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1

### AIPMT 2005

MCQ (Single Correct Answer)
Equilibrium constants K1 and K2 for the following equilibriam:

are related as
A
K2 = 1/K12
B
K2 = K12
C
K2 = 1/K1
D
K2 = K1/2

## Explanation

NO(g) + O2(g) ⇌ NO2(g) ,    K1

Reverse the above equation

NO2(g) ⇌ NO(g) + O2(g),     $${1 \over {{K_1}}}$$

Multiply the above equation by 2, we get

NO2(g) ⇌ NO(g) + O2(g),    $${\left( {{1 \over {{K_1}}}} \right)^2} = {K_2}$$
2

### 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.

3

### AIPMT 2005

MCQ (Single Correct Answer)
At 25oC, the dissociation constant of a base, BOH, is 1.0 $$\times$$ 10$$-$$12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be
A
1.0 $$\times$$ 10$$-$$5 mol L$$-$$1
B
1.0 $$\times$$ 10$$-$$6 mol L$$-$$1
C
2.0 $$\times$$ 10$$-$$6 mol L$$-$$1
D
1.0 $$\times$$ 10$$-$$7 mol L$$-$$1

## Explanation

BOH B+ + OH-
Initially C 0 0
At equilibbrium C - C$$\alpha$$ C$$\alpha$$ C$$\alpha$$

[OH-] = C$$\alpha$$

$$\Rightarrow$$ [OH-] = $$\sqrt {{K_b}C}$$

$$\Rightarrow$$ [OH-] = $$\sqrt {1 \times {{10}^{ - 12}} \times {{10}^{ - 2}}}$$

= 1.0 $$\times$$ 10$$-$$7 mol L$$-$$1
4

### 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

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