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NEET Chemistry MCQ
Quiz 6
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Q.1
At 90°C, pure water has H3O+ ion concentration of 10-6 mol/L. the Kw at 90°C is
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a) 10-12
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b) 10-14
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c)10-6
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d)10-8
Explanation
Kw=[ H+] [OH-]Kw=10-6 × 10⁻⁶=10-12 Answer:(a)
Q.2
The value of ionic product of water at 393K is [ Tamil Nadu CET 2002]
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a) Less than 1 × 10⁻¹⁴
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b) Greater than 1 × 10⁻¹⁴
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c)Equal to 1 × 10⁻¹⁴
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d)Equal to 1 × 10⁻⁷
Explanation
Degree of ionization increases with increase of temperature [H+] and [OH-] increases with the rise of temperature and ionic product becomes greater than 1 × 10⁻¹⁴Answer: (b)
Q.3
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.4
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.5
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.6
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.7
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.8
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.9
For reaction 2NO2(g) ↔ 2NO(g) + O2(g) Kc=1.8 ×10⁻⁶ at 185°C, the value of Kc for reaction NO + ½O2 ↔ NO2 is ..[MLNR 1993]
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a) 0.9 × 106
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b) 7.5 × 102
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c)1.95 × 10⁻³
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d)1.95 ×103
Explanation
Second reaction is reverse of first and half thus Equilibrium constant will be 1/√KConstant=1/( 1.8 ×10⁻⁶)-1/2=7.5 × 102 Answer:(b)
Q.10
If Kp for a reactionA(g) + 2B(g) ↔ 3C(g) + D(g) is 0.05 atm at 1000K. Its Kc in terms of R will be.. [CBSE PMT 1989]
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a) 20000R
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b) 0.02R
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c) 5×10⁻⁵R
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d) 5×10⁻⁵/R
Explanation
Δn g=4-3=1 Kp=Kc(RT)Δn 0.05=Kc(1000R) Kc=5×10⁻⁵/R Answer: (c)
Q.11
The KSP of Ag2CrOAgCl, AgBr and AgI are respectively 1.1×10⁻¹², 1.8×10⁻¹⁰, 5.0 ×10⁻¹³, 8.3×10⁻¹⁷ Which of the following salts will precipitate last if AgNO3solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4? [ AIPMT 2013]
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a) AgI
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b) AgCl
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c) AgBr
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d) Ag2CrO4
Explanation
Ag2CrO4 ⇒ 2Ag+ + CrO42- Solubility product KSP = (2S)2 ×S = 4s3 KSP = (1.1 × 10⁻¹²) (given) AgCl ⇒ Ag+ + Cl- KSP = s × s ( KSP = 1.8 × 10⁻¹⁰) s = √KSP = 1.34×10⁻⁵ AgBr ⇒ Ag+ Br- KSP = s × s ( KSP = 5 × 10⁻¹³) s = √KSP = 0.71×10⁻⁶ AgI ⇒ Ag+ + I- KSP = s × s ( KSP = 8.3 × 10⁻¹⁷) s = √KSP = 0.9×10⁻⁸ ∵ Solubility of Ag2CrO4 is highest So, it will precipitate last Answer: (d)
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