M, with respect to
acetic acid (pK = 4.76) would be nearly :
a
(A) 4.25 (B) 4.93 (C) 4.75 (D) 5.05
(B)
pH = pK – log [acid]
a
[salt]
0.3 M NaOH will react with acid to form 0.3 M CH COONa and therefore CH COOH concentration will be
33
reduced to 0.2 M.
pH = 4.76 – 0.2 4.93
log
0.3
Ex.4 Calculate the pOH and pH of a 0.1 M CH COO– solution (K = 1.8 × 10–5).
3a
CH COO– + H O CH COOH + OH–
32 3
(A) 6.12, 7.88 (B) 4.12, 9.88 (C) 5.13, 8.87 (D) none of the above
Sol. (C)
11
p H = 7 + pK + log C
2 a2
11
= 7 + × 4.74 + log (0.1)
22
pH = 8.87
pOH = 14 – 8.87 = 5.13
Ex.5 The pH of a solution of NH is 5.806. If its concentration is 0.95 M then what is the value of its dissociation
Sol. 3
constant ? :
(A) anti log [28 + log (0.95) – 23.242] (B) anti log [11.612 – log (0.95) – 28]
(C) anti log [11.612 – log (0.95) – 14] (D) anti log [14 + log (0.95) – 11.612]
(B)
11
Since pH = 14 – pOH and pOH = pK – log C
2b2
11
or pH = 14 – pK + log C
2 b2
1
or pK = 2 (14 + log C – pH)
b2
or K = antilog [11.612 – log (0.95) – 28]
b
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ABYSS Learning
Ex.6 The solubility product of BaSO is 1.5 × 10–9. The precipitation in a 0.01 M Ba2+ ions solution will start on
Sol. 4
Ex.7 adding H SO of concentration :
Sol. 24
Ex.8 (A) 10–9 M (B) 10–8 M (C) 10–7 M (D) 10–6 M
(D)
[Ba2+] [SO 2–] = 1.5 × 10–9 (K ) and [Ba2+] = 0.01 M
4 sp
so Required [SO 2–] = 1.5 10 9 = 1.5 × 10–7
4 0.01
so [H SO ] > 1.5 × 10–7 for precipitation of BaSO .
24 4
pH of a saturated solution of Ca(OH) is 12. Its solubility product is :
2
(A) 10–6 (B) 4 × 10–6 (C) 5 × 10–7 (D) None of these
(C)
pH = 12 so [OH–] = 10–2 M
Now Ca(OH) Ca2+ + 2OH–
2(s) 10–2 M
5 × 10–3M
so K = [Ca2+] [OH–]2
sp
= (5 × 10–3) (10–2)2 = 5 × 10–7
A sample of 100 ml of 0.10 M acid HA (K = 1 × 10–7) is titrated with standard 0.10 M KOH. How many mL
a
of KOH will have to be added when the pH in the titration flask will be 7.00 ?
(A) 0 (B) 10 (C) 100 (D) 50
Sol. (D) pH = pK + log [Salt]
a [Acid]
7 = 7 + log [N 2 V2 ]
[N1 V1 N 2 V2 ]
1 = 0.1 V2
0.1 100 0.1 V2
or 10 – 0.1 V = 0.1 V
22
or V = 50 mL
2
Ex.9 If the solubility of lithium sodium hexafluoroaluminate, Li Na (AlF ) is 's' mol L–1, its solubility product is equal to:
Sol. 33 62
(A) s8 (B) 12 s3 (C) 18 s3 (D) 2916 s8
(D)
Li Na (AlF ) 3Li+ + 3 Na+ + 2AlF 6–
33 6 2(s) 6
Let solubility of salt is 's' mol L–1
so [Li+] = 3s = [Na+]
[AlF 6–] = 2s
6
so K = [Li+]3 [Na+]3 [AlF 6–]2
sp 6
= (3s)3 (3s)3 (2s)2
= 2916 s8
E x . 1 0 pH of a buffer containing 6.0 g of CH COOH and 8.2 g of CH COONa in 1 L of water is (pK = 4.74) :
33 a
(A) 7.5 (B) 4.74 (C) 5.5 (D) 6.5
Sol. (B)
[CH COOH] = 6.0 × 1 = 0.1 M
3 60
[CH COONa] = 8.2 × 1 = 0.1 M
3 82
so [Salt] or pH = 4.74 + log 0.1 = 4.74
pH = pK + log 0.1
a [Acid]
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ABYSS Learning
E x . 1 1 For the hydrolytic equilibrium ; B+ + H O BOH + H+
2
K = 1 × 10–5. Calculate the hydrolysis constant :
b
(A) 10–5 (B) 10–19 (C) 10–10 (D) 10–9
Sol. (D)
Kh = Kw 10 14 = 10–9
Kb = 1 10 –5
E x . 1 2 The solution of a salt of a weak acid and weak base will have pH :
(K = 10–4 ; K = 10–6) (B) 8 (C) 6 (D) 4
ba
(A) 7
Sol. (B)
11
pH = 7 + 2 pK a 2 pK b = 7 + 3 – 2 = 8
E x . 1 3 Determine the effect on the concentrations of NH , NH + and OH– when small amounts of each of the
34
following is added to a solution of NH in water.
3
(A) HCl (B) NH (C) NaOH (D) NH Cl
3 4
(E) KNO
3
S o l . The equilibria of NH3 in water will be
NH + H O NH + (aq.) + OH– (aq.)
32 4
(A) If HCl is added it will furnish H+ and Cl– ions as it is a strong electrolyte. Now H+ will form H O with
2
OH– ions (neutralisation) so [NH ] will decrease, [NH +] and [OH–] increase since equilibria will shift in
34
forward direction.
(B) If more NH is added, then [NH ], [NH +] and [OH–] will increase.
3 34
(C) If NaOH is added it will furnish [Na+] and [OH–] ions since it is a strong electrolyte and [OH–] will
suppress the ionisation of weak base NH so [NH ] increases, [NH +] decreases and [OH–]increases.
33 4
(D) NH Cl will also apply common ion effect on NH OH but now [NH ] increases, [NH +] increases but
4 4 34
[OH–] decreases.
(E) NO effect expected.
E x . 1 4 The pH of 0.0516 M solution of HCN is 2.34, what is K for HCN ?
a
S o l . For weak acid.
11
pH = pK – log C
2 a2
11
2.34 = pK – log (0.0516)
2 a2
pK = 3.3926 or K = anti log (– 3.3926)
a a
Ka = 4.04 × 10–4
E x . 1 5 A buffer with pH 10 is to prepared by mixing NH Cl and NH OH. Calculate the number of moles of NH Cl
44 4
that should be added to one litre of 1 M NH OH. (K = 1.8 × 10–5) :
4b
S o l . Handerson equation for base buffer may be given as :
pOH [Salt]
= pK + log
b [Base]
14 – pH = pK + log [Salt] .......(i)
b
[Base]
Given, pH = 10
[Base] = [NH OH] = 1 M
pK 4
b = –log K
b
= – log [1.8 × 10–5] = 4.7447
Hence, from Eq. (i), we get
[Salt]
14 – 10 = 4.7447 + log
1
[Salt] = 0.18 M
No. of moles of NH Cl = 0.18
4
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