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NEET Chemistry MCQ
Quiz 2
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Q.1
Four species are listed below (i) HCO3- (ii) H+ (iii) HSO4- (iv) HSO3F Which one of the following is the correct sequence of their acid strength [ AIEEE 2008]
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a) iii < i < iv < ii
0%
b) iv < ii < iii < i
0%
c)ii < iii < i < iv
0%
d)i < iii < ii < iv
Explanation
HSO3F is super acid, hence strongest among all. We know weaker the conjugate base stronger the acid Answer: (d)
Q.2
Among the following acids, which has the lowest pKa value?
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a) CH3COOH
0%
b) HCOOH
0%
c)(CH3)2CHOOH
0%
d)CH3CH2COOH
Explanation
Strong acid have lower pKa valueAnswer: (b)
Q.3
Which of the following bases is the weakest?
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a) C9H7N : Kb=6.3 ×10⁻¹⁰
0%
b) C2H5NH2 : Kb=5.6 ×10⁻⁴
0%
c) C6H5NH2 : Kb=3.8 ×10⁻¹⁰
0%
d)NH4OH : Kb=1.6 ×10⁻⁶
Explanation
Greater the value of Kb of a base stronger the base Answer:(c)
Q.4
A solution which is 10-3 M each in Mn2+, Fe2+, Zn2+ and Hg2+ is treated with 10-16M sulphide ion. If Ksp of MnS, FeS, ZnS and HGS are 10-15,10-23, 10-20 and 10-54 respectively, which one will precipitate first.. [ IIT 2003]
0%
a) FeS
0%
b) MnS
0%
c) HgS
0%
d) ZnS
Explanation
MS ⇔ M2+ + S2- Lower the value of Ksp lower will be the solubility. Thus HgS with least value of Ksp will get precipitated first Answer: (c)
Q.5
On adding 0.1M solution each of [Ag+], [Ba2+], [Ca2+] in a Na2SO4 solution, species first precipitated is [Ksp of BaSO4=10-11, Ksp of CaSO4=10-6, Ksp of Ag2SO4=10-5]
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a)Ag2SO4
0%
b) BaSO4
0%
c)CaSO4
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d)All of these
Explanation
Salt with least value of Ksp will precipitate firstAnswer: (b)
Q.6
Precipitation takes place when the ionic product [ Roorkee 1983]
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a) Equals the solubility product
0%
b) Exceeds the solubility product
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c)Is less than the solubility product
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d)Is almost zero
Explanation
Answer: (b)
Q.7
At 25°C, the dissociation constant of Ch3COOH and NH4OH in aqueous solution are almost same ( 10-5). If pH of some acetic acid solution is 3.0 The pH of solution of NH4OH of same concentration at the same temperature would be [ Roorkee 1990]
0%
a) 3.0
0%
b) 4.0
0%
c)10.0
0%
d)11.0
Explanation
pH of CH3COOH=3 [H+]=10-3M or [CH3COOH]=10-3 M ∴ [OH-]=10-3 M But [H+] [ OH-]=10-14 ∴ [H+]=10-11 M ∴ pH of NH4OH solution=-log10-11=11 Answer:(d)
Q.8
Species acting both as Bronsted acid and base is
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a)HSO4-
0%
b) Na2CO3
0%
c)NH3
0%
d)OH-
Explanation
According to bronsted_lowry concept, an acid is a substance that can donate and a base is substance that can accept a proton from acid.HSO4- can accept a proton and form H2SO4 as well as can donate a proton and form SO42-Answer: (a)
Q.9
The pH of 0.1M aqueous solution of weak acid ( HA) isIts degree of dissociation is
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a) 1%
0%
b) 10%
0%
c)50%
0%
d)25%
Explanation
HA ⇔ H+ + A- Sin pH=3 [H+]=Cα=10-3 ∴ 0.1 × α=10-3 Or α=10-2 Or 1%Answer: (a)
Q.10
An acid buffer solution can be prepared by mixing equimolar amount of ..[ IIT 1981]
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a) Formic acid and sodium formate
0%
b) NH4Cl and NH4OH
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c)H2SO4 and NA2SO4
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d)CH3CONH2 and KOH
Explanation
Weak acid + salt of weak acid with strong base=acidic buffer Answer:(a)
Q.11
pKa of acetic acid is 4.The concentration of CH3COONa is 0.01 M. The pH of CH3COONa ia [ Orissa JEE 2004]
0%
a) 8.37
0%
b) 4.37
0%
c) 4.74
0%
d) 0.474
Explanation
CH3COONa is a salt of weak acid (CH3COOH) and strong base (NaOH) Applying pH=½ [ pKw + pKa + logC] pH=½ [ 14 +4.74 + log10-2]=8.37 Answer: (a)
Q.12
Which of the following salts undergoes anionic hydrolysis?
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a) CuSO4
0%
b) Na2CO3
0%
c)NH4Cl
0%
d)FeCl3
Explanation
CO3- + 2H2O → H2CO3 + OH-Answer: (b)
Q.13
An acidic buffer solution can be prepared by mixing solution of .. [ IIT 1981]
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a)Sodium acetate and acetic acid
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b) Ammonium chloride and ammonium hydroxide
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c)Sulphuric acid and sodium sulphate
0%
d)Sodium chloride and sodium hydroxide
Explanation
Weak acid + salt of weak acid with strong base=acidic bufferAnswer: (a)
Q.14
aht is the decreasing order of strength of the bases:
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a)
0%
b)
0%
c)
0%
d)
Explanation
Conjugate base of strong acid is weak while conjugate base of weak acid is strong. Acidic strength of conjugate acid is Thus Decreasing order of basic strength Answer:(a)
Q.15
Which of the following is Bronsted acid as well as a Bronsted base?
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a) BF3
0%
b) NH3
0%
c) H2O
0%
d) Na2CO3
Explanation
H2O is amphoteric and acts both as proton donor and proton acceptorH2O + H2O ⇔ H3O+ + OH- Answer: (c)
Q.16
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
0%
b) 9.58
0%
c)4.79
0%
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.17
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
0%
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.18
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
0%
b) 7.30
0%
c)10.30
0%
d)8.30
Explanation
Applying Since [Salt]=[Acid] pH=pKa=9.3 Answer:(a)
Q.19
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.20
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)
0%
b)
0%
c)
0%
d)
Explanation
Answer: (c)
Q.21
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
0%
b) 4.5
0%
c)2.5
0%
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.22
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.23
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.24
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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c) 1.0
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d) 2.0
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b) - log2×10⁻¹
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.25
The solubility of A2 X3 is y mol dm-Is solubility product is ..[ IIT 1997]
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a)6y4
0%
b) 64y4
0%
c)36y5
0%
d)108y5
Explanation
A2 X3 ⇔ 2A3+ + 3X2-Ksp=[A3+]2 + [X2-]3Since solubility=y mol dm-3Ksp=[2y]2 + [3y]3=108y5Answer: (c)
Q.26
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.27
Which of the following can not act a a Lewis acid or Bronsted acid?
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a)CCl4
0%
b) SnCl4
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c)AlCl3
0%
d)BF3
Explanation
CCl4can not accept electron pair due to the absence of d-orbitalAnswer: (a)
Q.28
pKa value of two acids A and B are 4 andThe strengths of these two acids are related as ..( Karnataka CET 2001]
0%
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.29
The pH of 0.1M solution of following increases in the order [ IIT 1999]
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a) NaCl < NH4Cl < NaCN
0%
b) HCl < NH4Cl < NaCl
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c) NaCN < NH4Cl < NaCl < HCl
0%
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.30
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)
0 h : 0 m : 1 s
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