Ostwald's dilution law is applicable only for weak electrolyte and does not hold good for concentrated solution.

The conjugate base of HF is F^-
For F^-, \({ K }_{ b }=\frac { { K }_{ w } }{ { K }_{ a } } =\frac { { 10 }^{ -14 } }{ { 6.8\times 10 }^{ -14 } } \)
= 1.47 x 10^-11.

pK_w = pH + pOH = 10^-12
For a neutral solution, pH = pOH
So pH = pOH = 6.

\(\underset{0.001M}{HCl}\overset{H_{2}O}{\rightleftharpoons }\underset{0.001M}{H_{3}O^{+}}+\underset{0.001M}{Cl^{-}}\)
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

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

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

(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: \( \mathrm{Cl}^{-} ; \mathrm{NH}_{3} ; \mathrm{H}_{2} \mathrm{O}\)

(i) An acid is defined as a substance that has a tendency to donate a proton to another substance and base is a substance that has a tendency to accept a proton form other substance.
(ii) In other words, an acid is a proton donor and a base is a proton acceptor.
(iii) When hydrogen chloride is dissolved in water, it donates a proton to the later. Thus, HCI behaves as an acid and H₂O is base. The proton transfer from the acid to base can be represented as
HCI + H₂O ⇌ H₃O⁺ + Cl^-
(iv) When ammonia is dissolved in water, it accepts a proton from water. In this case, ammonia (NH₃) acts as a base and H₂O is acid. The reaction is represented as
H₂O + NH₃ ⇌ NH₄⁺ + OH^-
(v) Let us consider the reverse reaction following equilibrium.
\(\underset { proton\ donar\\ \quad \quad \ (acid) }{ HCl } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad (base) }{ { H }_{ 2 }O } \leftrightharpoons \underset { Proton\ donar\\ \quad \quad \quad \quad \ (acid) }{ { H }_{ 2 }{ O }^{ + } } +\underset { Proton\ acceptor\\ \quad \quad \quad \quad \ (base) }{ { Cl }^{ - } } \) 
H₃O⁺ donates a proton to Cl^- to form HCI i.e., the products also behave as acid and base.
(vi) In general, Lowry - Bronsted (acid - base) reaction is represented as 
Acid₁ + Base₂ ⇌ Acid₂ + Base₁
(vii) The species that remains after the donation of a proton is a base (Base₁) and is called the conjugate base of the Bronsted acid (Acid₁). In other words, chemical species that differ only by a proton are called conjugate acid - base pairs.

If we do not consider [H₃O]⁺ from the ionisation of H₂O,
then [H₃O⁺] = [HCl] = 10^-7M
i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7M) Hence, the H₃O⁺ (10^-7M) formed due to the auto ionisation of water cannot be neglected.
so, in this case we should consider  [H₃O⁺] from ionisation of H₂O
[H₃O⁺] = 10^-7 (from HCl) + 10^-7 (from water)
= 10^-7 (1+1)
= \(2\times10^{-7}\)
pH = -log₁₀[H₃O⁺]
=\(-\log_{10}(2\times10^{-7})=-[\log2+\log_{10}10^{-7}]\)
=\(-\log2-(-7)\log_{10}^{10}\)
= 7-log²
= 7-0.3010 = 0.6990 = 6.70
= 6.70

\(\text { Normality }=\text { Molarity } \times \text { Basicity }  \)
\(=0.04 \times 1  \)
\({\left[\mathrm{H}_{3} \mathrm{O}\right]^{+}=0.04=4 \times 10^{-2} } \)
\(\mathrm{pH} =-\log _{10}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]  \)
\(=-\log \left[4 \times 10^{-2}\right]  \)               log₁₀ 10 =1
\(=-\left[\log _{10} 4+\log _{10} 10^{-2}\right]  \)
\(=-\left[\log _{10} 4-2 \log _{10} 10\right]=2-\log _{10} 4  \)
= 2 - 0.6021 
= 1.3979 \(\simeq\) 1.40