JEE Main 2020 (Online) 8th January Morning Slot | Ionic Equilibrium Question 67 | Chemistry

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively : JEE Main 2020 (Online) 8th January Morning Slot Chemistry - Ionic Equilibrium Question 67 English

JEE Main 2020 (Online) 8th January Morning Slot Chemistry - Ionic Equilibrium Question 67 English

XY, 2 × 10^–6 M³

XY₂, 1 × 10^–9 M³

XY₂, 4 × 10^–9 M³

X₂Y, 2 × 10^–9 M³

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

-1

The molar solubility of Cd(OH)₂ is 1.84 × 10^–5 M in water. The expected solubility of Cd(OH)₂ in a buffer solution of pH = 12 is :

2.49 × 10^–10 M

1.84 × 10^–9 M

6.23 × 10^–11 M

$${{2.49} \over {1.84}} \times {10^{ - 9}}M$$

JEE Main 2019 (Online) 12th April Evening Slot

MCQ (Single Correct Answer)

-1

The decreasing order of electrical conductivity of the following aqueous solutions is :

0.1 M Formic acid (A),

0.1 M Acetic acid (B),

0.1 M Benzoic acid (C)

A > B > C

C > B > A

A > C > B

C > A > B

JEE Main 2019 (Online) 12th April Morning Slot

MCQ (Single Correct Answer)

-1

What is the molar solubility of Al(OH)₃ in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)₃ = 2.4 × 10^–24 :

3 × 10^–22

3 × 10^–19

12 × 10^–21

12 × 10^–22