AIPMT 2004 | Ionic Equilibrium Question 49 | Chemistry

AIPMT 2004

MCQ (Single Correct Answer)

-1

The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentrations of the conjugate acid (HIn) and base (In^$$-$$) forms of the indicator by the expression

$$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = p{K_{In}} - pH$$

$$\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = p{K_{In}} - pH$$

$$\log {{\left[ {HIn} \right]} \over {\left[ {I{n^ - }} \right]}} = pH - p{K_{In}}$$

$$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {HIn} \right]}} = pH - p{K_{In}}$$

AIPMT 2004

MCQ (Single Correct Answer)

-1

The solubility product of a sparingly soluble salt AX₂ is 3.2 $$ \times $$ 10^$$-$$11. Its solubility (in moles/L) is

5.6 $$ \times $$ 10^$$-$$6

3.1 $$ \times $$ 10^$$-$$4

2 $$ \times $$ 10^$$-$$4

4 $$ \times $$ 10^$$-$$4

AIPMT 2003

MCQ (Single Correct Answer)

-1

The solubility product of AgI at 25^oC is 1.0 $$ \times $$ 10^$$-$$16 mol² L^$$-$$2. The solubility of AgI in 10^$$-$$4 N solution of KI at 25^oC is approximately (in mol L^$$-$$1

1.0 $$ \times $$ 10^$$-$$16

1.0 $$ \times $$ 10^$$-$$12

1.0 $$ \times $$ 10^$$-$$10

1.0 $$ \times $$ 10^$$-$$8

AIPMT 2002

MCQ (Single Correct Answer)

-1

Solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25. Then find out pK_b of NH₄OH

9.25

4.75

3.75

8.25

PREVIOUS NEXT

NEET Subjects

Biology

Botany

Zoology

Human Health and Diseases Body Fluids and Its Circulation Locomotion and Movement Neural Control and Coordination Reproduction in Organisms Reproductive Health Structural Organisation in Animals Digestion and Absorption Excretory Products and Their Elimination Chemical Coordination and Integration Human Reproduction Animal Kingdom Breathing and Exchange of Gases Evolution