1

### JEE Main 2017 (Online) 8th April Morning Slot

Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10$-$5, the ratio of salt to acid concentration in the buffer solution will be :
A
4 : 5
B
1 : 10
C
10 : 1
D
5 : 4

## Explanation

HA   $\rightleftharpoons$    H+ + A$-$

Ka = ${{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$ = 10$-$5

pH = pKa + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$\Rightarrow $$\,\,\, 6 = - log [10-5] + log {{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}} \Rightarrow$$\,\,\,$ 6 = 5 + log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$

$\Rightarrow$ log ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$ = 1

$\Rightarrow$$\,\,\,$ ${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$ = 10

$\therefore\,\,\,$ Salt : Acid = 10 : 1
2

### 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
3

### 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
4

### 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