QJEEM19Chad87: JEE Main 2019 Chemistry - Class 11 JEEM19Chad87 Solution: 20 mL of 0.1 M H 2 SO 4 solution is added to 30 mL of of 0.2 M…

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.

$\therefore$    P^OH = P^K_b + log ${{\left[ {Salt} \right]} \over {\left[ {base} \right]}}$

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

= 4.7 + log2

= 4.7 + 0.3

= 5

$\therefore$    P^H = 14 $-$ P^OH

= 14 $-$ 5

= 9