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### JEE Main 2017 (Online) 9th April Morning Slot

The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal :

Fe2O3(s) + 3 CO(g) $\rightleftharpoons$ 2 Fe(1) + 3 CO2(g)

Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
A
Removal of CO
B
Removal of CO2
C
D
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### JEE Main 2018 (Offline)

An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 $\times$ 10–10. What is the original concentration of Ba2+?
A
1.0 $\times$ 10–10 M
B
5 $\times$ 10–9 M
C
2 $\times$ 10–9 M
D
1.1 $\times$ 10–9 M

## Explanation

Let initially concentration of Ba+2 = x m.

After adding 50 ml Na2SO4 in Ba+2 solution final volume becomes 500 ml.

$\therefore\,\,\,$ Initial volume of Ba+2 solution

= (500 $-$ 50) ml = 450 ml

As at the begining of precipitation, ionic product = solubility product.

$\Rightarrow \,\,\,$ [Ba2+] [SO${_4^{ - 2}}$] = Ksp of BaSO4

$\Rightarrow \,\,\,$ [Ba2+] $\left( {{{50 \times 1} \over {500}}} \right)$ = 1 $\times$ 10$-$10

$\Rightarrow \,\,\,$ [Ba2+] = 10$-$9 M.

So, the concentration of Ba+2 in final solution is 10$-$9 M.

$\therefore\,\,\,$ concentration of Ba+2 in original solution,

M1 V1 = M2 V2

$\Rightarrow \,\,\,$ x $\times$ 450 = 10$-$9 $\times$ 500

$\Rightarrow \,\,\,$ x = 1.1 $\times$ 10$-$9 M
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### JEE Main 2018 (Offline)

An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the formation of HS from H2S is 1.0 $\times$ 10–7 and that of S2- from HS ions is 1.2 $\times$ 10–13 then the concentration of S2- ions in aqueous solution is
A
5 $\times$ 10–19
B
5 $\times$ 10–8
C
3 $\times$ 10–20
D
6 $\times$ 10–21

## Explanation

HCl $\to$ H+ + Cl$-$

H+ concentration is = 0.2 M.

H2S $\rightleftharpoons$ H+ + HS$-$; k1 = 1.0 $\times$ 10$-$7

HS$-$ $\rightleftharpoons$ H+ + S2$-$; K2 = 1.2 $\times$ 10$-$13

H2S $\rightleftharpoons$ S2$-$ + 2H+

K = K1 $\times$ K2 = 1.0 $\times$ 10$-$7 $\times$ 1.2 $\times$ 1.0$-$13 = 1.2 $\times$ 10$-$20

as K1 and K2 both are very low for this reaction so dissociation of H2S and HS$-$ will be very low so, the produced H+ from this reaction will also be very low.

So, we can say the concentration of H+ will be almost same as H+ in HCl.

$\therefore\,\,\,$ [ H+ ] = 0.2 M.

From the reaction, H2S $\rightleftharpoons\,$ 2H+ + S2$-$

We get [ H+ ]2 [ S2$-$] = K $\times$ [ H2 S ]

$\Rightarrow \,\,\,$ [ S2$-$ ] = ${{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}$

$\Rightarrow \,\,\,$ [ S2$-$ ] = 3 $\times$ 10$-$20 M
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### JEE Main 2018 (Offline)

Which of the following are Lewis acids?
A
BCl3 and AlCl3
B
PH3 and BCl3
C
AlCl3 and SiCl4
D
PH3 and SiCl4

## Explanation

The compound which have the ability to accepted at least are lone pair electron.

Structure of BCl3 is Here B is electron deficient atom, so it can accepted lone pair. So it is a lewis acid.

Structure of AlCl3 Here, Al also a electron deficient atom so it has vacant orbital and in that vacant orbital it can take lone pair. So it is also lewis acid. Here in PH3 there is vacant 3d orbital but it Can't take. Lone pair in 3d orbital because P is more electro-negative than H so around P atom negative charge density is created and tendency of accepting electron decreases. So PH3 is not lewis acid. Here octet of Si full ut in has a tendency of accepting lone pair in vacant 3d orbital.

This can be shown by following reaction. So, here option (A) and (C) both are correct. But as SiCl4 is not as strong lewis acid as BCl3 and AlCl3, So we can say option (A) is more correct.