1
MCQ (Single Correct Answer)

AIPMT 2004

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
2
MCQ (Single Correct Answer)

AIPMT 2004

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
3
MCQ (Single Correct Answer)

AIPMT 2003

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
4
MCQ (Single Correct Answer)

AIPMT 2003

The reaction quotient (Q) for the reaction

N2(g) + 3H2(g) $$\rightleftharpoons$$ 2NH3(g) is given by

$$Q = {{{{\left[ {N{H_3}} \right]}^2}} \over {\left[ {{N_2}} \right]{{\left[ {{H_2}} \right]}^3}}}$$.

The reaction will proceed from right to left if
A
Q = Kc
B
Q < Kc
C
Q > Kc
D
Q = 0

Explanation

For any reaction to process in forward direction the reaction quotient (Q) must be less than equilibrium constant KC.

Q < Kc

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