Acids and Bases

Question 1

Bronsted-Lowry Definition

Answer 1

• B-L Acid: Species that donates hydrogen ions.
• B-L Base: Species that accept hydrogen ions.
• Form Conjugate Acid-Base Pairs.

Question 2

Lewis Definition

Answer 2

• Lewis Acid: Electron pair acceptor.

* Lewis Base: Electron pair donor.

Question 3

Amphoteric and Amphiprotic

Answer 3

• Amphoteric: Species that reacts like an acid in a basic environment and like a base in an acidic environment.
• Amphiprotic: Amphoteric species in B-L sense that can both donate and accept hydrogen ions (protons).
• Water, bicarbonate ion, and amino acid zwitterion are amphoteric.

Question 4

Water

Answer 4

• Amphoteric water molecule can donate its hydrogen to another water molecule to produce Hydronium (H₃O+) and Hydroxide (OH-).
• Autoionization of water is reversible and proceeds to equilibrium and can be represented by Kw.
• Kw = [H₃O+][OH-] = 1.0 × 10⁻¹⁴ at 25°C (298K).
• [H₃O+] = 10⁻⁷ M.
• Water dissociation constant changes only when the temperature changes; isolated changes in concentration, pressure, volume will not affect Kw.

Question 5

pH and pOH

Answer 5

• pH = -log[H+].
• pOH = -log[OH-].
• pH + pOH = 14.
• For concentration n × 10⁻ᵐ.
• Log Estimation: p value = m - 0.n.

Question 6

Weak Acids and Bases

Answer 6

• Weak acid HA will dissociate partially in solution.
• HA + H₂O -> H₃O+ + A-.
• K(a) = [H₃O+][A-] / [HA], where K(a) is Acid Dissociation Constant.
• Weak acids have K(a) < 1.
• Weak base BOH will dissociate partially in solution.
• BOH + H₂O -> B+ + OH-.
• K(b) = [B+][OH-] / [BOH], where K(b) is Base Dissociation Constant.
• Weak bases have K(b) < 1.
• K(a,acid) × K(b,conjugate base) = K(w) = 10⁻¹⁴.
• K(b,base) × K(a,conjugate acid) = K(w) = 10⁻¹⁴.

Question 7

Titration

Answer 7

• Titrations are performed by adding small volumes of Titrant (solution of known concentration) to a known volume of Titrand (solution of unknown concentration).
• Equivalence Point is reached when the number of acid equivalents present in the original solution equals the number of base equivalents added, or vice-versa.
• At the Equivalence Point, the number of equivalents of acid and base are equal: N(a) V(a) = N(b) V(b).
• Titration Curve has pH of Titrand on y-axis and volume of Titrant added on x-axis.
• Indicator is a weak acid or base that changes color at Endpoint (not Equivalence Point). Good indicator will have Endpoint close to Equivalence Point.
• Strong Acid + Strong Base: pH = 7.
• Strong Acid + Weak Base: EP pH < 7.
• Weak Acid + Strong Base: EP pH > 7.

Question 8

Strong Acid + Strong Base

Answer 8

• Titration curve shows Strong Acid Titrand (initial pH ≪ 7) and Strong Base Titrant.
• Strong Base Titrand (initial pH ≫ 7) and Strong Acid Titrant would have an inverted curve.
• Equivalence Point at pH 7.

Question 9

Weak Acid + Strong Base

Answer 9

• Titration curve shows Weak Acid Titrand (initial pH < 7) and Strong Base Titrant.
• Equivalence Point at pH > 7, since conjugate base is stronger than conjugate acid, which results in excess hydroxide ions at equilibrium.

Question 10

Strong Acid + Weak Base

Answer 10

• Titration curve shows Weak Base Titrand (initial pH > 7) and Strong Acid Titrant.
• Equivalence Point at pH < 7, since conjugate acid is stronger than conjugate base, which results in excess hydronium ions at equilibrium.

Question 11

Buffers

Answer 11

• Acetic Acid (CH3COOH) and Sodium Acetate (NaCH3COO) Buffer: CH3COOH + H2O -> H3O+ + CH3COO-.
• Strong Base (NaOH) dissociates to produce OH-, which reacts with hydronium ions to be neutralized. As hydronium is used up, system shifts to the right to produce more hydronium and acetate to replenish lost hydronium. Hydroxide ion increase caused by increased acetate concentration is not as large as that caused by unbuffered NaOH, resulting in little change to pH.
• Strong Acid (HCl) dissociates to produce H3O+, which reacts with acetate ions to produce acetic acid. Hydroxide ion increase caused by increased acetic acid concentration is not as large as that caused by unbuffered HCl, resulting in little change to pH.
• Ammonia (NH3) and Ammonium Chloride (NH4Cl) Buffer: NH3 + H2O -> NH4+ + OH-.
• Strong Acid reduces [OH-], shifting system to the right.
• Strong Base raises [OH-], shifting system to the left.
• Buffering Capacity: pK(a) ± 1.
• Buffering capacity is optimal at half-equivalent point.
• Buffering capacity can be increased by increasing both acid and conjugate base to maintain same ratio.

Question 12

Henderson-Hasselbalch Equation

Answer 12

• To find pH in weak acid buffer system, pH = pK(a) + log [A-]/[HA].
• Half-Equivalence Point: When [A-] = [HA], pH = pK(a).
• To find pOH in weak base buffer system, pOH = pK(b) + log [B+]/[BOH].
• Half-Equivalence Point: When [B+] = [BOH], pOH = pK(b).