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1

### AIPMT 2011 Prelims

A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH+4 is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 $$\times$$ 10$$-$$5, what is the pH of this solution? (log 2.7 = 0.43)
A
9.08
B
9.43
C
11.72
D
8.73

## Explanation

pOH = pKb + log $${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$$

= – log Kb + log $${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$$

= – log 1.8× 10–5 + log $${{0.20} \over {0.30}}$$

= 5 – 0.25 + (–0.176) = 4.57

Now, pH = 14 – pOH = 14 – 4.57 = 9.43
2

### AIPMT 2011 Prelims

The value of $$\Delta$$H for the reaction
X2(g) + 4Y2(g) $$\rightleftharpoons$$ 2XY4(g)
is less than zero. Formation of XY4(g) will be favoured at
A
high temperature and high pressure
B
low pressure and low temperature
C
high temperature and low pressure
D
high pressure and low temperature

## Explanation

X2(g) + 4Y2(g) $$\rightleftharpoons$$ 2XY4(g)

$$\Delta$$ng = -ve and $$\Delta$$H = -ve

As $$\Delta$$H < 0 i.e., the given reaction is exothermic. According to Le-Chatelier principle, for exothermic reaction, forward reaction is favoured when temperature becomes low. Also, there are 5 gaseous moles on reactant side and 2 gaseous moles on products side. So, forward reaction is favoured when pressure of the reaction mixture becomes high. The reason is that at high pressure reaction tends to more in direction where there is lessee number of gaseous moles.
3

### AIPMT 2010 Mains

The reaction,
2A(g) + B(g) $$\rightleftharpoons$$ 3C(g) + D(g)
is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measuread and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression
A
[(0.75)3 (0.25)] $$\div$$ [(1.00)2 (1.00)]
B
[(0.75)3 (0.25)] $$\div$$ [(0.50)2 (0.75)]
C
[(0.75)3 (0.25)] $$\div$$ [(0.50)2 (0.25)]
D
[(0.75)3 (0.25)] $$\div$$ [(0.75)2 (0.25)]

## Explanation

2A(g) + B(g) &#8652; 3C(g) + D(g)
Initial mole 1 1 0 0
At equilibrium 1 - (2 $$\times$$ 0.25)
= 0.5
1 - 0.25
= 0.75
3 $$\times$$ 0.25
= 0.75
0.25

Equilibrium constant, K = $${{{{\left[ C \right]}^3}\left[ D \right]} \over {{{\left[ A \right]}^2}\left[ B \right]}}$$

$$\therefore$$ K = $${{{{\left[ {0.75} \right]}^3}\left[ {0.25} \right]} \over {{{\left[ {0.5} \right]}^2}\left[ {0.75} \right]}}$$
4

### AIPMT 2010 Prelims

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? Ka for CH3COOH = 1.8 $$\times$$ 10$$-$$5
A
3.5 $$\times$$ 10$$-$$4
B
1.1 $$\times$$ 10$$-$$5
C
1.8 $$\times$$ 10$$-$$5
D
9.0 $$\times$$ 10$$-$$6

## Explanation

CH3COOH and CH3COONa constitute to form an acidic buffer.

$$\Rightarrow$$ pH = pKa + log$${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

pH = –log(1.8 × 10–5) + log $${{\left( {0.20} \right)} \over {\left( {0.10} \right)}}$$

= 4.74 + log 2

= 4.74 + 0.3010 = 5.041

Now, pH = – log[H+]

$$\Rightarrow$$ 5.041 = – log[H+]

$$\Rightarrow$$ [H+] = 10–5.041 = 9.0 × 106 mol L–1

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