CIE – 1
IONIC EQUILIBRIUM
C1A Ionic equilibrium is the study of equilibrium among ions in the aqueous solution. The process of
splitting of a molecules into its ions is known as ionization.
C1B
There are two types of electrolytes :
(i)
strong electrolyte : are converted into ions element completely.
(ii)
weak electrolyte : dissociation is incomplete. In aqueous solution of these compounds only
slight amount of electrolyte ionises and there exist an equilibrium between ionized
molecules and unionized molecules. for e.g.
(a)
weak acids : CH3COOH, HCN etc.
(b)
weak bases : NH4OH, C5H5N etc.
(c)
weak salts : AgCl, PbCl2, AgCrO4, MnS, H2S etc.
C1C STRENGTH OF ACIDS AND BASES AND pH SCALE
Concept of Acids and Bases
(a)
According to Arrhenius an acid is a substance that dissociates in water to give hydrogen ions, and
base is a substance that dissociates in water to give hydroxyl ion.
(b)
According to Bronsted-Lowry Concept, acid is a substance that is capable of donating a hydrogen
ion (H+) and bases are substances capable of accepting a hydrogen ion (H+). In short, acids are
proton donors and bases are proton acceptors.
(c)
According to Lewis, acid is defined as a species which accepts electron pair and base which donates
an electron pair.
Acidic strength means the tendency of an acid to give H+ ions in water and basic strength means the
tendency of a base to give OH– ions in water. So more the tendency to give H+ or OH– ions more will
be the acidic or basic strength of acid or base. In 1909 S.P.L. Sorrenson introduced a scale called pH
scale or pOH scale to measure the acidic or basic strength of acids and bases
pH = –log[H–]
pOH = –log[OH–]
When [H+] of a solution is more than [OH+] the solution is called to be acidic. When [H+] of a solution
is less than [OH+] the solution is called to be basic. If [H+] and [OH–] of a solution are equal the
solution is called neutral.
Practice Problems :
1.
Water has pKw = 13.26 at 500C. Its pH will be
(a)
6.0
(b)
7.0
(c)
6.63
(d)
13.26
[Answers : (1) c]
C2A SELF IONISATION OF WATER
Pure water act as a very weak electrolyte i.e. it gets ionised very weakly to give H+ and OH– ions, that
is H2O
H+ + OH–. Now, applying law of chemical equilibrium.
K
[H  ][OH  ]
[ H 2 O]
where K is equilibrium constant
K[H2O] = [H+] [OH–]
Kw = [H+] × [OH–]
where Kw is ionic product of water
Since the concentration of pure water is constant, so it multiplied by K will also be a constant (Kw)
called ionic product of water which is constant at constant temperature. The value of K w is
approximately 10–14 at 298 K.
[H+] [OH–] = 10–14 at 298 K
Therefore,
As water gives equal amount of H+ and OH–, we can say
[H+] [H+] = 10–14
Einstein Classes,
or
[OH–] [OH–] = 10–14
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CIE – 2
+
[H ] = 10
–7
pH = –log 10
–7
pH = 7
–
–7
or
[OH ] = 10
or
pOH = –log 10–7
or
pOH = 7
For some other temperature
[H+]2 = [OH–]2 = Kw
[H+] = [OH–] =
Kw
pH = pOH = – log
Kw
For pure water [H+] is always equal to [OH–] so it is called neutral. For acidic or basic solution the
[H+] and [OH–] are not equal but their product is always equal to Kw at that temperature.
C2B
COMMON ION EFFECT
If to an ionic equilibrium, we add any ion which appears in the equilibrium reaction, the
equilibrium will shift in a direction opposite to that in which that ion appears.
So dissociation of weak acid and bases is suppressed in presence of strong acids and bases
respectively. Similarly the solubility of salts decreases in presence of any common ion of that salt.
C2C DETERMINATION OF pH OF ACIDS AND BASES
(i)
Strong Acid : A strong acid is that which ionises completely to give maximum H+ ions. For e.g.
HCl  H+ + Cl–
Let the concentration of HCl is c. So the [H+] coming from acid is also c, but some [H+] is also coming
due to self ionisation of water (say x)
H2O
H+ + OH–
Now pH = –log [H+]total = –log([H+]acid + [H+]water)
But [H+]total [OH–]total = Kw
(c + x) (x) = Kw
If we know c and Kw we can calculate x.
Now pH = –log (c + x)
But in most of the cases if c is more than 10–6, the value of x will be negligible and we can take
pH = – log c
(ii)
(as x << c)
Weak acids : Weak acids are those which dissociates partially in water at equilibrium as
CH3COOH
CH3COO– + H+
The equilibrium constant is called Ka (dissociation constant)
Ka 
[CH 3COO  ][H  ]
[CH 3COOH]
Let the [H+] coming from water is x and the degree of dissociation of weak acid is .
CH3COO– + H–
CH3COOH
initial
c
at equilibrium
c – c
H2O
Ka 
+
H + OH
0
0
c
c
–
  (c  x )
1 
Kw = (c + x) (x)
...(i)
...(ii)
Now if we know c, Ka and Kw we can calculate  and x.
pH = –log [H+]total = –log (c + x)
+
If [H ]water is very less (i.e. c is is more than 10–6)
Einstein Classes,
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CIE – 3
c 2
1 
Ka 
pH = –log (c)
If  < < 1 then 1 –   1 (i.e., if  is less than 0.1)
Ka = c2

or
Ka
c

K a 
  log K a c
pH = –log c = –log  c 

c 

(iii)
Strong acid + Weak acid
Suppose we want to calculate pH of a solution containing a strong acid (HX) and a weak acid (HY)
whose concentrations are c1 and c2 respectively. Dissociation constant of weak acid is Ka.
HX  H+ + A–
c1
Ka 
c1
+
HY
H + Y–
c2(1 – )
c2 c2
[H  ]total [Y  ]total [c1  c 2 ][c 2 ]

[HY]total
c 2 (1   )
Ka 
[c1  c 2 ]
1 
From above equation we can calculate 
pH = –log [H+]total = –log (c1 + c2)
Note : In this case we have neglected H+ coming from water.
(iv)
Two weak acids
Suppose we want to calculate pH of a mixture of two weak acids HX and HY whose concentrations
are c1 and c2 and their dissociation constants are K a1 and K a 2 respectively..
HA
H + + A–
c1(1 – 1)
c11 c11
HB
H+ + B –
c2(1 – 2)
c22 c22
K a1 
(c11  c 2  2 )(c1 1 ) (c1 1  c 2  2 ) 1

c1 (1   1 )
(1   1 )
...(1)
K a2 
(c1 1  c 2  2 )(c 2  2 ) (c1 1  c 2  2 ) 2

c1 (1   2 )
(1   2 )
...(2)
For equation (1) and (2) we can calculate a1 and a2
pH = –log [H+]total = –log (c11 + c22)
Note : In this case we have neglected [H+] coming from water.
(v)
Dibasic (Dipotic) and Polyprotic weak acids
Dibasic acid is that acid which have the tendency to give two H+ ions per molecule of that acid e.g.
H2SO4, H2S etc. Let us calculate pH of a diprotic acid H2A whose dissociation constants for first and
second proton are K a1 and K a 2 respectively and concentration is c.
H 2A
Einstein Classes,
H+
+
HA–
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CIE – 4
c(1 – 1)
HA
c1
–
H
c1(1 – 2)
K a1 
+
c1
+
A2–
c12
c12
[H  ]total [HA  ]total (c 1  c 1 2 )[c 1 (1   2 )]

[H 2 A]total
c(1   1 )

K a2 
+
( 1   1 2 )[c 1 (1   2 )]
(1  1 )
[H  ]total [ A 2 ]total

[HA ]total

(c1  c1 2 )(c1 2 ) (c1  c1 2 ) 2

c1 (1   2 )
(1   2 )
From equation (1) and (2) we can calculate 1 and 2
pH = –log[H+]total = –log (c1 + c12)
Note : In this case we have neglected [H+] coming from water
pH OF BASES
The pH of bases can also be calculated as we have done for acids. In case of a base, instead of H+ we
take OH– ions, instead of Ka we take Kb and instead of pH we calculate pOH. Then pH can be
calculated as pH = 14 – pOH.
NOTE :
1.
Strong acid – [H+] = Normality
2.
Strong Base – [OH–] = Normality
3.
pH of mixture of strong acids / strong bases : We calculate the normality of final solution.
4.
pH of mixture of strong acids and strong bases : we calculate normality of final solution.
5.
i)
If equivalents of acids > eq. of base. Final solution will be acidic and normality = [H+]
ii)
If eq. of base > eq. of acid. Final solution will be alkaline and normality = [OH–]
iii)
If eq. of acid = eq. of base, final solution will be neutral and pH = 7 at 250C.
pH of weak monobasic acid or weak monoacidic base [H+] =
K a  C [OH  ]  K b  C
Here   K a / C
Note :
6.
i)
In above formula for , we have assumed  is very small compared to one and
hence neglected compared to one.
ii)
In case when we use above formula and  > 0.1, we do not apply above
approximation and if   0.1, approximation is valid.
pH of mixture of two weak acids : we must consider ionisation of two acid separtely in which
Total [H+] = [H+] produced from acid (1) and [H+] produced from acid (II) = C11 + C22
Where C1, C2 are concentration of two acids and 1 and 2 are degree of dissociation of acids in
presence of each other.
Practice Problems :
1.
The pH of a solution obtained by dissolving 5 × 10–4 moles of Ca(OH)2 (strong electrolyte) to 100 ml
solution at 298 K will be
(a)
2.
11
(b)
Consider the reaction A— + H3O+
value of K for this reaction.
(a)
1.0 × 106
Einstein Classes,
(b)
12
(c)
9.8
(d)
2
HA + H2O. The Ka value for acid HA is 1.0 × 10–6. What is the
1.0 × 10–8
(c)
1.0 × 108
(d)
1.0 × 10–6
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CIE – 5
3.
–4
The ionisation constant of HCO2H is 1.8 × 10 . What is the percent ionization of a 0.001 M
solution ?
(a)
4.
66%
The pH of a 10
(a)
(b)
–10
42%
(c)
34%
(d)
58%
(c)
4
(d)
– 10
M NaOH solution is nearest to
10
(b)
7
[Answers : (1) b (2) a (3) c (4) b]
C3
SALT HYDROLYSIS
Reaction in which cation or anion of the salt react with water to convert water acidic or basic in
nature, is known as salt hydrolysis.
1.
Salt of a Strong Acid and
Salt of Weak acid and
Salt of Weak acid and
Weak Base e.g. NH4Cl
Strong Base e.g. CH3COONa
Weak Base e.g. CH3COONH4
NH4+ + H2O
NH4OH + H+ CH3COO– + H2O
NH4+ + CH3COO– + H2O
CH3COOH + OH–
i)
KH = KW / K b
ii)
h
KH
c
i)
KH = Kw / Ka
iii)
[H+] = Ch
ii)
h
iv)
[H  ] 
iii)
[H  ] 
v)
pH = (1/2)
iv)
pH = 1/2
Kw c
Kb
[pKw – log c – pKb].
vi)
their solution is acidic
pH< 7
CH3COOH + NH4OH
i) KH = Kw / Ka . Kb
KH
c
ii) h  K H
Kw  Ka
c
iii) [H+] = hKa, [OH–] = hKb
iv) [H  ] 
K wK a
Kb
(pKw + log c + pKb)
v)
their solution is basic
v) pH = 1/2 [pKw + pKa – pKb]
pH > 7
vi) solution is acidic if
Ka > Kb
solution is basic if
Kb > Ka.
solution is neutral if
Ka = Kb
NOTE :
1.
All the formulae for salt hydrolysis are for univalent salts. The term ‘c’ in the above equations
however represents the concentration of ion that undergoes hydrolysis.
2.
In all the formulae mentioned above we have neglected h compared to one.
Practice Problems :
1.
pH of 0.01 M HS— will be :
(a)
pH  7 
(c)
pH 
Einstein Classes,
pK a log c

2
2
pK1  pK 2
2
pK b log c

2
2
(b)
pH  7 
(d)
 pK a  pK b 
pH  7  

2


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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 6
2.
3.
4.
5.
The degree of hydrolysis of anilinium acetate is
(a)
independent of initial concentration
(b)
directly proportional to initial concentration
(c)
inversely proportional to initial concentration
(d)
inversely proportional to square root of initial concentration
The compound whose 0.1 M solution is basic is
(a)
Ammonium acetate
(b)
Ammonium chloride
(c)
Ammonium sulphate
(d)
Sodium acetate
Which of the following when mixed, will give a solution with pH greater than 7 ?
(a)
0.1 M HCl + 0.2 M NaCl
(b)
100 ml of 0.2 M H2SO4 + 100 ml of 0.3 M NaOH
(c)
100 ml of 0.1 M CH3COOH + 100 ml of 0.1 M KOH
(d)
25 ml of 0.1 M HNO3 + 25 ml of 0.1 M NH3
The correct order of increasing [H3O+] in the following aqueous solutions is
(a)
0.01 M H2S < 0.01 M H2SO4 < 0.01 M NaCl < 0.01 M NaNO2
(b)
0.01 M NaCl < 0.01 M NaNO2 < 0.01 M H2S < 0.01 M H2SO4
(c)
0.01 M NaNO2 < 0.01 M NaCl < 0.01 M H2S < 0.01 M H2SO4
(d)
0.01 M H2S < 0.01 M NaNO2 < 0.01 M NaCl < 0.01 M H2SO4
[Answers : (1) a (2) a (3) d (4) c (5) c]
C4
BUFFER SOLUTION
There are three types of buffers :
(i)
Acidic buffer : This consists of solution of a weak acid and its conjugate base (i.e., its salt with weak
or strong base). Let us consider an acid buffer containing acetic acid and sodium acetate (conjugate
base is CH3COO–).
CH3COOH
CH3COO– + H+
CH3COONa  CH3COO– + Na+
Now if we add any acid (H+) in this buffer it will shift the above equilibrium in backward direction
and hence some H+ will be consumed to resist the change in pH. If we add base (OH–) it will consume
the H+ from solution, but the above equilibrium will shift in forward direction and so the decrease in
H+ are compensated by forward shift of equilibrium to some extent. If we add water (i.e., we dilute
the solution) the concentration of H+ will decrease which will shift the equilibrium in forward
direction to give more H+ and so compensate the decrease in H+ concentration.
CH3COONa  CH3COO– + Na+
CH3COOH
CH3COO– + H+
Now due to the presence of CH3COO– ions of salt the dissociation of CH3COOH is suppressed due to
common ion effect of CH3COO– ions. So we can assume that the total [CH3COO–] is almost equal to
(CH3COO–) from salt.
Ka 
[CH 3COO  ]total [H  ]total
[CH 3COOH]
[H  ]  K a
[CH 3 COOH ]
[CH 3 COO  ]
pH   log H    log K a  log
Einstein Classes,
[CH 3COOH ]
[CH 3 COO  ]
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CIE – 7
(ii)
pH  pK a  log
[CH 3COO  ]
[CH 3COOH]
pH  pK a  log
[Salt ]
[ Acid]
Basic buffer : It is a solution of a weak base and its conjugate acid (salt with weak or strong acid). Let
us consider a buffer containing NH4OH and NH4Cl
NH4Cl  NH4+ + Cl–
NH4+ + OH–
NH4OH
The buffer action may be explained as in the case of acid buffer. The pH is given by
pOH  pK b  log
[Salt ]
[Base]
pH = 14 – pOH
(iii)
Salt buffer : A solution of a salt of weak acid with weak base (CH3COONH4) also act as a buffer.
CH3COONH4  CH3COO– + NH4+
When an acid is added to it, the H+ combine with CH3COO– to give CH3COOH.
CH3COO– + H+
CH3COOH
–
When a base is added, the OH combine with NH4+ to give NH4OH.
NH4+ + OH–
NH4OH
The pH of this type of solution has been discussed already in the case of hydrolysis of such salts.
Practice Problems :
1.
The pH of mixture of CH3COONa + CH3COOH after adding water shows........value.
(a)
2.
(b)
Decreased
(c)
Constant
(d)
—
All
A certain buffer solution contains equal concentration of X and HX. The Kb for X is 10–10. The pH
of the buffer is
(a)
3.
Increased
4
(b)
7
(c)
—
10
(d)
14
One litre of a buffer solution containing 0.01 M NH4Cl and 0.1 M NH4OH having pKb of 5 has pH of
(a)
9
(b)
10
(c)
4
(d)
6
[Answers : (1) c (2) a (3) b]
C5
SOLUBILITY PRODUCT
Insoluble substances like AgCl, BaSO4, PbCl2, etc., are infact not completely insoluble when present
in an aqueous medium. A very small amout of these dissolves and is present as ions. Further, there
exists an equilibrium between the undissolved and the dissolved salt. For AgCl, the equilibrium
equation may be written as,
AgCl(s)
Ag+(aq) + Cl–(aq)
Applying the law of mass action, K 
[ Ag  ][Cl  ]
[ AgCl ]
[AgCl] is assumed to be constant because of the fact that very little of this solid dissolved in aqueous
solution (by definition)
Ksp = [Ag+][Cl–]
NOTE :
1.
Let the solubility of salt of weak acid and strong base is s1 in pure water, s2 in basic buffer and s3 in
acidic buffer then
s3 > s1 > s2
Einstein Classes,
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 8
2.
For Preciptitation, Ksp < Kip
Mutual solubility of two springly soluble salts
Let Ksp(AgCl) = x
x and y are close
Ksp(AgBr) = y
[Ag+] [Cl–] = x
+
...(i)
–
[Ag ] [Br ] = y
+
...(ii)
–
–
[Ag ] = [Cl ] + [Br ]
...(iii)
If salts were Ag2A and Ag2B then
1
[ Ag  ]  [ A]  [B]
2
Divide equation (i) by (ii), we get,
[Cl  ]
[Br  ]

x
y
...(iv)
Divide equation (iii) by [Cl–]
[ Ag  ]

[Cl ]
 1
[Br  ]

[Cl ]
 1
y x y

x
x
...(v)
From equation (i) and (v), we get
[Ag+]2 = x
[Ag+] =
(x  y )
x
xy
Knowing [Ag+], we can calculate [Cl–] and [Br–] in the solution which will be the mutual solubility of
AgCl and AgBr respectively.
Practice Problems :
1.
2.
3.
Ksp of CaSO4 is 4 × 10–12. CaSO4 is precipitated on mixing equal volumes of the following solutions :
(a)
3 × 10–6 M CaCl2 and 3 × 10–6 M (NH4)2SO4
(b)
4 × 10–6 M CaCl2 and 3 × 10–6 M (NH4)2SO4
(c)
6 × 10–6 M CaCl2 and 3 × 10–6 M (NH4)2SO4
(d)
In all cases.
The precipitate of Ag2CrO4 (Ksp = 1.9 × 10–12) is obtained when equal volumes of the following are
mixed
(a)
10–4 M Ag+ + 10–4 M Cro42–
(b)
10–2 M Ag+ + 10–3 M Cro42–
(c)
10–5 M Ag+ + 10–3 M Cro42–
(d)
10–4 M Ag+ + 10–5 M Cro42–
M(OH)x has (KSP 4 × 10–12) and solubility 10–4 M. Then the value of x is
(a)
1
(b)
2
(c)
3
(d)
–4
[Answers : (1) d (2) b (3) b]
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 9
SINGLE CORRECT CHOICE TYPE
1.
A solution is a mixture of 0.05 M NaCl and 0.05 M
NaI. The concentration of iodide ion in the solution
when AgCl just starts precipitating is equal to :
10.
pH of water is 7.0 at 250C. If water is heated to 700C,
the
(a)
pH will decrease and solution becomes
acidic
(b)
pH will increase and solution becomes
basic
(c)
pH will remain constant as 7
(d)
pH will decrease but solution will be
neutral
(Ksp AgCl = 1 × 10–10 M2 ; Ksp AgI = 4 × 10–16 M2)
(a)
(c)
2.
3.
4.
5.
6.
7.
8.
9.
4 × 10–6 M
(b)
–7
2 × 10 M
(d)
2 × 10–8 M
8 × 10
–15
M
Silver iodide is used in cloud seeding to produce
rain AgI
Ag+(aq) + I— (aq); Ksp = 8.5 × 10–7.
AgNO3 and KI are mixed to give [Ag+]  0.010 M;
[I—]  0.015 M. Will AgI precipitate ?
(a)
yes
(b)
(c)
cant’ say
(d)
this depends on [NO3—] and [K+]
The conjugate acid of NH
—
2
no
11.
F— + H2O
is
H3O+ + F—
Given, HF + H2O
HF + OH—
Which relation is correct
(a)
NH3
(b)
NH2OH
(a)
(b)
NH4+
Kb = Kw
Kb = 1/Kw
(c)
(d)
N2H2
(c)
Ka × Kb = Kw
(d)
Ka/Kb = Kw
Which does not act as Bronsted acid
12.
(a)
NH4+
(b)
CH3COO—
(c)
HCO3—
(d)
HSO3—
Conjugate base of hydrazoic acid is
(a)
HN3—
(c)
—
N
(b)
N3 —
(d)
—
N2
13.
Which one is the strongest acid
(a)
ClO3(OH)
(b)
ClO2(OH)
(c)
SO(OH)2
(d)
HCOO—
The heat liberated when one mole of water is formed
by combining sulphuric acid and sodium
hydroxide is
(a)
34 kcal
(b)
13.7 kcal
(c)
8.5 kcal
(d)
25.5 kcal
The pH of solution A, B, C, D are 9.5, 2.5, 3.5 and
5.5 respectively. The most acidic solution is
(a)
A
(b)
B
(c)
C
(d)
D
(a)
6y4
(b)
64y4
The solubility product of PbBr2 is 8 × 10–5. If the
salt is 80% dissociated in saturated solution, the
solubility of the salt is [Atomic weights of Pb and
Br are 208 and 80 a.m.u. respectively]
(c)
36y5
(d)
108y5
(a)
12.486 g/lt
(b)
6.25 × 10–3
(c)
5.25
(d)
9
The solubility of A2X3 is y mol dm–3. Its solubility
product is
Which equilibrium can be described as Lewis acidbase reaction but not Bronsted acid-base reaction
(a)
H2O + CH3COOH
CH3COO—
(b)
2NH3 + H2SO4
(c)
NH3 + CH3COOH
CH3COO—
(d)
[Cu(H2O)4]2+ + 4NH3
+ 4H2O
(c)
Less
Equally
(b)
(d)
15.
The extent of ionisation increases
H3O+ +
(a)
with the increase in the concentration of
the solute
2NH4+ + SO42—
(b)
on addition of excess of water to the
solution
NH4+ +
(c)
on decreasing the temperature of the
solution
(d)
on stirring the solution vigorously
[Cu(NH3)4]2+
16.
A solution of pH 8 is.....basic than a solution of pH
12
(a)
14.
The hydrogen ion concentration in weak acid of
dissociation constant Ka and concentration c is
nearly equal to
More
None
(a)
(c)
Einstein Classes,
Ka
c
Ka c
(b)
(d)
c
Ka
Kac
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 10
17.
18.
19.
20.
21.
22.
23.
24.
0
+
–6
–1
At 90 C, pure water has [H3O ] = 10 mol litre .
The value of Kw at 900C is
25.
In which of the following solvents with AgBr have
highest solubility ?
(a)
10–6
(b)
10–8
(a)
10–3 M NaBr
(b)
10–3 M NH4OH
(c)
10–12
(d)
10–14
(c)
pure water
(d)
10–3 M HBr
An acid solution of pH 6 is diluted hundred times.
The pH of solution becomes approximately
(a)
6.95
(b)
6
(c)
4
(d)
9
26.
For pure water
(a)
pH increases and pOH decreases with
increase in temperature
(b)
pH decreases and pOH increases with
increase in temperature
About buffer solution which is correct ?
(a)
It contains a weak acid and its conjugate
base
(c)
Both pH and pOH increases with increase
in temperature
(b)
It contains a weak base and its conjugate
acid
(d)
Both pH and pOH decrease with increase
in temperature
(c)
It shows little change in pH on adding
small amount of acid or base
(d)
None of the above
27.
An acidic buffer solution can be prepared by
mixing solution of
1 M NH4OH and 1 M HCl are mixed to make total
volume of 300 mL. If pH of the mixture is 9.26 and
pKa (NH4+) = 9.26 then volume ratio of NH4OH and
HCl will be :
(a)
1:1
(b)
1:2
(c)
2:1
(d)
3:1
(a)
Sodium acetate and acetic acid
(b)
Ammonia and ammonium hydroxide
(c)
Sulphuric acid and sodium sulphate
[Ag(CN)2]—
(d)
Sodium chloride and sodium hydroxide
constant at 250C is 4.0 × 10–19, then the silver iron
concentration in a solutionwhich was originally 0.1
molar in KCN and 0.03 molar in AgNO3 is
28.
A mixture of weak acid (say acetic acid) and its salt
with a strong base (say sodium acetate) is a buffer
solution. Which other pair of substances from the
following may have a similar property ?
(a)
HCl and NaCl
(b)
NaOH and NaNO3
(c)
KOH and KCl
(d)
NH4OH and NH4Cl
29.
When 1.0 ml of dilute HCl is added to 100 ml of
buffer solution of pH 4, the pH of the solution
(a)
Becomes 7.0
(b)
Remain almost same
(c)
Becomes 2.0
(d)
Becomes 10.0
Kc for the reaction
(a)
7.5 × 1018
(b)
7.5 × 10–18
(c)
7.5 × 1019
(d)
7.5 × 10–19
The pKa of acetylsalicylic acid (aspirin) is 3.5. The
pH of gastric juice in human stomach is about 2 – 3
and the pH in the small intestine is about 8. Aspirin
will be
(a)
Unionized in the small intestine and in
the stomach
(b)
Completely ionized in the small intensine
and in the stomach
(c)
Ionized in the stomach and almost
unionized in the small intensine
(d)
Ionized in the small intensine and almost
unionized in the stomach
A precipitate is formed when
(a)
the solution becomes saturated
(b)
the ionic product is less than the
solubility product
(c)
the ionic product is nearly equal to the
solubility product
(d)
the ionic product exceeds the solubility
product
Solubility product of A2B is 4 × 10–9 (mol/lit)3. Its
solubility is
(a)
10–3 M
(b)
41/3 × 10–3 M
(c)
10–4 M
(d)
2 × 10–5 M
Einstein Classes,
30.
A certain weak acid has a dissociation constant of
1.0 × 10–4. The equilibrium constant for its reaction
with a strong base is
(a)
(c)
31.
Ag+ + 2CN—, the equilibrium
1.0 × 10–4
1.0 × 1010
(b)
(d)
1.0 × 10–10
1.0 × 1014
Solubility of BaF2 in a solution of Ba(NO3)2 will be
represented by the concentration term
(a)
(c)
[Ba2+]
½ [F—]
(b)
(d)
[F—]
2[NO3—]
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 11
32.
33.
Solubility products of Al(OH)3 and Zn(OH)2 are
8.5 × 10–23 and 1.8 × 10–4 respectively. If equal moles
of Al3+ and Zn2+ ions are present in a solution, which
one will be precipitated first on addition of
NH4OH ?
EXCERCISE BASED ON NEW
PATTERN
COMPREHENSION TYPE
Comprehension-1
(a)
Al(OH)3
(b)
Zn(OH)2
(c)
Both of them
(d)
None of them
0
At – 50 C, the self-ionization constant (ion
product) is


Given that 2.0 × 10–4 mol each of Mn2+ and Cu2+
was contained in 1.0 L of a 0.0030 M HClO4 solution, and this solution was saturated with H2S. The
solubility if H2S, 0.10 mol/L, is assumed to be
independent of the presence of other materials in
the solution. (Given K a (H 2 S =10 –21 ),
Ksp(MnS = 2.5 × 10–10)

K NH3  NH4 NH2  1030 . How
many amide ions are present per mm3 of pure
liquid ammonia ?
34.
35.
36.
(a)
600 ions/mm3
(b)
6 × 106 ions/mm3
(c)
6 × 104 ions/mm3
(d)
60 ions/mm3
1.
The pH of a solution is 5.0. To this solution
sufficient acid is added to decrease the pH to 2.0.
The increase in hydrogen ion concentration is
(a)
100 times
(b)
1000 times
(c)
2.5 times
(d)
10 times
2.
A solution contains 10 ml of 0.1 N NaOH and 10 ml
of 0.05 N H2SO4, pH of this solution is
(a)
1
(b)
(c)
Greater than 7 (d)
Less than 7
Zero
3.
Which of the following solutions will have pH close
to 1.0 ?
(a)
100 ml of M/10 HCl + 100 ml of M/10
NaOH
(b)
55 ml of M/10 HCl + 45 ml of M/10 NaOH
(c)
10 ml of M/10 HCl + 90 ml of M/10 NaOH
(d)
75 ml of M/5 HCl + 25 ml of M/5 NaOH
Determine whether or not each of the ions Mn2+ and
Cu2+ will precipitate as the sulfide.
(a)
both MnS and CuS will be precipitated
(b)
CuS is precipitated but MnS will not
precipitate
(c)
MnS is precipitated but CuS will not
precipitate
(d)
both will remain in the solution
How much percentage Cu 2+ ions escapes
precipitations
(a)
10–19 %
(b)
4 × 10–13 %
(c)
2 × 10–4 %
(d)
9.8 × 10–19 %
If the original solution is made neutral by
buffering the [H+] at 10–9 will MnS precipitate
(a)
Precipitation takes place
(b)
No precipitation takes place
(c)
Ionic product is less than Ksp
(d)
Ionic product is equal to the Ksp
Comprehension-2
A buffer solution was prepared by dissolving
0.0200 mol propionic acid and 0.05 mol sodium
propionate in enough water to make 1.00 L of
solution. [Ka for propionic acid is 1.34 × 10–5].
ANSWERS
(SINGLE CORRECT CHOICE TYPE)
1.
c
11.
c
21.
d
31.
c
2.
a
12.
b
22.
b
32.
a
(a)
3.748
3.
a
13.
b
23.
d
33.
a
(c)
7.5
4.
b
14.
b
24.
a
34.
b
The pH change if 1.0 × 10 mol HCl was added to
10 mL of the buffer is
5.
b
15.
b
25.
b
35.
c
(a)
–0.051
(b)
(d)
16.
d
26.
d
d
0.0035
a
36.
(c)
6.
7.
d
17.
c
27.
c
8.
d
18.
a
28.
b
9.
a
19.
c
29.
d
10.
d
20.
a
30.
b
Einstein Classes,
4.
5.
6.
The pH of buffer is
(b)
4.7
(d)
5.7
–5
0.051
–0.0035
–5
The change in pH if 1.0 × 10 mol NaOH were
added to 10 mL of the buffer is
(a)
–0.051
(b)
0.051
(c)
0.0035
(d)
–0.0035
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 12
MATRIX-MATCH TYPE
MULTIPLE CORRECT CHOICE TYPE
Matching-1
1.
Column - A
Column - B
+
–7
(A)
Ka(CH3COOH) =
1.8 × 10–5 normality
is N/100
(p)
[H ] = 1.1 × 10
(B)
250 cc N/10 HCl
solution is mixed with
250 cc M/20 H2SO4
solution
(q)
[H+] = 2.3 × 10–9
250 cc N/10 HCl
(r)
(C)
[H+] = 0.1
2.
solution is mixed with
100 cc N/10 HCl
solution
(D)
10–8 M HCl solution
(s)
[H+] = 23.33 ×
10–14
3.
Matching-2
Column - A
(A)
(B)
Column - B
0.05 M of Na2SO4
(p)
Ksp (BaSO4) = 1.5 × 10–9
Buffer solution can be prepared from a mixture of
(a)
sodium acetate and acetic acid in water
(b)
sodium acetate and hydrochloric acid in
water
(c)
ammonia and ammonium chloride in
water
(d)
ammonia and sodium hydroxide in
water
Which of the following sets represents the
conjugate acid-base pair (s)
(a)
H2O, H3O+
(b)
H2SO4, SO42–
(b)
CH3COOH, CH3COO–
(d)
H3PO4, PO43–
For which of the solution PH = pka
(a)
100 ml of 0.1 M CH3COOH + 100 ml of
0.1 M CH3COONa
(b)
100 ml of 0.1 M CH3COOH + 50 ml of
0.1 M NaOH
(c)
100 ml of 0.1 M CH3COOH + 100 ml of
0.1 M NaOH
(d)
100 ml of 0.1 M CH3COOH + 100 ml of
0.1 M NH3
–8
1.8 × 10 mol/L
–8
Ksp for PbBr2 is
8 × 10–5 salt is 80%
dissociated
(q)
3 × 10 mol/L
(C)
0.01 M of AgNO3
Ksp (AgCl) = 1.8 × 10–10
(r)
3.4 × 10–2 mol/L
(D)
0.01 M of KCl
Ksp (AgCl) = 1.8 × 10–10
(s)
1 × 10–5 mol/L
4.
Which of the following are Lewis bases ?
(a)
Ag+
(b)
CH4
(c)
H2 O
(d)
CN–
Matching-3
(A)
(B)
Column - A
Column - B
[(CH3)3NH+] in an aq.
solution that is 1.02 M
(CH3)3 N is
(Kb = 6.02 × 10–5)
(p)
For an 0.025 M solution (q)
of the weak diprotic acid
H2CO3, the [H3O+] conc. is
(given
5.
8.0 × 10–3 M
1.025 × 10–4 M
6.
K1 = 4.2 × 10–7
–11
K2 = 5.6 × 10 )
(C)
(D)
The conc. of
(r)
CH3CH2COONa should
be added to 1 L of aq.
solution containing 0.02 mol
of CH3CH2COOH to
obtain a buffer solution of
pH 4.75
1.52 × 10–2 M
Conc. of HCN and NaCN (s)
in a solution is 0.01 M each.
Ka of HCN is 7.2 × 10–10 the
conc. of OH– ion is
1.39 × 10–5 M
Einstein Classes,
7.
PH of the following solution(s) is not affected by
dilution
(a)
0.01 M CH3COONa
(b)
0.01 M NaHCO3
(c)
0.01 M CH3COONH4
(d)
buffer of 0.01 M CH3COONa and
0.01 M CH3COOH
Which of the following are conjugate acid-base
pair(s)
(a)
H3O+, O–H
(c)
H3O+, CH3COO– (d)
(b)
HS–, S2–
HNO2, NO2–
Which of the following reaction (s) represent self
ionisation of water
H+ + OH–
(a)
H2O
(b)
2H2O
H3O+ + OH–
(c)
5H2O
H9O4+ + OH–
(d)
none of these
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 13
8.
9.
10.
11.
Which of the following will suppress the ionization
of phthalic acid in an aqueous solution ?
(a)
KCl
(b)
H2SO4
(c)
HNO3
(d)
NaOH
Which does not react with NaOH or which is not
acid salt :
13.
(c)
[Ag ] = 7.6 × 10 mol/L
(d)
[IO3–] = 3.125 × 10–10 mol/L
A solution contains 0.10 M H2S and 0.3 M HCl.
Given that for H 2 S K 1 = 1 × 10 –7 and
K2 = 1.3 × 10–13. Which are correct
(a)
[HS] = 3.33 × 10–8M
NaH2PO2
(b)
Na2HPO3
(b)
[HS] = 3 × 10–8M
(c)
Na2HPO2
(d)
NaHCO3
(c)
[S2–] = 1.44 × 10–10M
(d)
[S2–] = 13 × 10–13 M
The dissociation constant of a weak acid HA is
4.9 × 10–8. Then which are correct
(a)
percentage ionisation is 7 × 10–4
(b)
percentage ionisation is 7 × 10–2
(c)
concentration of [OH–] is
1.43 × 10–10 mole/litre
(d)
concentration of [H+] is
49 × 10–5 mole/litre
16.
–log (1.8 × 10–5) = 4.74
log(0.9230) = 0.04
(a)
The hydrolysis constant is 1.5 × 10–11
(b)
The degree of hydrolysis is 6.05 × 10–7
(c)
Concentration of [H+] is 2.583 × 10–8
The hydrolysis constant is 3 × 10
A buffer solution is prepared by mixing 6g of acetic
acid and 13.6 g of sodium acetage
(CH3COONa.3H2O) and making the total volume
to 250ml. Which are correct.
Given that Ka = 1.8 × 10–5
Given that the 0.01 M solution of KF and
Ka(HF) = 6.6 × 154. Then which are correct
(a)
pH of the solution is 4.74
(b)
pH of the solution is 5.74
(c)
pH change on addition of 1 ml of 1m-HCl
is 0.04
(d)
pH change on addition of 1 ml of 1M-HCl
is zero
–11
The solubility of CaF 2 in water at 18 0 C is
2.04 × 10–4 mole/litre. Which are correct
(a)
Ksp is 8.48 × 10–12
(b)
Ksp is 3.4 × 10–11
(c)
For precipitation ionic product > Ksp
(d)
For precipitation ionic product < Ksp
17.
The dissociation constant for NH4OH is 4 × 10–5 and
that of CH3COOH is 1.8 × 10–5. Then which are
correct.
log(2) = 0.3010, log(3) = 0.4771
Solid AgNO3 in added to a solution which is 0.1M
in Cl– and 0.1 M in CrO42–. If Ksp values for AgCl
and Ag2CrO4 are 1.7 × 10–12 respectively. Then
which are correct
18.
(a)
Degree of hydrolysis is 63%
(b)
Degree of hydrolysis is 37%
(c)
pH is 11.57
(d)
pH is 7.17
Freshly prepared Al(OH)3 and Mg(OH)2 are stirred
vigorously in a buffer solution containing 0.25 mol/
litre of NH4Cl and 0.05 moles/litre of NH 4OH.
Which are correct
Given that, KbNH4OH = 1.8 × 105
(a)
If AgCl precipitate then [Ag+] is
1.7 × 10–9 M
(b)
If Ag2CrO4 precipitate then [Ag+] is
4.34 × 10–6 M
KspMg(OH)2 = 6.0 × 10–10
(a)
[OH–] = 3.6 × 10–5
If Ag2CrO4 precipitate then [Cl–] is
3.9 × 10–5 M
(b)
[OH–] = 0.36 × 10–5
(c)
[Mg2+] = 46.3
(d)
[Mg2+] = 4.63
(c)
(d)
14.
–4
(a)
(d)
12.
15.
+
–
Al(OH)3 = 6.0 × 10–23
–5
If precipitate then [Cl ] is 2.87 × 10 M
0.01 mole of AgNO 3 is added to one litre of a
solution which is 0.1 M in Na2CrO4 and 0.005 M in
NaIO3. Then which are correct.
Given that
Ksp (Ag2CrO4) = 10–8
Ksp (AgIO3) = 10–13
(a)
[Ag+] = 3.2 × 10–4 mol/L
(b)
[CrO42–] = 9.7 × 10–3 mol/L
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 14
Assertion-Reason Type
2.
Each question contains STATEMENT-1 (Assertion)
and STATEMENT-2 (Reason). Each question has
4 choices (A), (B), (C) and (D) out of which ONLY
ONE is correct.
1.
(A)
Statement-1 is True, Statement-2 is True;
Statement-2 is a correct explanation
for Statement-1
(B)
Statement-1 is True, Statement-2 is True;
Statement-2 is NOT a correct
explanation for Statement-1
STATEMENT-1 : M(OH)X has Ksp 4 × 10–12 and
solubility 10–14M. The value of x may be 3.
STATEMENT-2 : Ksp as the product of ionic
concentrations with suitable powers of a sparingly
soluble in its saturated solution.
3.
STATEMENT-1 : Ka for formic acid and acetic acid
are 2.1 × 10 –4 and 1.1 × 10 –5 respectively. The
relative strength of acid is 4.36 : 1.
STATEMENT-2 : Relative strength of strong acid
is K1/K2
4.
(C)
Statement-1 is True, Statement-2 is False
(D)
Statement-1 is False, Statement-2 is True
STATEMENT-1 : Ka for butric acid is 2 × 10–5. The
pH of concentration of 0.2 M aqueous. Solution of
sodium butyrate is 9.5
STATEMENT-2 : pH in weak acid and strong base
STATEMENT-1 : 10 ml of 10–6 M HCl solution is
mixed with 90 ml H2O. pH will change nearly by
0.7 unit.
is

1 kw
p  p ka  log C
2

STATEMENT-2 : CH3COONH4 and NH4CN is
example of salt of weak acid and weak base in
water.
(Answers) EXCERCISE BASED ON NEW PATTERN
COMPREHENSION TYPE
1.
b
2.
b
3.
a
2.
4.
b
5.
a
6.
b
[A-q; B-p; C-r; D-p]
3.
[A-p; B-q; C-r; D-s]
MATRIX-MATCH TYPE
1.
[A-q; B-s; C-r; D-p]
MULTIPLE CORRECT CHOICE TYPE
1.
a, c
2.
a, c
3.
a, b
4.
c, d
5.
c, d
6.
a, c
7.
a, b, c
8.
b, c
9.
a, b, c
10.
b, c
11.
a, c
12.
b, c
13.
a, b, c
14.
a, d
15.
a, c
16.
a, c
17.
b, d
18.
b, c
3.
B
4.
D
ASSERTION-REASON TYPE
1.
B
2.
Einstein Classes,
D
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 15
INITIAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
(a)
Find the pH of 10–7 M solution of HCl at
298 K
(b)
Calculate the pH of 10–6 m CH3COOH.
ka(CH3COOH) = 1.8 × 10–5.
Liquid ammonia ionizes to a slight extent. At – 500C,
its ion product is
4.
5.
How many moles of sodium hydroxide can be added
to 1.00 L of a solution 0.100 M in NH3 and 0.100 M
in NH4Cl without changing the pOH by more than
1.00 unit ? Assume no change in volume.pK b
(NH3) = 4.75.
12.
The acid ionization (hydrolysis) constant of Zn2+ is
1.0 × 10 –9. (a) Calculate the pH of a 0.0010 M
solution of ZnCl2. (b) What is the basic dissociation
constant of Zn(OH)+ ?
13.
Calculate the ammonia concentration of a solution
prepared by dissolving 0.150 mol of NH4C2H3O2 in
sufficient water to make 1.0 L of solution.
Ka = Kb = 1.8 × 10–5.
14.
K1 and K2 for oxalic acid, H2C2O4, are 5.6 × 10–2
and 5.4 × 10–5. What is [OH—] in a 0.0050 M
solution of Na2C2O4 ?
15.
Bromphenol blue is an indicator with a Ka value of
5.84 × 10–5. What percentage of this indicator is in
its basic form at a pH of 4.84 ? Ka(HBb) = 5.84 ×
10–5 (HBb  representing bromophenol blue).
16.
0.15 mole of pyridinium chloride has been added
into 500 cm3 of 0.2 M pyridine solution. Calculate
pH and hydroxyl ion concentration in the resulting
solution assuming no change in volume. (Kb for
pyridine = 1.5 × 10–9 M)
17.
Calculate simultaneous solubility of AgCNS and
AgBr in water. KSP of AgBr = 5 × 10–13 and KSP of
AgCNS = 1 × 10–12
K NH3  [ NH 4 ][ NH 2 ]  1030 .
How many amide ions, NH2—, are present per mm3
of pure liquid ammonia ?
3.
11.
When 0.100 mol of ammonia, NH3, is dissolved in
sufficient water to make 1.00 L of solution, the
solution is found to have a hydroxide ion
concentration of 1.34 × 10–3 M. Calculate Kb for
ammonia.
At 250C, a 0.0100 M ammonia solution is 4.1%
ionized. Calculate (a) the concentration of the OH—
and NH4+ ions (b) the concentration of molecular
ammonia (c) the ionisation constant of aqueous
ammonia (d) [OH—] after 0.0090 mol of NH4Cl is
added to 1.00 L of the above solution (e) [OH—] of a
solution prepared by dissolving 0.010 mol of NH3
and 0.0050 mol of HCl per L.
Find the value of [OH—] in a solution made by
dissolving 0.0050 mol each of ammonia and
pyridine (C5H5N) in enough water to make 200 cm3
of solution. What are the concentrations of
ammonium and pyridinium ions ? Kb(NH3) = 1.8 ×
10–5, Kb (C5H5N) = 1.52 × 10–9.
6.
Calculate [H+] in a solution that is 0.100 M HCOOH
and 0.100 M HOCN. Ka(HCOOH) = 1.8 × 10–4,
Ka(HOCN) = 3.3 × 10–4.
18.
Calculate the amount of (NH4)2SO4 in grams which
must be added to 500 ml of 0.200 M NH3 to yield a
solution with pH 9.35 (Kb for NH3 = 1.78 × 10–5)
7.
What is the pH of a solution containing 0.010 mol
HCl/L ? Calculate the change in pH if 0.020 mol
NaC 2H 3O 2 is added to 1.0 L of this solution.
Ka(C2H4O2) = 1.8 × 10–5.
19.
CH3COOH (50 mL, 0.1 M) is titrated against 0.1 m
NaOH solution. Calculate the pH at the addition of
0 mL, 20 mL, 25 mL, 40 mL, 50 mL and 60 mL of
NaOH.
8.
Calculate the concentration of H 3O + ion in a
solution prepared by dissolving 0.20 mol of
HC2H3O2 and 0.050 mol of Mg(C2H3O2)2 in enough
water to make 1.0 L of solution.
KaCH3COOH = 1.8 × 10–5.
20.
Calculate the percentage of hydrolysis of 0.003 M
aqueous of NaOCN.
(Ka for HOCN = 3.33 × 10–4 M)
21.
When 25 ml of 0.20 M acetic acid is neutralised with
25 ml 0.20 M NaOH, the resulting solution is slightly
alkaline, calculate the pH of the resulting solution
(Ka for acetic acid = 1.8 × 10–5).
9.
Calculate the pH of a solution made by mixing 50.0
mL of 0.200 M NH4Cl and 75.0 mL of 0.100 M
NaOH. Kb(NH3) = 1.8 × 10–5.
22.
10.
Determine the pH of a solution after 0.10 mol of
NaOH is added to 1.00 L of a solution containing
0.15 M HC2H3O2 and 0.20 M NaC2H3O2. Assume
no change in volume. Ka = 1.8 × 10–5.
What volume of 0.10 M sodium formate solution
should be added to 50 ml of 0.05 M formic acid to
produce a buffer solution of pH 4.0 ? pK a for
formic acid is 3.80.
23.
Calculate the change in pH of one litre of buffer
solution containing 0.10 mole each of NH 3 and
NH4Cl upon addition of
Einstein Classes,
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CIE – 16
(a)
0.02 mole of HCl
(b)
0.02 mole of NaOH
30.
Given Kb of NH3 = 1.8 × 10–5
24.
25.
The solubility product of PbBr2 is 8 × 10–5. If the
salt is 80% dissociated in saturated solution, find
the solubility of the salt. Atomic weights of Pb and
Br are 208 and 80 a.m.u. respectively.
Given : Ka = 1.8 × 10–5.
31.
Given that Ksp of CaF2 is 1.7 × 10–10, how many
grams of it can be dissolved in (a) one litre 0.001 M
NaF, (b) one litre 0.01 M CaCl2.
26.
The solubility products of BaSO4 and BaCrO4 at
250C are 1 × 10–10 and 2.4 × 10–10 respectively.
Calculate the simultaneous solubility of BaSO4 and
BaCrO4.
27.
One litre of a buffer solution has 0.13833 moles of
acetic acid and 0.1951 moles of sodium acetate.
Calculate :
29.
Calculate at 250C the hydrolysis constant and its
degree of hydrolysis in 0.10 M solution of :
(a) sodium acetate, and (b) sodium carbonate. What
will be the pH values ?
(Given : Ka(HAc) = 1.8 × 10–5 M and Ka(HCO3–) =
4.7 × 10–11 M).
32.
It is found that 0.1 M solution of three sodium salts
NaX, NaY and NaZ have pHs 7.0, 9.0 and 11.0,
respectively. Arrange the acids HX, HY and HZ in
order of increasing strength. Where possible,
calculate the ionization constants of the acids.
33.
(a)
pH of solution
Calculate the hydrolysis constant and its degree of
hydrolysis in 10–2 M solution of NH4Cl. What will
be its pH value ?
(b)
the change in pH on addition of 0.70 ml
of 1 M HCl
Given : Kb (NH4OH) = 1.8 × 10–5 M.
(c)
the change in pH on addition of 0.70 ml
of 1 M NaOH
34.
(a) Determine the pH of a 0.10 M solution of
pyridine, C5H5N. (b) Predict the effect of addition
of pyridinium ion,C5H5NH+, on the position of the
equilibrium. Will the pH be raised or lowered ? (c)
Calculate the pH of 1.00 L of 0.10 M pyridine
solution to which 0.15 mol of pyridinium chloride,
C5H5NH+Cl—, has been added, assuming no change
in volume. Kb[C5H5N] = 1.52 × 10–9.
Calculate the value of K h,  h and pH of the
following solutions at 250C.
(i)
Neglect volume change, Ka = 1.8 × 10–5
28.
What is the pH of a 1.0 M solution of acetic acid ?
To what volume must one litre of this solution be
diluted so that the pH of the resulting solution will
be twice the original value ?
0.1 M ammonium acetate
Ka = Kb = 1.8 × 10–5 M
(ii)
0.1 M anilinium acetate
Ka = 1.8 × 10–5 M; Kb = 4.6 × 10–10 M
(iii)
0.1 M ammonium carbonate
K a1 = 4.5 × 10–7 M; K a 2 = 4.7 × 10–111 M
35.
Calculate the pH of an aqueous solution of 1.0 M
ammonium formate assuming complete
dissociation. (pka of formic acid = 3.8 and pkb of
ammonia = 4.8)
Calculate the pH of a solution made by adding 0.001
mole of NaOH to 100 cm3 of a solution which is
0.50 M in CH3COOH and 0.50 M in CH3COONa.
FINAL STEP EXERCISE
(SUBJECTIVE)
1.
2.
The self-ionization constant for pure formic acid,
K = [HCOOH2+][HCOO—], has been estimated as
10–6 at room temperature. What percentage of
formic acid molecules in pure formic acid, HCOOH,
are converted to formate ion ? The density of
formic acid is 1.22 g/cm3.
When 0.100 mol of NH3 is dissolved in sufficient
water to make 1.00 L of solution, the solution is
found to have a hydroxide ion concentration of 1.33
× 10–3 M. (a) What is the pH of the solution ? (b)
What will be the pH of the solution after 0.100 mol
of NaOH is added to it ? (Assume no change in
volume). (c) Calculate Kb for ammonia. (d) How
will NaOH added to the solution affect the extent
of dissociation of ammonia ?
Einstein Classes,
3.
Calculate the pH at which Mg(OH)2 begins to
precipitate from a solution containing 0.10 M Mg2+
ions. KSP for Mg(OH)2 = 1.0 × 10–11
4.
A solution contains 0.10 M H2S and 0.3 M HCl.
Calculate the concentration of S2– and HS— ins in
the solution. For H2S
K a1  1.0 107
5.
K a 2  1.3 1013
Calcium lactate is a salt of a weak organic acid and
represented as Ca(Lac)2. A saturated solution of
Ca(Lac)2 contains 0.13 mol of this salt in 0.50 litre
solution. The pOH of this solution is 5.60.
Assuming a complete dissociation of the salt,
calculate Ka of lactic acid.
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road
New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 17
6.
Calculate the amount of NH3 and NH4Cl required
to prepare one litre buffer solution of pH 9.0, when
total concentration of buffering reagents is 0.6 mol
L–1. pKb for NH3 = 4.7.
7.
Calculate the solubility of AgCN is a buffer
solution of pH = 3.0 Ksp (AgCN) = 1.2 × 10 –16,
Ka(HCN) = 4.8 × 10–10. There is no CN— or Ag+ ion
in the buffer previously.
8.
A solution of weak base BOH was titrated with HCl.
The pH of the solution was found to be 10.04 and
9.14 after the adding of 5 ml and 20 ml of the acid
respectively. Find the dissociation constant of the
base.
9.
A sample of solid AgCl was treated with 5.00 ml of
1.5 M Na 2CO 3 solution to give Ag 2CO 3. The
remaining solution contained 0.0026 g of Cl— per
litre. Calculate the solubility product of AgCl
(Ksp Ag2CO3 = 8.2 × 10–12).
3+
2+
K for the reaction of Fe to give Fe(OH) and H is
6.5 × 10–3. What is the maximum pH value which
could be used so that at least 95% of the total iron
(III) in a dilute solution exists as Fe3+ ?
11.
At what minimum pH will 10–3 mol of Al(OH)3 go
into solution (V = 1L) as Al(OH)4— and at what
maximum pH it will dissolve as Al3+ ?
Given :
3+
Al(OH)
Al + 4OH
–34
Keq = 1.3 × 10
13.
14.
15.
Calculate the [NH4+] ion concentration needed to
prevent Mg(OH)2 from precipitating in a litre of
solution which contains 0.01 mole NH3 and 0.001
mole of Mg 2+, K b (NH 3 ) = 1.8 × 10 –5 , K sp
(Mg (OH)2) = 1.12 × 10–11.
The pH of blood is maintained by a proper balance
of H 2CO 3 and NaHCO 3 concentrations. What
volume of 5M NaHCO3 solution should be mixed
with a 10 ml of sample of blood which is 2 M in
H 2 CO 3 in order to maintain a pH of 7.4 ?
Ka(H2CO3) = 7.8 × 10–7.
A certain acid – base indicator is red in acid
solution and blue in basic solution. At pH = 5, 75%
of the indicator is present in the solution in its blue
form. Calculate Ka for the indicator and pH range
over which the indicator changes from 90%
red – 10% blue to 90% blue – 10% red.
A weak base BOH was titrated against a strong acid.
The pH at 1/4th equivalence point was 9.24. Enoght
strong base (6 meq.) was now added to consume
the salt completely. The total volume was 50 ml.
Find the pH at this point.
Einstein Classes,
17.
An acid type indicator, HIn differs in colour from
its conjugate base (In — ). The human eye is
sensitive to colour differences only when the ratio
(In—)/(HIn) is greater than 10 or smaller than 0.1.
What should be the minimum change in the pH of
the solution to observe a complete colour change
(Ka = 1.0 × 10–5).
18.
A solution of weak acid HA was titrated with base
NaOH. The equivalence point was reached when
36.12 ml of 0.1 M NaOH has been added. Now 18.06
ml of 0.1 M HCl were added to titrated solution,
the pH was found to be 4.92. What will be the pH
of the solution obtained by mixing 10 ml of 0.2 M
NaOH and 10 ml of 0.2 M HA.
19.
A solution contains a mixture of Ag+ (0.10 M) and
Hg2+2 (0.10 M) which are to be separated by selective precipitation. Calculate the maximum concentration of iodide ion at which one of them gets precipitated almost completely. What % of that metal
ion is precipitated ? (Ksp of AgI = 8.5 × 10–17 and Ksp
of Hg2I2 = 2.5 × 10–26)
20.
The average concentrations of SO 2 in the
atmosphere over a city on a certain day is
10 ppm, when the average temperature is 298 K.
Given that the solubility of SO2 in water at 298 K is
1.3653 moles litre–1 and the pKa of H2SO3 is 1.92,
estimate the pH of rain on that day.
21.
The solubility of Pb(OH)2 in water is 6.7 × 10–6M.
—
Al(OH)3
Al3+ + 3OH—
Ksp = 5.0 × 10–33
12.
Solid AgNO 3 is gradually added to a solution
containing equimolar concentration of Cl— and I—.
If Ksp of AgCl and AgI are 1.7 × 10–10 and 1.5 × 10–
16
respectively, which one will precipitate first ? Also
find the relative concentration of I— and Cl— just
before the precipitation of AgCl.
+
10.
—
4
16.
Calculate the solubility of Pb (OH)2 in a buffer
solution of pH = 8.
22.
For the reaction
[Ag (CN)2]—
Ag+ + 2CN—,
the equilibrium constant at 25 0C is 4.0 × 10 –19.
Calculate the silver ion concentration in a solution
which was originally 0.10 molar in KCN and 0.03
molar in AgNO3.
23.
An aqueous solution of a metal bromide MBr2
(0.05M) is saturated with H 2 S. What is the
minimum pH which MS will precipitate ?
24.
The solubility product (Ksp) of Ca(OH)2 at 250C is
4.52 × 10 –5. A 500 ml of saturated solution of
Ca(OH)2 is mixed with equal volume of 0.4 M
NaOH. How much Ca(OH) 2 in milligrams is
precipitated ?
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New Delhi – 110 018, Ph. : 9312629035, 8527112111
CIE – 18
ANSWERS SUBJECTIVE (INITIAL STEP EXERCISE)
1.
3.
5.
8.
11.
13.
15.
16.
18.
20.
22.
24.
25.
26.
27.
28.
31.
32.
33.
(a)
6.79
(b)
6.01
2.
600 ions/mm3
1.81 × 10–5
4.
(a) 0.00041 (b) 0.0096 (c)1.8 × 10–5 (d) 2.0 × 10–5 (e) 1.8 × 10–5
5.7 × 10–8
6.
7.1 × 10–3
7.
2.74
–5
3.6 × 10 M
9.
pH = 9.73
10.
pH = 5.5
0.082 mol of NaOH can be added
12.
(a) 6.0 (b) 1.0 × 10–5
–4
[NH3] = 8.3 × 10
14.
9.6 × 10–7 M
80%
5
17.
Solubility of AgCNS = 8.16 × 10–7 mol lt.–1
5.28 gm
19.
(i)
2.85
(ii)
4.0969
(iii) 4.5229
(iv)
4.699
(v)
5.3011
(vi)
8.699
(vii) 11.9586
0.01 %
21.
8.37
39.625 ml
23.
(a)
pH decrease by 0.1761
(b)
pH increase by 0.1761
12.486 g l–1
(a)
1.326 × 10–2 g,
(b)
5.08 × 10–3 g
5.423 × 10–6, 1.3 × 10–5 M
(a)
4.894
(b)
0.0038 less
(c)
0.0038 more
(a)
9.09
(b)
5.01
(a)
0.55 × 10–9, 7.44 × 10–5, 8.9
(b)
2.127 × 10–4, 0.046, 11.66
HX > HY > HZ, Ka(HY) = 10–5, Ka(HZ) = 10–9
5.6 × 10–10, 5.63
34.
(i)
0.3086 × 10–4, 0.553 × 10–2, 7
(ii)
1.208, 0.52, 4.71
(iii) 11.82, 0.775, 9.79
35.
4.76
ANSWERS SUBJECTIVE (FINAL STEP EXERCISE)
1.
2.
3.
5.
6.
7.
8.
11.
14.
0.004 %
(a) 11.12 (b) 13 (c) 1.79 × 10–5 (d) The OH— of the NaOH represses the dissociation of
NH3
9
4.
[HS—] = 3.33 × 10–8 M; [S2–] = 1.44 × 10–20 M
9, 10–5 M
NH3 = 0.2 moles, NH4Cl = 0.4 moles
1.58 × 10–5 M each
1.8 × 10–5
9.
1.71 × 10–10
10.
0.908
9.415, 4.233
12.
1.7 × 10–3
13.
78.37 ml
Ka = 3 × 10–5, 3.56 to 5.47
15.
11.22
16.
AgI;
17.
19.
Minimum change in pH = 2
5 × 10–13, 99.83%
I   8.8 10
Cl 

7

Einstein Classes,
18.
20.
8.96
4.865
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