[H₃O⁺] = 4 \(\times\) 10^-5M
pH = - log₁₀[H₃O⁺]
pH=-log₁₀[4 \(\times\) 10^-5]
pH = -log₁₀[4] - log₁₀[10^-5]        log₁₀ 10 = 1
pH = -log 4 + 5log₁₀10
= 5 - log 4
= 5 - 0.6021
=4.3979
Since pH is less than 7, the solution is acidic.

1. If \(\left[\mathrm{OH}^{-}\right]>1 \times 10^{-7} \mathrm{M}\), the solution is basic. \(2.5 \times 10^{-6} \mathrm{M}>1 \times 10^{-7} \mathrm{M}\)
2. \(\therefore\) The solution is basic.

(i)  According to Lewis theory a species that donates a pair of electron is called Lewis base.
(ii) Nitrogen more in NH₃ is less electro negative than oxygen in water. So the non - bonded electron pair on nitrogen is more available for sharing than a non - bonded electron pair on oxygen atom. So NH₃ is a stronger lewis base than H₂O.

\(\mathrm{HClO}_{4} \rightleftharpoons \mathrm{H}^{+}+\mathrm{ClO}_{4}^{-}\)
Bronsted Acid Proton Conjugate Base
HClO₄ can donate a proton. Therefore HClO₄ is an acid. Its conjugate base is ClO₄^-

Conjugate Pairs:
\((a) \mathrm{HS}_{\text {(aq) }}^{-} \& \mathrm{H}_{2} \mathrm{~S}_{\text {(aq) }} \) \((b) \mathrm{HF}_{\text {(aq) }} \& \mathrm{~F}_{\text {(aq) }}^{-} \)
\((a) \mathrm{HPO}_{4}^{2-} \& \mathrm{PO}_{4}^{3-} \) \((b) \mathrm{SO}_{3}^{2-} \& \mathrm{HSO}_{3}^{-} \)
\((a) \mathrm{NH}_{4}^{+} \& \mathrm{NH}_{3} \) \((b) \mathrm{CO}_{3}^{2-} \& \mathrm{HCO}_{3}^{-}\)