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NEET Chemistry MCQ
Chemical And Ionic Equilibrium Mcq
Quiz 3
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Q.1
Which of the following is Bronsted acid as well as a Bronsted base?
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a) BF3
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b) NH3
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c) H2O
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d) Na2CO3
Explanation
H2O is amphoteric and acts both as proton donor and proton acceptorH2O + H2O ⇔ H3O+ + OH- Answer: (c)
Q.2
The pKa of weak acid, HA is 4.The pKb of a weak base, BOH is 4.The pH of an aqueous solution of the corresponding salt, BA will be [ AIEEE 2008]
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a)9.22
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b) 9.58
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c)4.79
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d)7.01
Explanation
Aqueous solution of weak acid with weak base produces salt of weak acid with weak base for which Answer: (d)
Q.3
The molar solubility ( in molL-1) of a parinly soluble salt MX4 is 's" . The corresponding solubility product is Ksp, .'s' is given in terms of Ksp by the relation
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a) s=(256Ksp)1/5
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b) s=(128Ksp)1/4
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c) s=(Ksp/128)1/4
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d)s=(Ksp/256)1/5
Explanation
MX4 ⇔ M4+ + 4X- If solubility is 's' then, Ksp=[M4+] [X-]4 Ksp=(s)(4s)4=256s5 s=(Ksp/256)1/5Answer: (d)
Q.4
The pKa of HCN is 9.The pH of solution prepared by mixing 2.5 moles of KCN and 2.5 moles of HCN in water and making up total to 500 ml is ..[ IIT 1992]
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a)9.30
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b) 7.30
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c)10.30
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d)8.30
Explanation
Applying Since [Salt]=[Acid] pH=pKa=9.3 Answer:(a)
Q.5
An aqueous solution of sodium carbonate has a pH greater than 7 because
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a) Carbonate ions react with H2O
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b) Na+ ions react with water
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c) It contains more hydroxide ions than carbonate ions
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d) It contains more carbonate ions than H2O molecules
Explanation
Answer: (c)
Q.6
The precipitate of CaF2 ( Ksp=1.7 ×10-10) is obtained when equal volumes of the following are mixed
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a)10-4 M Ca2+ + 10-4 M F-
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b) 10-2 M Ca2+ + 10-3 M F-
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c)10-5 M Ca2+ + 10-3 M F-
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d)10-3 M Ca2+ + 10-5 M F-
Explanation
When ionic product is greater than solubility product then precipitation occurs CaF2 ⇔ Ca2+ + 2 F- In option (b) ionic product=10-2 × ( 10-3)2=1 × 10-8 Here ionic product exceeds solubility product, hence precipitation occurs.Answer: (b)
Q.7
The solubility product of salt having general formula MX2 in water is , 4 ×10-The concentration of M2+ ions in the aqueous solution of the salt is [ AIEEE 2005]
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a) 2.0 × 10-6 M
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b) 1.0 × 10-4 M
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c)1.6 × 10-4 M
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d)4.0 × 10-10 M
Explanation
MX2 ⇔ M2+ + 2X- Let solubility is 's' Then Ksp=[M2+][X-]2 4 ×10-12=[s] [2s]2s=10-4Answer: (b)
Q.8
hydrogen ion concentration in mol/L in solution of pH=5.4 will be [ AIEEE 2005]
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a) 3.98 × 108
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b) 3.88 × 106
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c)3.68 ×10-6
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d)3.98 × 10-6
Explanation
pH=- log[H+]Taking antilog we get H+=3.98 ×10-6 Answer:(d)
Q.9
In the hydrolysis of salt of weak acid and weak base, the hydrolysis constant Kh is equal to [ orissa JEE 2003]
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a)
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b)
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c)
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d)
Explanation
Answer: (c)
Q.10
The pKa of a weak acid (HA) is 4.The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is [ AIEEE 2007]
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a)7.0
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b) 4.5
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c)2.5
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d)9.5
Explanation
For acidic buffer When acid is 50% ionised [A-]=[HA] or pH=pKa + log 1 pH=pKa=4.5 pH + pOH=14 ∴ pOH=9.5Answer: (d)
Q.11
1M NaCl and 1M HCl are present in a na aqueous solution . the solution is [ AIEEE 2002]
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a) Not a buffer solution with pH < 7
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b) Not a buffer solution with pH >7
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c)A buffer solution with pH < 7
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d)A buffer solution with pH > 7
Explanation
Strong acid and salt of strong acid and strong base does not form a bufferAnswer: (a)
Q.12
Compound whose 0.1M solution is basic is .. [ IIT 1986]
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a) Ammonium acetate
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b) Ammonium chloride
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c)Ammonium sulphate
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d)Sodium Acetate
Explanation
Sodium acetate on ionistaion gives weak acid and strong base thus resulting solution is basicCH3COONa + H2O ⇔ CH3COOH + NaOH NaOH is trong base while CH3COOH is a weak acid NaOh → Na+ + OH- Answer:(d)
Q.13
The pH of a solution obtained by mixing 50ml of 0.4M HCl with 50ml of 0.2 M NaOH is
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a) - log2
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b) - log2×10-1
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c) 1.0
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d) 2.0
Explanation
Millimoles of NaOH=50 × 0.4=20 Millimoles of NaOH=50 × 0.2=10 Millimoles of HCl left=10, Volume of solution=100 ml ∴ Moles of HCl per litre= concentration of H3O+=0.1 mol/L Now pH=- log[H3O+] pH=- log(0.1)=1 Answer: (c)
Q.14
The solubility of A2 X3 is y mol dm-Is solubility product is ..[ IIT 1997]
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a)6y4
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b) 64y4
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c)36y5
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d)108y5
Explanation
A2 X3 ⇔ 2A3+ + 3X2-Ksp=[A3+]2 + [X2-]3Since solubility=y mol dm-3Ksp=[2y]2 + [3y]3=108y5Answer: (c)
Q.15
The pH at the equivalence point of titration may differ from 7.0 because of
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a) The self ionization of water
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b) Hydrolysis of the salt formed
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c)The indicator used
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d)The concentration of the standard solution
Explanation
Salt hydrolysis affects the pH of the solution. Hence pH at the equivalence point of titration may differ from 7 due to salt hydrolysisAnswer: (b)
Q.16
The pKb for fluoride ion at 25°C is 10.83, the ionization constant of hydrofluoric acid at this temperature is ... [ IIT 1997]
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a) 1.74 × 10-2
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b) 3.52 × 10-5
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c)6.57 ×10-4
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d)5.38 ×10-2
Explanation
Applying K=Kw / Kb pKb=-logKb=10.83, ∴ Kb=1.479 ×10-11 K=10-14 / 1.479 ×10-11 K=6.76 × 10-4 Answer:(c)
Q.17
Hydrogen ion concentration of an aqueous solution is 1 × 10-4M. The solution is diluted with equal volume of water. Hydroxyl ion concentration of the resultant solution in terms of mol dm-3 is [ Karnataka CET 2001]
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a) 1 × 10-8
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b) 1 × 10-6
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c) 2 × 10-10
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d) 0.5× 10-10
Explanation
Applying [H+][OH- ]=10-14 to aqueous solution [OH-]=10-14 / 10-4=10-10 M By diluting the solution with equal volume of water, OH- ion concentration gets halved resulting [OH-]=10-10 /2=0.5 ×10-10 M Answer: (d)
Q.18
Which of the following can not act a a Lewis acid or Bronsted acid?
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a)CCl4
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b) SnCl4
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c)AlCl3
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d)BF3
Explanation
CCl4can not accept electron pair due to the absence of d-orbitalAnswer: (a)
Q.19
Solid Ba(NO3)2 is gradually dissolved in a 1.0 × 10- M, Na2CO3 solution. At what concentration of Ba2+ will a precipitate being to form ? Ksp for BaCO3=5.1 × 10-9) [ AIEEE 2009]
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a) 4.1 × 10-5M
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b) 5.1 × 10-5M
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c) 8.1 × 10-8M
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d) 8.1 × 10-7M
Explanation
Ksp for BaCO3=[ Ba2+][ CO32-] Given, [ CO32-]=1 ×1 ×10-4 ( from Na2CO3) Ksp=5.1 × 10-9 ∴ 5.1 × 10-9=[ Ba2+] ×[10-4 ⇒ [ Ba2+]=5.1 × 10-5 M Thus, when [ Ba2+]=5.1 × 10-5 M, BaCO3 precipitate will begin to formAnswer: (b)
Q.20
pKa value of two acids A and B are 4 andThe strengths of these two acids are related as ..( Karnataka CET 2001]
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a) Acid A is 10 times stronger than acid B
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b) Strength of acid A : strength of acid B=4 :5
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c)The strength of two acids cannot be compared
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d)Acid B is 10 times stronger than acid A
Explanation
pKa=- log KbFor A : pKa=4 ⇒ Ka=10-4For B : pKa=5 ⇒ Ka=10-5 Thus Ka for A is 10 times more than that of B Answer:(a)
Q.21
The pH of 0.1M solution of following increases in the order [ IIT 1999]
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a) NaCl < NH4Cl < NaCN
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b) HCl < NH4Cl < NaCl
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c) NaCN < NH4Cl < NaCl < HCl
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d) HCl < NaCl < NaCn < NH4Cl
Explanation
HCl is strong acid and NaCl is neutral salt. NaCN undergoes anionic hydrolysis and gives basic solution. Hence its pH > 7. Whereas NH4Cl undergoes cation hydrolysis and gives acidic solution, thus its pH < 7Therefore option (b) is correct Answer: (b)
Q.22
For reaction H2(g) + I2(g) ↔ 2HI(g) the equilibrium constant Kp changes with [ IIT 1981]
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a)Total pressure
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b) Catalyst
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c)The amount of H2 and I2 present
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d)Temperature
Explanation
Equilibrium constant depends only on temperatureAnswer: (d)
Q.23
in the equilibrium reaction involving the dissociation of CaCO3(s) ↔ CaO(s) + CO2(s)the equilibrium constant is given by ...[Pb. CET 1991]
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a)
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b)
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c)
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d)
Explanation
Answer: (d)
Q.24
The yield of the product in the reactionA2(g) + 2B(g) ↔ C(g) + Q kJ would be higher at [ MLNR 1988]
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a) High temperature and high pressure
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b) High temperature and low pressure
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c)Low temperature and high pressure
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d)Low temperature and low pressure
Explanation
Reaction is exothermic and in forward direction number of moles of gas molecules decreases Thus at low temperature and high pressure reaction will move in forward direction Answer:(c)
Q.25
The equilibrium constant for the reaction H2(g) + I2(g) ↔ 2HI9g) at a certain temperature isthen the equilibrium constant for the reaction HI9g) ↔ ½ H2(g) + ½I2(g) at the same temperature is ..[Pb. CET1990]
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a) 7
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b) 1/7
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c) 24.5
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d) 98
Explanation
Reaction is reversed and half K=√(1/49)=1/7 Answer: (b)
Q.26
A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal .. [ IIT 1984]
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a)Intermolecular force
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b) Potential energy
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c)kinetic energy
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d)both kinetic and potential energy
Explanation
Since temperature and vapour and liquid is same kinetic energy is sameAnswer: (c)
Q.27
Select the reaction for which the equilibrium constant is written as [MX3]2=K [ MX2]2[X2]
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a) MX3 ↔ MX2 + ½ X2
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b) 2MX3 → 2MX2 + X2
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c)2MX2 + X2 ↔ 2MX3
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d)MX2 + ½X2 ↔ MX3
Explanation
Answer: (c)
Q.28
The equilibrium constant for the reaction H2(g) + I2(g) ↔ 2HI(g) is 32 at given temperature. the equilibrium concentrations of I2 and HI are 0.5×10-3 and 8×10-3 M respectively. the equilibrium concentration of H2 is ..[Pb CET sample paper 1988]
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a) 1 ×10-3 M
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b) 0.5 ×10-3 M
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c)2 ×10-3
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d)4 ×10-3
Explanation
Answer:(d)
Q.29
For which of the following systems at equilibrium at constant temperature, doubling the volume will cause a shift to the right? [ KCET 1992]
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a) H2(g) + Cl2(g) ↔ 2HCl(g)
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b) 2CO(g) + O2(g) ↔ 2CO2(g)
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c) N2(g) + 3H2(g) ↔ 2NH3(g)
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d) PCl5(g) ↔ PCl3 + Cl2(g)
Explanation
In option 'd' on doubling volume pressure will decrease to increase pressure reaction will shift right to increase number of gas molecules resulting in increase in pressure Answer: (d)
Q.30
Which of the following will not change the concentration of ammonia in the equilibrium N2(g) + 3H2(g0 ↔ 2NH3(g); ΔH=-x kJ [ Pb. CET 1985]
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a)Increase of pressure
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b) Increase of temperature
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c) Decrease of volume
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d)Addition of catalyst
Explanation
Answer: (d)
0 h : 0 m : 1 s
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