1

### NEET 2013 (Karnataka)

Accumulation of lactic acid (HC3H5O3), a monobasic acid in tissues leads to pain and a feeling of fatigue. In a 0.10 M aqueous solution, lactic acid is 3.7% dissociates. The value of dissociation constant, Ka, for this acid will be
A
1.4 $\times$ 10$-$5
B
1.4 $\times$ 10$-$4
C
3.7 $\times$ 10$-$4
D
2.8 $\times$ 10$-$4

## Explanation

$\alpha = \sqrt {{{{K_a}} \over c}}$

$\Rightarrow$ 0.037 = $\sqrt {{{{K_a}} \over {0.10}}}$

$\Rightarrow$ Ka = (0.037)2 $\times$ 0.10

= 1.37 ×10–4

$\simeq$ 1.4 $\times$ 10$-$4
2

### NEET 2013

Which of these is least likely to act as a lewis base ?
A
BF3
B
PF3
C
CO
D
F$-$

## Explanation

BF3 acts as Lewis acid. It is electron pair acceptor.
3

### NEET 2013

KMnO4 can be prepared from K2MnO4 as per the reaction,

3MnO42$-$ + 2H2O $\rightleftharpoons$ 2MnO4$-$ + MnO2 + 4OH$-$

The reaction can go to completion by removing OH$-$ ions by adding
A
CO2
B
SO2
C
HCl
D
KOH

## Explanation

KOH is the base thus, it gives OH ions thus it cannot remove OH ions from reaction mixture but it adds on the concentration of OH ions.

So, an acid must be added but if a strong acid is added to the reaction mixture then in acidic condition the MnO4 formed reduces to give Mn2+ thus, HCl which is a strong acid and SO2 which on treating with water forms a strong H2SO4 cannot be used for this purpose.

Thus, CO2 which forms H2CO3 a weak acid reacts to remove OH but not that much acidic that MnO4 undergo reduction. Thus, CO2 is used for this reaction for completion.
4

### AIPMT 2012 Mains

Given the reaction between 2 gases represented by A2 and B2 to give the compound AB(g),

A2(g) + B2(g) $\rightleftharpoons$ 2AB(g)

At equilibrium, the concentration of
A2 = 3.0 $\times$ 10$-$3 M, of B2 = 4.2 $\times$ 10$-$3 M, of AB = 2.8 $\times$ 10$-$3 M
If the reaction takes place in a sealed vessel at 527oC, then the value of Kc will be
A
2.0
B
1.9
C
0.62
D
4.5

## Explanation

A2(g) + B2(g) $\rightleftharpoons$ 2AB(g);

Kc = ${{{{\left[ {AB} \right]}^2}} \over {\left[ {{A_2}} \right]\left[ {{B_2}} \right]}}$

= ${{{{\left( {2.8 \times {{10}^{ - 3}}} \right)}^2}} \over {3 \times {{10}^{ - 3}} \times 4.2 \times {{10}^{ - 3}}}}$

= 0.62