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