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Test: JEE Previous Year Questions- Ionic Equilibrium - NEET MCQ


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15 Questions MCQ Test Chemistry Class 11 - Test: JEE Previous Year Questions- Ionic Equilibrium

Test: JEE Previous Year Questions- Ionic Equilibrium for NEET 2024 is part of Chemistry Class 11 preparation. The Test: JEE Previous Year Questions- Ionic Equilibrium questions and answers have been prepared according to the NEET exam syllabus.The Test: JEE Previous Year Questions- Ionic Equilibrium MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: JEE Previous Year Questions- Ionic Equilibrium below.
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Test: JEE Previous Year Questions- Ionic Equilibrium - Question 1

The solubility of Mg(OH)2 is x mole/ltr. then its solubility product is - 

[AIEEE-2002]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 1

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 2

The solubility in water of a sparingly soluble salt AB2 is 1.0 × 10-5 mol L-1 . Its solubility product will be -

[AIEEE-2003]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 2

AB2                A       +     2B
S                     S                2S
Ksp = [A][B]2
= s (2s)2
1.0×10-5×(2×1.0×10-5)2
= 4*10-15

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 3

The molar solubility (in mol L-1) of a sparingly soluble salt MX4 is `s'. The corresponding solubility product is KSP. `s' is given in terms of Ksp by the relation -

[AIEEE-2004]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 3

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 4

The solubility product of a salt having general formula MX2, in water is 4x10-12. The concentration of M2+ ions in the aqueous solution of the salt is - [AIEEE-2005]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 4


Test: JEE Previous Year Questions- Ionic Equilibrium - Question 5

The first and second dissociation constants of an acid H2A are 1.0 x 10-5 & 5.0 x 10-10 respectively. The overall dissociation constant of the acid will be -

[AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 5


Test: JEE Previous Year Questions- Ionic Equilibrium - Question 6

The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is-          [AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 6

For buffer solution

as HA is 50% ionized so [Salt] = [Acid]
pH = 4.5 pH + pOH = 14
⇒ pOH = 14 – 4.5 = 9.5

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 7

In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molecular mass = 283) the equilibrium which sets in is -

AgIO3(s) Ag+(aq) + IO3¯(aq)

If the solubility product Ksp of AgIO3 at a given temperature is 1.0 x 10-8, what is the mass of AgIO3 contained in 100 ml of its saturated solution ?   

[AIEEE-2007]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 7

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 8

The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be -  

[AIEEE-2008]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 8

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 9

Solid Ba(NO3)2 is gradually dissolved in a 1.0 x 10-4 M Na2CO3 solution. At what concentration of Ba2+ will a precipitate begin to form ? (Ksp for Ba CO3 = 5.1 x 10-9)  

 [AIEEE-2009]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 9

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 10

In aqueous solution the ionization constants for carbonic acid are K1 = 4.2 x 10-7 and K2 = 4.8 x 10-11 Selection the correct statement for a saturated 0.034 M solution of the carbonic acid. 

 [AIEEE-2010]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 10

Carbonic acid is a weak acid it dissociates as follows 

Since both acids are weak cannot be 0.034 M as there is no complete dissociation. 

Total hydrogen ion concentration is approximately equal to concentration of hydrogen ion in first reaction as K1 is larger. This is approximately equal to concentration of 

The concentration of  depends on  and that of  on . But . So The concentration of  is not double that of 

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 11

Solubility product of silver bromide is 5.0 x 10-13. The quantity of potassium bromide (molar mass taken as 120 g mol-1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of AgBr is   

[AIEEE-2010]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 11

For precipitation

[Ag+][Br-] > Ksp[AgBr]

[Br-]min = (5 x 10-13)/0.05 = 10-11 M

Mass of potassium bromide needed = 10-11 x 120

= 1.2 x 10-9 g

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 12

Three reactions involving are given below:                         

(i)

(ii)

(iii)

In which of the above does act as an acid?

[AIEEE-2010]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 12

Acids are those substances which gives H+. Here in reaction ii only, H2PO4- acts as an acid.

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 13

At 25°C, the solubility product of Mg(OH)2 is 1.0 ×10-11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ions ? 

[AIEEE-2010]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 13

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 14

The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is -

[AIEEE-2012]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 14

Test: JEE Previous Year Questions- Ionic Equilibrium - Question 15

How many litres of water must be added to litre of an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2? 

 [AIEEE-2013]

Detailed Solution for Test: JEE Previous Year Questions- Ionic Equilibrium - Question 15

Volume of the original solution = 1 L

pH of the new solution = 2

Volume of the new solution = ?
As per volumetric principle

That means the volume of water added to the original solution of 1 L = 10 −1 = 9 L

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