AIPMT 2005

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MCQ (Single Correct Answer)

H₂S gas when passed through a solution of cations containing HCl precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group. It is because

presence of HCl decreases the sulphide ion concentration

solubility product of group II sulphides is more than that of group IV sulphates

presence of HCl increases the sulphide ion concentration

sulphides of group IV cations are unstable in HCl.

Explanation

H₂S ⇌ H⁺ + HS^–

HCl ⇌ H⁺ + Cl^–

In presence of HCl this ionization of H₂S is suppressed due to the presence of extra H⁺ ions from HCl and produces less amount of sulphide ions due to common ion effect, thus HCl decreases the solubility of H₂S which is sufficient to precipitate II^nd group radicals.

AIPMT 2004

MCQ (Single Correct Answer)

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

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

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

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

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

Explanation

For an acid-base indicator

H_In ⇌ H⁺ + In^-

K_in = $${{\left[ {{H^ + }} \right]\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ $$\left[ {{H^ + }} \right] = {{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

Take – log on both sides

$$ - \log \left[ {{H^ + }} \right] = - \log \left( {{{{K_{in}}\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}} \right)$$

$$ \Rightarrow $$ pH = –log K_In + $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ pH = pK_In + $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

$$ \Rightarrow $$ $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$ = pH - pK_In

AIPMT 2004

MCQ (Single Correct Answer)

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

Explanation

AX₂	⇌	A²⁺	+	2X^-
s		s		2s

K_sp = = [A²⁺] [X^– ]² = s × (2s)² = 4s³

$$ \Rightarrow $$ 3.2 $$ \times $$ 10^$$-$$11 = 4s³

$$ \Rightarrow $$ s³ = 8 × 10^–12

$$ \Rightarrow $$ s = 2 × 10^–4 mol L^–1

AIPMT 2003

MCQ (Single Correct Answer)

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

Explanation

AgI	⇌	Ag⁺	+	I^-
s		s		s

K_sp = s²

$$ \Rightarrow $$ 1.0 × 10^–16 = s²

$$ \Rightarrow $$ s = 1.0 × 10^–8 mol L^–1

$$ \therefore $$ [Ag⁺] = 1.0 × 10^–8 mol L^–1

Also, in 10^–4 N KI solution,

[I^–1] = (10^–4 + 1.0 × 10^–8) mol L^–1

$$ \Rightarrow $$ [I^–1] = (10^–4) mol L^–1

[As 1.0 × 10^–8 mol L^–1 << 1.0 × 10^–4 mol L^–1]

$$ \therefore $$ K_sp of AgI = [Ag⁺][I^–]

= (1.0 × 10^–8)(10^–4)

= 1.0 × 10^–12 mol L^–1