12th Grade > Chemistry
IONIC EQUILIBRIUM MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option C. -> α = √KaC
:
C
According to the Ostwald's dilution formula α2−Ka(1−α)C. But for weak
electrolytes αis very small. So that (1−α) can be taken as 1. So that α=√KaC.
:
C
According to the Ostwald's dilution formula α2−Ka(1−α)C. But for weak
electrolytes αis very small. So that (1−α) can be taken as 1. So that α=√KaC.
Answer: Option A. -> It gives OH− ion.
:
A
NaOH completely dissociates to give Na+and OH−ions in water. Hence it is considered as a Strong base.
:
A
NaOH completely dissociates to give Na+and OH−ions in water. Hence it is considered as a Strong base.
Answer: Option C. -> 1.0 × 109
:
C
HA⇋H++A−;Ka=[H+][A−][HA]….(i)
neutralization of the weak acid with strong base is
HA+OH−⇋A−+H2O ….(ii)
K=[A−][HA][OH−]
dividing (i) by (ii) KaK=[H+][OH−]=Kw=10−14
K=KaKw=10−510−14=109
:
C
HA⇋H++A−;Ka=[H+][A−][HA]….(i)
neutralization of the weak acid with strong base is
HA+OH−⇋A−+H2O ….(ii)
K=[A−][HA][OH−]
dividing (i) by (ii) KaK=[H+][OH−]=Kw=10−14
K=KaKw=10−510−14=109
Answer: Option A. -> CO
:
A
CO doesn't have a vacantd-orbital.
:
A
CO doesn't have a vacantd-orbital.
Answer: Option A. -> 4 × 10−15
:
A
AB2⇋A++2B−
At equilibrium, [A+]=1×10−5,[B−]=2×10−5
Ksp=[1×10−5][2×10−5]2=4×10−15
:
A
AB2⇋A++2B−
At equilibrium, [A+]=1×10−5,[B−]=2×10−5
Ksp=[1×10−5][2×10−5]2=4×10−15
Answer: Option A. -> Basic
:
A
HCl = d
NaOH = 1.5 d
HCl in 10 ml of it = 10d36.5
NaOH in 10 ml of it =10×1.5d40= 15d40
NaOH> HCl
:
A
HCl = d
NaOH = 1.5 d
HCl in 10 ml of it = 10d36.5
NaOH in 10 ml of it =10×1.5d40= 15d40
NaOH> HCl
Answer: Option C. -> Na2CO3
:
C
Na2CO3+2H2O⇌2NaOH+H2CO3
:
C
Na2CO3+2H2O⇌2NaOH+H2CO3
Answer: Option D. -> Is a electron pair acceptor
:
D
Lewis acids accept electron pairs
:
D
Lewis acids accept electron pairs
Answer: Option C. -> HClO4
:
C
It is a HClO4
:
C
It is a HClO4
Answer: Option B. -> A is stronger than B
:
B
The value of pKafor strong acid is less.
:
B
The value of pKafor strong acid is less.
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