JEE Main 2019 — Ionic Equilibrium Question with Solution

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 $$\frac{\left[Salt\right]}{\left[base\right]}$$

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

= 4.7 + log2

= 4.7 + 0.3

= 5

$$\therefore$$   P^H = 14 $$-$$ P^OH

= 14 $$-$$ 5

= 9