1

### JEE Main 2018 (Online) 15th April Evening Slot

Following four solutions are prepared by mixing different volumes of NaOH and HCl of different concentrations, pH of which one of them will be equal to 1 ?
A
100 mL ${M \over {10}}$ HCl + 100 mL ${M \over {10}}$ NaOH
B
75 mL ${M \over {5}}$ HCl + 25 mL ${M \over {5}}$ NaOH
C
60 mL ${M \over {10}}$ HCl + 40 mL ${M \over {10}}$ NaOH
D
55 mL ${M \over {10}}$ HCl + 45 mL ${M \over {10}}$ NaOH

## Explanation

(a)    100 mL  ${M \over {10}}$ NaOH will nutalise

100 mL   ${M \over {10}}$ HCl, so number extra

HCl   will remain.

This will be neutral solution

$\therefore\,\,\,$ pH = 7

(b)    Here 25 mL  ${M \over 5}$ NaOH   nutralise

25 mL  ${M \over 5}$ HCl

$\therefore\,\,\,$ Extra HCl   =   75 $-$ 25   =   50 mL

Total volume   =   75 + 25   =   100 mL

$\therefore\,\,\,$ Milimole of HCl   =   ${{50} \over 5}$ = 10

$\therefore\,\,\,$ Concentration of HCl  =  ${{10} \over {100}}$  = 0.1

$\therefore\,\,\,$ pH  =  $-$ log[H+]  =  $-$ log(0.1)  =  1

(c)    HCl left  =  60 $-$ 40  =  20 mL

$\therefore\,\,\,$ milimole of HCl   =   ${{20} \over {10}}$   =  2

$\therefore\,\,\,$ Concentration of HCl   =   ${2 \over {100}}$   =  0.02 M

$\therefore\,\,\,$ pH  =  $-$ log (0.02)   =   1.69

(d)    HCl left  = 55 $-$ 45  =  10 mL

$\therefore\,\,\,$ milimole of HCl   =  ${{10} \over {10}}$  = 1

$\therefore\,\,\,$ Concentration of HCl  = ${{1} \over {100}}$   =  0.01 M

$\therefore\,\,\,$ pH = $-$ log (0.01) = 2
2

### JEE Main 2018 (Online) 15th April Evening Slot

At a certain temperature in a $5$ $L$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,
CO + Cl2 $\rightleftharpoons$ COCl2
At equilibrium, if one mole of CO is present then equilibrium constant (Kc) for the reaction is :
A
2
B
2.5
C
3
D
4

## Explanation

CO + Cl2 $\rightleftharpoons$ COCl2
Initially number of moles 2 3 0
At equilibrium number of moles 1 2 1

The equilibrium constant,

Kc = ${{\left[ {COC{l_2}} \right]} \over {\left[ {CO} \right]\left[ {C{l_2}} \right]}}$

=  ${{\left( {{1 \over 5}} \right)} \over {\left( {{1 \over 5}} \right) \times \left( {{2 \over 5}} \right)}}$

=   ${5 \over 2}$

=  2.5
3

### JEE Main 2018 (Online) 16th April Morning Slot

The gas phase reaction 2NO2(g) $\to$ N2O4(g) is an exothermic reaction. The decomposition of N2O4, in equilibrium mixture of NO2(g) and N2O4(g), can be increased by :
A
lowering the temperature.
B
increasing the pressure.
C
addition of an inert gas at constant volume.
D
addition of an inert gas at constant pressure.

## Explanation

2NO2 (g) $\buildrel \, \over \longrightarrow$ N2O4 (g); $\Delta$H = $-$ Ve

At equilibrium, N2O4 $\rightleftharpoons$ 2NO2 ; $\Delta$H = + Ve

Decomposition N2O4 is endothermic

Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.

So, it will move towards backward direction which lead to formation of N2O4 from NO2.
4

### JEE Main 2019 (Online) 9th January Morning Slot

20 mL of 0.1 M H2SO4 solution is added to 30 mL of of 0.2 M NH4OH solution. The pH of the resultant mixture is : [pkb of NH4OH = 4.7].
A
5.2
B
9.0
C
5.0
D
9.4

## Explanation

H2SO4 + 2NH4OH $\to$ (NH4)2SO4 + H2O

Initially,

H2SO4 present = 20 $\times$ 0.1 $\times$ 2 = 4 miliequivalent

NH4OH present = 30 $\times$ 0.2 = 6 miliequivalent

Here H2SO4 is the limiting reagent,

So, finally. H2SO4 present = 0

and NH4OH present = (6 $-$ 4) = 2

and (NH4)2SO4 produced = 4 miliequivalent.

As in the solution there is (NH4)2 SO4 present so it a basic buffer.

$\therefore$   POH = PKb + log ${{\left[ {Salt} \right]} \over {\left[ {base} \right]}}$

= 4.7 + log ${4 \over 2}$

= 4.7 + log2

= 4.7 + 0.3

= 5

$\therefore$   PH = 14 $-$ POH

= 14 $-$ 5

= 9