AIPMT 2004 | Ionic Equilibrium Question 49 | Chemistry

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

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