Define the following terms. i) Lewis base ii) Basic buffer [2] Calculate the pH of solution obtained by mixing equal volumes of two solutions having pH = 3 and pH = 5 . [

i) Lewis Base

A Lewis base is a chemical species that donates an electron pair to form a coordinate covalent bond. Unlike Brønsted–Lowry bases (which donate protons), Lewis bases focus on electron pair donation. They contain lone pairs of electrons or π-electrons that can be shared with a Lewis acid (an electron pair acceptor).

Examples:

• Ammonia (NH₃): Donates a lone pair on nitrogen to a proton (H⁺) or metal ion (e.g., Ag⁺).
• Water (H₂O): Acts as a Lewis base by donating lone pairs to H⁺ (forming H₃O⁺) or to BF₃ (forming H₂O·BF₃).
• Carbon monoxide (CO): Donates π-electrons to metal centers in coordination complexes (e.g., Ni(CO)₄).

Key Point: Lewis bases are not limited to hydroxide ions (OH⁻) or proton acceptors; they include molecules/ions with available electron pairs, even if they lack acidic hydrogens.

---

ii) Basic Buffer

A basic buffer is a solution that resists changes in pH when small amounts of acid or base are added, with a pH > 7. It consists of:

1. A weak base (e.g., NH₃, CH₃NH₂) and its conjugate acid (e.g., NH₄⁺, CH₃NH₃⁺), or
2. A strong base (OH⁻) combined with its conjugate weak acid (e.g., CH₃COO⁻ from CH₃COOH).

Mechanism:

• Resists acid addition: The weak base neutralizes added H⁺ (e.g., NH₃ + H⁺ → NH₄⁺).
• Resists base addition: The conjugate acid neutralizes added OH⁻ (e.g., NH₄⁺ + OH⁻ → NH₃ + H₂O).

Example: A buffer made from NH₃ (weak base) and NH₄Cl (source of NH₄⁺) maintains pH ~9.25 (pK_{a} of NH₄⁺ = 9.25).

Comparison with Acidic Buffer:

---

Calculation: pH of Mixed Solutions (pH = 3 and pH = 5)

Given:

• Equal volumes of solutions with pH = 3 and pH = 5 are mixed.

Step 1: Relate pH to [H⁺] $$\text{pH}=-\log [{\text{H}}^{+}]⟹[{\text{H}}^{+}]=10^{-\text{pH}}$$

• For pH = 3: $[{\text{H}}^{+}{]}_1=10^{-3}\text{M}$
• For pH = 5: $[{\text{H}}^{+}{]}_2=10^{-5}\text{M}$

Step 2: Assume Volume = 1 L (for simplicity)

• Moles of H⁺ from pH 3 solution: $n_1=10^{-3}\text{mol}$
• Moles of H⁺ from pH 5 solution: $n_2=10^{-5}\text{mol}$

Step 3: Total Volume and Moles After Mixing

• Total volume = 1 L + 1 L = 2 L
• Total moles of H⁺ = $n_1+n_2=10^{-3}+10^{-5}=1.01\times 10^{-3}\text{mol}$

Step 4: Calculate New [H⁺] $$[{\text{H}}^{+}{]}_{\text{mix}}=\frac{1.01\times 10^{-3}\text{mol}}{2\text{L}}=5.05\times 10^{-4}\text{M}$$

Step 5: Compute pH $$\text{pH}=-\log (5.05\times 10^{-4})\approx 3.30$$

Final Answer: The pH of the mixed solution is 3.30.