1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
Addition of CO2
D
Addition of Fe2O3
3
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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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