1

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

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pKb of ammonia solution is 4.75, the pH of the mixture will be :
A
3.75
B
4.75
C
8.25
D
9.25

## Explanation

NH3 + HCl $\to$ NH4Cl

moles of HCl = 0.2 M × 25 × 10–3 L = 0.005 moles HCl (total consumed)
moles of NH3 = 0.2 M × 50 × 10–3 L = 0.01 moles HCl
excess NH3 = 0.01 – 0.005 = 0.005 moles

From reaction, 1 mole ammonia = 1 mole NH4Cl

$\therefore$ 0.005 NH3 = 0.005 NH4Cl

Total Volume = VHCl + VNH3 = 25 + 50 = 75 mL

[NH3] = [NH4Cl] = ${{0.005} \over {75 \times {{10}^{ - 3}}}}$ = 0.066 M

pOH = pKb + log ${{\left[ {salt} \right]} \over {\left[ {base} \right]}}$

= pKb + log${{\left[ {N{H_4}Cl} \right]} \over {\left[ {N{H_3}} \right]}}$

= 4.75 + log${{\left[ {0.066} \right]} \over {\left[ {0.066} \right]}}$

$\Rightarrow$ pOH = 4.75

$\therefore$ pH = 14 – 4.75 = 9.25
2

### 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
3

### 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
4

### 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