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1

### AIPMT 2005

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
2

### 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
3

### 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
4

### 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

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