pH=-log[H⁺]
For weak acids,
\(= \sqrt{k_a \times C}\)
=\(\sqrt{1.8\times10^{-5}\times0.1}\)
=\(1.34 \times10^{-3}\) M
\(pH=-\log(1.34\times10^{-3})\)
= 3-log1.34
= 3-0.1271
= 2.8729 \(\simeq\) 2.87

Given that \(\alpha=1.20\)%=\(\frac{1.20}{100}\times1.2\times10^{-2}\)
\(K_{a}=\alpha^{2}c\)
\(=(1.2\times10^{-2})^{2}(0.1)=1.44\times10^{-4}\times10^{-1}\)
=\(1.44\times10^{-5}\)

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.
Hence [H₃O⁺] = 0.001 mol dm^-3
pH = -log₁₀ [H₃O⁺]
= -log₁₀^(0.001)
= -log₁₀(10^-3) = 3

Given that H₃O⁺ = \(2\times10^{-3}M\)
\(K_{w}=[H_{3}O^{+}][OH^{-}]\)
\(\therefore [OH^{-}]=\frac{K_{w}}{[H_{3}O^{+}]}=\frac{1\times10^{-14}}{2\times10^{-3}}=0.5\times10^{-11}M\)
\(2\times10^{-3} >>0.5\times10^{-11}\)
i.e., [H₃O⁺]>>[OH^-], hence the juice is acidic in nature

K_sp = 1.8 \(\times\)10^-10, [AgNO₃]= 1 M
\(\mathrm{AgCl}_{(\mathrm{s})} \rightleftharpoons \mathrm{Ag}_{(\mathrm{aq})}^{+}+\mathrm{Cl}_{(\mathrm{aq})}^{-} \)
                    s            s
\(\mathrm{AgNO}_{3(\mathrm{aq})} \rightleftharpoons \mathrm{Ag}_{(\text {aq })}^{+}+\mathrm{NO}_{3_{(\text {aq })}}^{-}\\ 1 \mathrm{M} \quad \quad \quad \quad \quad 1 \mathrm{M} \quad \quad 1 \mathrm{M} \)
\(\left[\mathrm{Ag}^{+}\right]=(\mathrm{s}+1) \approx 1 \quad(\therefore \mathrm{s}<<1) \)
\(\left[\mathrm{Cl}^{-}\right]=\mathrm{s} \)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \)
\(1.8 \times 10^{-10}=(1)(s) \)
\(\therefore \mathrm{s}=1.8 \times 10^{-10} \mathrm{M}\)

(i) p^H = -log₁₀ [H₃O⁺]
(ii) The pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

The solubility product of a compound is defined as the product of the molar concentration of the constituent ions, each raised to the power of its stoichiometric co - efficient in a balanced equilibrium equation.
X_mY_n (s) \(\rightleftharpoons\) mXⁿ⁺ _(aq) + nY^m-_(aq)
K_sp=[Xⁿ⁺]^m[Y^m-]ⁿ

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

(i) Lewis acid: It is a species that accepts an electron pair. Eg: \(\mathrm{Ag}^{+} ; \mathrm{BF}_{3} ; \mathrm{A} / \mathrm{Cl}_{3}\)
(ii) Lewis base: It is a species that donates an electron pair. Eg: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)