1

### AIPMT 2010 Prelims

In which of the following equilibrium Kc and Kp are not equal?
A
2NO(g) $\rightleftharpoons$ N2(g) + O2(g)
B
SO2(g) + NO2(g) $\rightleftharpoons$ SO3(g) + NO(g)
C
H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)
D
2C(s) + O2(g) $\rightleftharpoons$ 2CO2(g)

## Explanation

As we know, Kp = Kc × (RT)$\Delta$ng

So, for reaction having same number of gaseous moles on reactants and products side will have same value of Kc and Kp otherwise their values are different.

For reaction, 2NO(g) $\rightleftharpoons$ N2(g) + O2(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, SO2(g) + NO2(g) $\rightleftharpoons$ SO3(g) + NO(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, H2(g) + I2(g) $\rightleftharpoons$ 2HI(g)

$\Delta$ng = 2 – 2 = 0

$\Rightarrow$ Kp = Kc × (RT)0

$\therefore$ Kp = Kc

For reaction, 2C(s) + O2(g) $\rightleftharpoons$ 2CO2(g)

$\Delta$ng = 2 – 3 = -1

$\Rightarrow$ Kp = Kc × (RT)-1

$\therefore$ Kp $\ne$ Kc
2

### 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
3

### AIPMT 2010 Prelims

In a buffer solution containing equal concentration of B$-$ and HB, the Kb for B$-$ is 10$-$10. The pH of buffer solution is
A
10
B
7
C
6
D
4

## Explanation

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

$\Rightarrow$ pOH = - log Kb + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$\Rightarrow$ pOH = –log10–10 + log 1

[As conc. of HB and B are same]

$\Rightarrow$ pOH = 10

$\Rightarrow$ pH = 14 – pOH = 14 – 10 = 4
4

### AIPMT 2010 Prelims

If pH of a saturated solution of Ba(OH)2 is 12, the value of its Ksp is
A
4.00 $\times$ 10$-$6 M3
B
4.00 $\times$ 10$-$7 M3
C
5.00 $\times$ 10$-$6 M3
D
5.00$\times$ 10$-$7 M3

## Explanation

Ba(OH)2 Ba2+ + 2OH
At equilibrium x 2x

pH = – log[H+]

12 = – log [H+]

$\Rightarrow$ [H+] = 10–12

As, [H+][OH ] = 10–14

10–12 [OH ] = 10–14

$\Rightarrow$ [OH ] = 10–2

As [OH ] = 2x = 10–2 then x = 5.0 × 10–3

Now, Ksp = [Ba2+][OH ]2

Ksp = (5 × 10–3) (10–2)2 = 5.0 × 10–7