Solved Ionic Equilibrium By NKB.CO.IN
1
Multiple Choice Questions
1.
The conjugate base of HCO3 is
(A) H2CO3
(B) CO32 
(C) CO2
(D) H2O
Sol.
(B). HCO3 will release one H+ to form CO32 .
2.
Stronger lewis acid is
(A) Na+
(C) Rb+
(B) K+
(D) Cs+
Sol.
(A). The species which can accept electrons are called lewis acid.
3.
The correct order of conjugate base strength is
(A) H 3O   H 2O  OH   O 2 
(B) O 2   OH   H 2O  H 3O 
(C) H 2O  H 3O   O 2   OH 
(D) H 3O   H 2O  O 2   OH 
Sol.
(B).
4.
Amphoteric behaviour is shown by
(A) H2CO3 and Al2O3
(C) HCO3– and H3O+
Sol.
(B).
5.
The pH of an acidic buffer mixture is
(A)  7
(C)  7
Sol.
(D). pH  pK a  log
6.
The pH of aqueous solution of sodium acetate is
(A) 7
(B) < 7
(C) > 7
(D) None of these
Sol.
(C). CH3COONa  H2O 
CH3COOH  NaOH
(B) HCO3– and H2O
(D) H2CO3 and H2O
(B)  7
(D) depends on K a of the acid
[salt]
[acid]
The solution will be basic due to presence of strong base.
7.
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) 4
(B) 7
(C) 10
(D) 14
Sol.
(A). pOH  pK b  log
8.
[X  ]
 10. Hence pH = 14 –10 = 4
[HX]
In which of the following cases the acid strength is highest
(A) Ka = 10–6
(B) pKa = 5
(C) pKb = 10
(D) Kb = 10–11
Sol.
(D). Smaller the Kb value lesser is the basicity and more is the acidity.
9.
Which is acidic salt?
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Solved Ionic Equilibrium By NKB.CO.IN
(A) KHF2
(C) Na2HPO4
2
(B) NaH2PO4
(D) All of these
Sol.
(D). All contain H+ ions.
10.
The number of H+ ions present in 1ml of a solution whose pH is 13
(A) 6 × 107
(B) 6 × 1010
23
(C) 6 × 10
(D) 6 × 1022
Sol.
(A). pH = –log [H+], hence [H+] = 10–13 in one litre and 10–16 in one ml.
Hence, number of H+ ions  6.023 1023  1016  6.02  107
11.
The pH of 0.001 M HCl is
(A) 1
(C) 5
(B) 3
(D) 10
Sol.
(B). 10–3 = [H+];  pH = 3
12.
Among the following statements which is true?
(A) K w is independent of temperature
(B) At 65°C pH of water is not equal to 7
(C) pK w increases with temperature
(D) pH + pOH = 14, at all temperatures
Sol.
(B). pH of water = 7 at 25°C. With increase in temperature Kw value increases (i.e. pKw decreases) and pH of water
becomes lesser than 7.
13.
The pKa of HCN is 9.30. The pH of a solution prepared by mixing 2.5 mole of KCN & 2.5 mole HCN in water and
making the total volume 500 ml is
(A) 9.30
(B) 7.30
(C) 10.30
(D) 8.30
Sol.
(A). pH  pK a  log
14.
In which case pH will not change on dilution
(A) 0.01M CH3COONa + 0.01M CH3COOH buffer
(B) 0.01M CH3COONH4
(C) 0.01M NaH2PO4
(D) in all the above cases
Sol.
(D). Mixture of sodium acetate and acetic acid is a buffer of pH value equal to pKa so its buffer capacity is
maximum and hence its pH will not change significantly while CH3COONH4 is a salt of weak acid CH3COOH and
weak base NH4OH whose magnitude of Ka and Kb are equal. So its pH does not depend upon concentration.
Further more, NaH2PO4 is, in fact, a single solute buffer.
The hydroxide having the lowest value of K sp at 25 C is
15.
[KCN]
 9.30
[HCN]
(A) Mg( OH )2
(B) Ca( OH )2
(C) Ba( OH )2
(D) Be( OH )2
Sol.
(D). Be(OH)2 is the weakest base in this group hence it has least solubility.
16.
If 0.1 mol of a salt is added to 1.0 L of water, which of these salts is expected to produce the most acidic solution?
(A) NaC2H3O2
(B) NH4NO3
(C) CuSO4
(D) AlCl3
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Solved Ionic Equilibrium By NKB.CO.IN
Sol.
3
(D). AlCl3  H2O 
3HCl  Al(OH)3
Since more acid is produced hence, more acidic solution is obtained.
17.
The sweetener, saccharin, is a weak monoprotic acid with Ka = 2.1  10–12. Calculate the H3O+ concentration in a
solution that contains 1.0  10–2 mole of saccharin in 1.00 L of otherwise pure water.
(A) 1.4  10–7
(B) 1.8  10–7
–12
(C) 2.1  10
(D) 2.1  10–14
Sol.
(A). [H+] =
=
Ka  C
2.1  1012  1  102 = 1.4  10–7
18.
The pH of 0.5 M solution of NaHCO3 is almost equal to
(For H2CO3; K1 = 2  10–7 & K2 = 4  10–11)
(A) 10.52
(B) 9.8
(C) 8.55
(D) 7.2
Sol.
(B). pH =
19.
Solubility of Ag is decreased in
(A) HCl (B)
(C) NH3
pK a1  pK a2
2
.
CH3COOH
(D) CHCl3
Sol.
(A). Common ion effect.
20.
Dissociation constant of two acids HA & HB are respectively 4  10–10 &
higher.
(A) HA
(B) HB
(C) Both same
(D) Can’t say
Sol.
(A). Lesser the acidity more is pH.
21.
o
The H ionisation
for HCN and CH 3COOH are 45.2 and 2.1 kJ/mol respectively which of the following is correct
(A) pK aHCN  pK aCH COOH
(B) pK aHCl  pK aCH COOH
3
(C) pK a
HCN 
Sol.
1.8  10–5M whose pH value will be
3
 pK aCH COOH
(D) None of these
3
(A).More is heat of ionisation of acid, more is stability of acid or lesser is dissociation, or
Ka( CH COOH)  Ka(HCN) thus, pKa( HCN )  pKa( CH COOH)
3
3
–3
22.
At 25°C the pH of 10 M NaOH is 11. Now let 10–3M solution of NaOH is warmed to 45°C without changing the
volume of solution and hence molarity of solution. Choose the correct statement regarding pH and pOH of the
solution after heating.
(A) pH will decrease while pOH will remain constant
(B) pH will increase while pOH will remain constant
(C) pH will increase while pOH will decrease
(D) Both pH and pOH will decrease
Sol.
(A). Kw will increase and hence, pH + pOH will decrease. Molarity of NaOH solution remaining constant, pOH will
remain constant but pH will decrease.
23.
A buffer solution can be prepared from a mixture of
(1) Sodium acetate and acetic acid in water
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Solved Ionic Equilibrium By NKB.CO.IN
4
(2) Sodium acetate and hydrochloric acid in water
(3) Ammonia and ammonium chloride in water
(4) Ammonia and sodium hydroxide in water
(A) 1, 3, 4
(B) 1, 3
(C) 1, 2, 4
(D) 2, 4
Sol.
(B). Buffer is a solution of a weak acid or base and its salt with strong base or strong acid respectively.
24.
The solubility of CH3COOAg would be least amongst the following solvent in
(A) a solution of pH = 4
(B) pure water
(C) neutral solution of pH=7
(D) a solution of pH=9
Sol.
(D). The solution of pH = 9 is basic and it will least dissolve CH3COOAg as CH3COO– produced is also basic.
25.
The precipitate of CaF2(Ksp = 1.710–10) is obtained when equal volumes of the following are mixed
(A) 10–4 M Ca2+ + 10–4 M F–
(B) 10–2 M Ca2+ + 10–3 M F–
(C) 10–5 M Ca2+ + 10–3 M F–
(D) 10–3 M Ca2+ + 10–5 M F–
Sol.
(B). [Ca2+][F–]2
2
102  V  103  V 
= 


 2V   2V 
1
=  108 , which is more than 1.710–10
8
26.
A weak monobasic (0.1 M) has a pH of 3 at a particular temperature (25°C). When this acid is neutralised by strong
base (NaOH), what is the value of equilibrium constant at equivalent point at 25°C.
(A) 10 5
(B) 10 4
(C) 10 7
(D) 10 14
Sol.
(B). Let the weak acid be HA
HA
0.1 (1   )
H
0.1 

A
0.1 
0.1   103
  1  104
0.1  0.1
 Ka 
 0.12 = 109
0.1(1   )
At equivalence point,
A   H2O
Kh 
27.
HA  OH
K w 1014

 105
Ka
109
The pH of a saturated solution of Ca(OH)2 in water at 25°C is 12.67. What is the concentration of OH– in the
saturated solution, in moles of OH– per litre?
(A) 10–14/12.67
(B) 10–14 10–12.67
(C) 10–14/10–12.67
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(D)
10 14
10 12.67
Solved Ionic Equilibrium By NKB.CO.IN
5
[1014 ]
[OH]
Sol.
(D). pH   log[H ]   log
28.
The molarity of pure water is
(A) 55.6
(C) 100
(B) 50
(D) 18
1000
 55.6
18
Sol.
(A). Molarity of pure water 
29.
The pH of a buffer is 4.325. When 0.01 mole of NaOH is added to 1 litre of it, the pH changes to 4.625. It buffer
capacity is
(A) 30
(B) 0.0307
(C) 2.3
(D) 1
Sol.
(B). Buffer capacity 
30.
A certain weak acid has a dissociation constant of 1  10–4. The equilibrium constant for its reaction with a strong
base is
(A) 1  10–4
(B) 1  10–10
10
(C) 1  10
(D) 1  10–14
Sol.
(C). Equilibrium constant = Dissociation constant in this case.
31.
Separate solution of four sodium salts NaW, NaX, NaY and NaZ had pH 7.0, 9.0, 10.0 and 11.0 respectively. When
each solution was 0.1M, the strongest acid is
(A) HW
(B) HX
(C) HY
(D) HZ
Sol.
(A). As pH of NaW is least, it will produce weakest acid on hydrolysis.
32.
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) Unionised in the small intestine and in the stomach
(B) Completely ionized in the small intestine and in the stomach.
(C) Ionised in the stomach and almost unionized in the small intestine
(D) Ionised in the small intestine and almost unionsied in the stomach
Sol.
(D). Since pH in small intestine is more, the conditions are basic and aspirin will dissociate.
33.
Match list–I with listII and select the correct answer using the codes given below
Number of moles of acid or base added in one litre
Change in pH
ListI
(a)
(b)
(c)
(d)
ListII
CH3COONa
NH4Cl
Bi2S3
CdS
Codes
(a)
(b)
(1)
(2)
(3)
(4)
(c)
(d)
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Strong electrolyte with pH > 7
Strong electrolyte with the pH < 7
Weak electrolyte with Ksp = s2
Weak electrolyte with Ksp = 108s5
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(A)
(B)
(C)
(D)
2
1
1
1
3
2
3
3
1
4
2
4
6
4
3
4
2
Sol.
(B). Ksp of CdS = s2 and Ksp of Bi2S3 = 108s5
34.
What is the total volume solution (in ml), when 1 M NaOH is required to be added in 100 ml of 1M CH 3COOH (Ka =
10–5) solution so that its pH becomes 6
(A) 190.91
(B) 90.9
(C) 100
(D) 135.5
Sol.
(A). This is a acidic buffer solution, let the volume of NaOH added = V ml
V
 pH  pK a  log
100  V
V
6  5  log
100  V
V

 10
100  V
1000
V
 90.909
11
 VT  VNaOH  VCH3COOH
= 90.909 + 100
= 190.91
35.
Addition of hydrochloric acid to a saturated solution of cadmium hydroxide Cd(OH)2, Ksp = 2.5 x 10–14) in water
would cause
(A) the solubility of cadmium hydroxide to decrease
(B) the OH– concentration to decrease and the Cd2+ concentration to increase
(C) the concentrations of both Cd2+ and OH– to decrease
(D) the concentrations of both Cd2+ and OH– to increase
Sol.
(B). H+ from HCl will neutralize OH–. To maintain the following equilibrium constant value of constant [Cd2+] will
have to increase.
Cd(OH)2(s)
36.
Cd2+(aq.) + 2OH–(aq.)
Which of the following best describes what will happen when a solution of AgNO3 is slowly added to a saturated
solution of silver acetate, CH3COOAg, without changing the volume significantly?
CH3COOAg (s)
CH3COO– (aq) + Ag+ (aq)
(A) Some of the solid silver acetate will dissolve
(B) The concentration of acetate ion will increase
(C) Some solid silver acetate will precipitate
(D) The concentrations of acetate ion and silver ion will both increase
Sol.
(C). Due to common ion effect.
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Solved Ionic Equilibrium By NKB.CO.IN
37.
7
Let the solubility of AgCl in H2O, 0.01 M CaCl2, 0.01 M NaCl and 0.05 M AgNO3 be S1, S2, S3 and S4 respectively.
What is the correct relationship between these quantities?
Sol.
(A) S1  S2  S3  S4
(B) S1  S2 = S3  S4
(C) S1  S3  S2  S4
(D) S4  S2  S3  S1
(C). Since the concentration of common ion 
1
so lub ility
Hence, the solubility follows the order
S1  S3  S2  S4
38.
An acid–base indicator has a pH range of 8.3 – 9.9. The indicator will not be a suitable indicator for the titration
between
(B) H2SO4 and NaOH
(C) CH3COOH and NaOH
(D) HCl and NH4OH
(A)
(B)
Conductivity
(D). Solution of HCl and NH4OH will have less pH than the indicator range.
Which graph best represents the electrical conductivity behaviour that occurs when an aqueous solution of acetic
acid is titrated with an aqueous solution of NaOH.
Conductivity
Sol.
39.
(A) HCl and NaOH
mL of NaOH
(D)
mL of NaOH
Conductivity
(C)
Conductivity
mL of NaOH
mL of NaOH
Sol.
(C). After equivalence point pH increases sharply as NaOH is a strong base.
40.
Percentage hydrolysis of 0.1 M CH3COONH4 when Ka = Kb = 1.810–5 is
(A) 0.5510–2
(B) 0.55
(C) 7.63
(D) 1.810–2
Sol.
(B). h 
Kw
1014

Ka  K b
(1.8  105 )2
% Hydrolysis  h 100  0.55%
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Solved Ionic Equilibrium By NKB.CO.IN
8
41.
The percentage error in hydronium ion concentration made by neglecting the ionisation of water in 1  10 6 NaOH
is
(A) 1%
(B) 2%
(C) 3%
(D) 4%
Sol.
(A). % error 
42.
A solution is a mixture of 0.1 M NaCl and 0.1 M NaI. The minimum concentration of iodide ion in the solution when
AgCl just starts precipitating is equal to
( K sp of AgCl = 1  10 10 M 2 ; K sp AgI = 4  10 -16 M 2 )
1014
 100  1%
1012
(A) 2  10 7
(C) 4  10 26
Sol.
(B). When AgCl starts precipitating then [Ag+] =
At that time [I ] 
43.
(B) 4  10 7
(D) 4  10 6
1010
 109
0.1
4  1016
 4  107
9
10
N 
30 ml of   CH 3COOH is added in NaOH solution, then at the end point of the reaction, the equilibrium
5 
constant of the reaction at 25°C
(given that K a( CH3COOH )  10 5 ) will be
(A) 10 9
(C)
1
 10 9
5
(B) 10 9
(D) None of these
Ka
Kw
(A).At the end point of the reaction, the equilibrium constant =
44.
K a for HCN is 5  10 10 at 25°C. For maintaining a constant pH = 9, the volume of 5 M KCN solution required to be
added to 10 ml of 2 M HCN solution is
(A) 4 ml
(C) 2 ml
Sol.
(B) 7.95 ml
(D) 9.3 ml
(C). Let the volume of KCN = x ml, then for acidic buffer
[salt]
pH  pK a  log
[acid]
5x
 9  10  log 5  log
20
4
 9  10  log 5  log
x
20
 9  10  log
x
20
 log
1
x
20

 10
x
x = 2 ml
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=
105
 109
1014
Sol.
Solved Ionic Equilibrium By NKB.CO.IN
9
45.
1.0 M solution of a weak base, BOH is only 0.5% ionised. If 2 mL, 1 M solution of BOH is mixed with 30 mL water,
the degree of dissociation of the base in the resulting solution and H+ ion concentration of the solution will be
respectively
(A) 0.02 and 1.25  10–3M
(B) 0.08  3.5  10–3M
–4
(C) 0.04 and 3.0  10 M
(D) 0.02 and 8.0  10–12M
Sol.
(D).  =
=
0.5
= 5  10–3
100
Kb
C
1
C
Dilution is made 16 times so concentration will decrease 16 times and  will increase 4 times.
 = 5  4  10–3 = 2  10–2
1 1
  102 M
[OH] = C = 2  102 
16 8
8  10 14
[H+] =
= 8  1012 M
102
i.e.  
46.
50 mL of a weak acid, HA required 30 mL, 0.2 M NaOH for the end point. During titration, the pH of acid solution is
found to be 5.8 upon addition of
20 mL of the above alkali. The pKa of the weak acid is
(A) 6.3
(B) 5.5
(C) 6.1
(D) None of these
Sol.
(B). pH = pKa + log
5.8 = pKa + log
[salt]
[acid]
20
10
 pKa = 5.5
47.
In a buffer solution consisting of a weak acid and its salt, the ratio of concentration of salt to acid is increased 10
fold then the pH of the solution will
(A) increase by one
(B) increase ten fold
(C) decrease by one
(D) decrease ten fold
Sol.
(A). pH  pK a  log
48.
To a mixture of NH4Cl and NH3 in water, a further amount of NH4Cl is added. The pH of the mixture
(A) Increase
(B) Decrease
(C) Remains same
(D) Be unpredictable
Sol.
(B). pOH  pK a  log
49.
If solubility product for the salts AB5, AB4C, AB3C2 are same then the solubility is maximum for
(A) AB5
(B)
AB4C
(C) AB3C2
(D) All are having same solubility
Sol.
(C). Ksp for (C) is 108s5. Since Ksp < 1, hence s is maximum in this case.
[salt]
[acid]
[salt]
, hence, pOH increases and pH decreases.
[base]
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Solved Ionic Equilibrium By NKB.CO.IN
10
50.
The ionization constant of NH4+ in water is 5.6  10–10 at 25°C. The rate constant for the reaction of NH4+ and OH–
to form NH3 and H2O at 25°C is
3.4  1010 L mol–1 s–1. The rate constant for proton transfer from water to
NH3 is
(A) 6.07  105 s–1
(B) 6.07  1010 s–1
–5 –1
(C) 6.07  10 s
(D) 6.07  1010 s–1
Sol.
(A). K(ionisation) 
51.
What is the pH value at which Mg(OH)2 begins to precipitate from a solution containing 0.2 M Mg+2 ions? Ksp of
Mg(OH)2 is 2  10–13 M3?
(A) 8
(B) 9
(C) 6
(D) 7
Sol.
(A). Ksp = [Ag2+] [OH–]2 hence, [OH–] = 10–6 and [H+] = 10–8
52.
The solubility products of MA, MB, MC and MD are 1.8  10 10 , 4  10 -3 , 4 10 -8 and 6  10 5 respectively. If a
0.01 M solution of MX is added drop wise to a mixture containing A  , B - , C - and D  ions, then the one to be
Kf
Kf
, hence K b 
Kb
K(ionisation)
precipitated first will be
(A) MA
(C) MC
(B) MB
(D) MD
Sol.
(A). Compound having lowest solubility product is precipitated first.
53.
A weak base B( OH )2 has dissociation constant 10 8 . The equilibrium constant for its reaction with excess of
strong acid will be
(A) 10 20
(C) 10 20
Sol.
(A). B(OH)2  2H
K
(B) 10 6
(D) 10 16
B2  2H2O
1
Kh
K h for hydrolysis of B2
=
K 2w
[B(OH)2 ][H ]2
 2

[OH
]

Kb
[B2 ][OH ]2
Kh 
1028
 1020
108
1
 1020
Kh
54.
A student determines the concentration of a solution of NaOH. She pipets a sample of the NaOH solution into a
flask and adds two drops of an indicator. Standardized acid is added from a burette until the indicator changes
colour. Which device may contain residual water without affecting the results of the titration?
(1) Burette; (2) Flask; (3) Pipette
(A) 1 only
(B) 2 only
(C) 1 and 3 only
(D) 2 and 3 only
Sol.
(B). Titrated solution is present in flask.
55.
Which acid, together with its sodium salt, would be best for preparing a buffer with a pH = 4.5?
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Solved Ionic Equilibrium By NKB.CO.IN
(A) HCN–(Ka = 4.9 x 10¯10)
(C) HF–(Ka = 6.8 x 10¯4)
Sol.
56.
11
(B) C3H5O2H–(Ka = 1.3 x 10¯5)
(D) ClC2H2O2H–(Ka = 1.4 x 10¯3)
(B). The acid having pKa  4.5 will be the best.
pH of a saturated solution of M( OH )2 is 13. Hence, K sp of M( OH )2 is
(A) 5  10 6
(C) 5  10 4
(B) 10 3
(D) 2.5  10 4
Sol.
(C). pH = 13 = –log [H+]. Hence, [H+] = 10–13 and [OH–] = 10–1 = 25
Ksp = [M2+] [OH]2 = (s) (2s)2 = 5 × 10–4
57.
The sodium salt of which of the following four weak acids will be most hydrolysed?
(A) HA ( K a  10 8 )
(B) HB (K a  2  10 6 )
(C) HC (K a  3  10 7 )
(D) HD (K a  4  10 10 )
Sol.
(D). The salt which produces weak acid is hydrolysed most.
58.
The pH of 0.5 M solution of NaHCO3 (for H2CO3: k1 = 2.0  107 and
(A) 10.52
(B) 9.8
(C) 8.55
(D) 7.2
Sol.
(C). For the solution of salt containing amphiprotic anion and unhydrolysed cation the pH is,
pk1  pk2 6.7  10.4 17.1
pH =
= 8.55


2
2
2
59.
10 ml of 0.2 M acid is added to 250 ml of a buffer solution with pH = 6.34.
6.32. The buffer capacity of the solution is
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
Sol.
(A). Buffer capacity 
60.
Which of the following solution(s) have pH between 6 and 7?
1. 2  10 6 M NaOH
2. 2  10 6 M HCl
3.
k2 = 4.0  10–11) is almost equal to
The pH of the solution becomes
Number of moles of acid or base added in one litre of buffer
Change in pH
10 8 M HCl
4.
(A) 1, 2
(C) 3, 4
10 13 M NaOH
(B) 2, 3
(D) 2, 3, 4
Sol.
(B).
61.
The ratio of dissociation constant of two weak acids HA and HB is 4.0. At what molar concentration ratio, the two
acids have the same pH.
(A) 2
(B) 0.5
(C) 4
(D) 0.25
Sol.
(D). [H ]  C  Ka  C
For same pH, 4C1 = C2 or
62.
C1
 0.25
C2
A weak acid HA after treatment with 12 ml of 0.1 M strong base BOH has a pH = 5. At the end point, the volume of
the base required is 26.6 ml K a of acid is
(A) 1.8  10 5
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(B) 8.2  10 6
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(C) 8.5  10 6
12
(D) 8.2  10 5
Sol.
(C).
63.
The solubility of CsBr3 (MW = 373) in water is 746 ppm. The solubility product of CsBr3 is, therefore
(A) 1.6  10 11 M 4
(C) 8  10 9 M 3
Sol.
(B) 4  10 6 M 2
(D) 3.2  10 14 M 5
746
 2 mole of CsBr3
373
d = 1) is dissociated with 2 103 mole of CsBr3
(B). 106 g water is associated with 746 g, i.e.
103 g 1 L (
Thus, solubility of CsBr3  2  103 M
Cs  Br3
CsBr(s)
s
s
Ksp  S  4  106 M2
2
64.
The dissociation constants of formic and acetic acids are 1.77  10 4 and 1.77  10 5 respectively. The relative
strengths of the two acids is
(A) 3.18
(C) 6.36
(B) 100
(D) 5.0
Sol.
(A).
65.
H – A and H – B are two weak monobasic acids. If their
K a1
K a2
 9 , then at same pH their molar concentration ratio
will be
(A) 0.111
(C) 4
Sol.
1
2
(D) 0.25
(B)
(A). [H ]  Ka  C
Hence 9C1 = C2
C1 1

C2 9
66.
A solution is a mixture of 0.05 M NaCl and 0.05M NaI. The concentration of iodide ion in the solution when AgCl
just starts precipitating is equal to
Ksp(AgCl) = 1  10–10, Ksp(AgI) = 4  10–16)
(A) 4  10–6
(B) 2  10–8
–7
(C) 2  10
(D) 8  10–15
Sol.
(C). Ksp(AgCl) = [Ag+] [Cl–]
1010
 [Ag  ] 
 2  109
0.05
Ksp (AgI) = [Ag+] [I–]
4  1016
 [I ] 
 2  107
2  109
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Solved Ionic Equilibrium By NKB.CO.IN
67.
13
Liquid NH3 ionises to a slight extent. At –60C, its ionic product
K NH3  [ NH 4 ][ NH 2 ]  10 30
The number of NH2– ions present per c.c. of pure liquid NH3 is
(A) 3105
(B) 5102
5
(C) 610
(D) 6103
Sol.
(C). Ksp = 10–30  [NH4 ] [NH2 ]  x2
 x  1015
Number of [NH2 ] present per cc 
6.02  1023  1015
1000
68.
The concentration of [OH–] left in a solution after mixing 50 ml of 0.2 M ACl2 with 50 ml 0.4 m NaOH. [Ksp of A[OH]2
= 1.2  10–11)
(A) 2.8  10–4 m/L
(B) 1.4  10–4 m/L
(C) 2  10–2 M
(D) 4  10–2 M
Sol.
(A). ACl2 + 2NaOH
A(OH)2 + 2NaCl
10
20
0
0
0
0
10
20
– 2
+ 2
[OH ] [A ] = Ksp
4s3 = Ksp
1
s  3 1.2  1011  1.4  104
4
[OH–] = 2.8  10–4
69.
Initially
After reaction
A student wants to prepare a saturated solution of Ag+ ion. He has got three samples AgCl(Ksp = 10–10), AgBr(Ksp =
10–13) and Ag2CrO4(Ksp = 10–12). Which of the above compound will be used by him using minimum weight of
prepare the saturated solution
(A) AgCl
(B) AgBr
(C) Ag2CO2
(D) All of these
Sol.
(B). AgBr is least soluble.
70.
Solution of ammonium chloride is X due to hydrolysis of X, we get Y, X and Y are
(A) basic NH4+
(B) acidic, NH4+
–
(C) basic, Cl
(D) acidic Cl–
Sol.
(B).
71.
0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between ¼
and ¾ stages of neutralization of acid?
(A) 2 log 3/4
(B) 2 log 1/4
(C) log 1/3
(D) 2 log 3
Sol.
(D). pH1/ 4  pK a  log
1/4
3/4
3/4
1/4
 2 log3
pH3 / 4  pK a  log
 pH1/ 4  pH3/ 4
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Solved Ionic Equilibrium By NKB.CO.IN
14
72.
10 ml of 0.2 M acid is added to 250 ml of a buffer solution with pH = 6.34 and the pH of the solution becomes 6.32.
The buffer capacity of the solution is
(A) 0.1
(B) 0.2
(C) 0.3
(D) 0.4
Sol.
(D). Buffer capacity =
73.
Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation.( pK a of
formic acid = 3.8 and pK b of ammonia = 4.8).
moles of acid or base added per litre
change in pH
0.2  10 1000

250  0.4
Buffer capacity = 1000
6.34  6.32
(A) 3.8
(C) 4.8
(B) 6.5
(D) 8.6
Sol.
(B).
74.
One litre of saturated solution of CaCO3 is evaporated to dryness when 7.0 mg of residue is left. The solubility
product for CaCO3 is
(A) 4.9  10–8
(B) 4.9  10–5
–9
(C) 4.9  10
(D) 4.9  10–7
Sol.
(C). Solubility is 7.0 mg/litre
or, 7.0  10–3 g/litre
7.0  103
moles/litre
or,
100
= 7.0  10–5 moles/litre
Ksp = s2= 49  10–10
= 4.9  10–9
75.
50 litres of 0.1 M HCl are mixed with 50 litres of 0.2 M NaOH. The pOH of the resulting solution is
(A) 12.70
(B) 12.34
(C) 8.7
(D) 4.2
Sol.
76.
(A).
The following reactions are known to occur in the body
CO2 + H2O
H2CO3
H+ + HCO3–
If CO2 escapes from the system
(A) pH will decrease
(B) Hydrogen ion conc. will diminish
(C) H2CO3 conc. will remain unchanged
(D) The forward reaction will be favoured
Sol.
(B). If CO2 escapes the reaction shifts to backward direction.
77.
An acid indicator by HIn (Ka =10–5). The range of change of colour for the indicator is
(A) 3–5
(B) 4–6
(C) 5–7
(D) 6–8
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Solved Ionic Equilibrium By NKB.CO.IN
Sol.
(B). Range of an indicator  pKa  1
78.
At Isoelectric point, zwitter ion has
15
( )
(A) pH = 7
(B) [ R  COO  ]  [ R  NH 3 ]
(C) zwitter ion has +ve change
(D) zwitter ion has –ve change
Sol.
(B).
79.
50% neutralisation of a solution of formic acid (Ka = 2  10–4) with NaOH would result in a solution having a
hydrogen ion concentration of
(A) 2  10–4
(B) 3.7
(C) 2.7
(D) 1.85
Sol.
(A). At isoelectric point positive and negative charge is equal.
80.
The M(OH)4 has Ksp 4  10–42 and x then solubility is
(A) 1
(B) 2
(C) 3
(D) None of these
Sol.
(D). Ksp  s  (4s)4  256s5
Solved Ionic Equilibrium By NKB.CO.IN