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1

AIPMT 2005

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
A
presence of HCl decreases the sulphide ion concentration
B
solubility product of group II sulphides is more than that of group IV sulphates
C
presence of HCl increases the sulphide ion concentration
D
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 IInd group radicals.

2

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

Explanation

For an acid-base indicator

HIn ⇌ H+ + In-

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

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

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 KIn + $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$

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

$$ \Rightarrow $$ $$\log {{\left[ {I{n^ - }} \right]} \over {\left[ {{H_{in}}} \right]}}$$ = pH - pKIn
3

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
A
5.6 $$ \times $$ 10$$-$$6
B
3.1 $$ \times $$ 10$$-$$4
C
2 $$ \times $$ 10$$-$$4
D
4 $$ \times $$ 10$$-$$4

Explanation

AX2 A2+ + 2X-
s s 2s


Ksp = = [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
4

AIPMT 2003

MCQ (Single Correct Answer)
The solubility product of AgI at 25oC is 1.0 $$ \times $$ 10$$-$$16 mol2 L$$-$$2. The solubility of AgI in 10$$-$$4 N solution of KI at 25oC is approximately (in mol L$$-$$1
A
1.0 $$ \times $$ 10$$-$$16
B
1.0 $$ \times $$ 10$$-$$12
C
1.0 $$ \times $$ 10$$-$$10
D
1.0 $$ \times $$ 10$$-$$8

Explanation

AgI Ag+ + I-
s s s


Ksp = 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]

$$ \therefore $$ Ksp of AgI = [Ag+][I]

= (1.0 × 10–8)(10–4)

= 1.0 × 10–12 mol L–1

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