1
MCQ (Single Correct Answer)

### 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.
2
MCQ (Single Correct Answer)

### 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
3
MCQ (Single Correct Answer)

### AIPMT 2012 Mains

Given that the equilibrium constant for the reaction,
2SO2(g) + O2(g) $\rightleftharpoons$ 2SO3(g)
has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature ?
SO3(g) $\rightleftharpoons$ SO2(g) + ${1 \over 2}$ O2(g)
A
1.8 $\times$ 10$-$3
B
3.6 $\times$ 10$-$3
C
6.0 $\times$ 10$-$2
D
1.3 $\times$ 10$-$5

## Explanation

2SO2(g) + O2(g) $\rightleftharpoons$ 2SO3(g),    K

SO3(g) $\rightleftharpoons$ SO2(g) + ${1 \over 2}$ O2(g),     K' = $\sqrt {{1 \over K}}$

$\therefore$ K' = $\sqrt {{1 \over {278}}} =$ 6.0 $\times$ 10$-$2
4
MCQ (Single Correct Answer)

### AIPMT 2012 Prelims

Buffer solutions have constant acidity and alkalinity because
A
these give unionised acid or base on reaction with added acid or alkali
B
acids and alkalies in these solutions are shielded from attack by other ions
C
they have large excess of H+ or OH$-$ ions
D
they have fixed value of pH

## Explanation

For this example,

CH3COOH ⇌ CH3COO + H+ ;

CH3COONa ⇌ CH3COO + Na+

when few drops of HCl are added to this buffer, the H+ of HCl immediatly combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no H+ ions to combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH ions will combine with H+ ions to form unionised water molecule. Thus pH of solution will remain constant.

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