Acid/Base Theory
Define a Brønsted-Lowry acid.
A proton (H⁺) donor.
What is the formula for pH?
pH = –log[H⁺]
What does Ka measure?
The strength of an acid (how much it dissociates).
What is the equivalence point?
When moles of acid = moles of base added.
What is a buffer?
Solution that resists pH change when small amounts of acid/base are added.
What’s the conjugate base of HSO₄⁻?
SO₄²⁻
If [OH⁻] = 1.0 × 10⁻⁶ M, what is the pOH and pH?
pOH = 6, pH = 8 (since 14 – 6 = 8)
What does a small Ka value mean?
Weak acid; low degree of ionization.
True/False: pH = 7 at equivalence for all titrations.
False — only for strong acid–strong base.
What two components make up a buffer?
Weak acid + conjugate base (or weak base + conjugate acid)
Identify the acid and base: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
Base: NH₃; Acid: H₂O
A solution has pH = 3. Is it acidic or basic?
Acidic (pH < 7)
Write the Ka expression for HF ⇌ H⁺ + F⁻
Ka = [H⁺][F⁻] / [HF]
Sketch pH curve for weak acid + strong base. What happens at equivalence?
pH > 7 at equivalence; curve rises gradually, then sharp jump.
Which pair is a buffer: HCl + NaCl or CH₃COOH + CH₃COONa?
CH₃COOH + CH₃COONa
How does a Lewis acid differ from a Brønsted acid?
Lewis acid accepts electron pairs; Brønsted acid donates H⁺.
Calculate [H⁺] from pH = 5.2
[H⁺] = 10⁻⁵.² ≈ 6.3 × 10⁻⁶ M
If Ka of HA = 1.0 × 10⁻⁵, what’s the pKa?
pKa = –log(Ka) = 5.0
During titration of NH₃ with HCl, what is the initial pH like?
Basic (NH₃ is a weak base).
What is the Henderson-Hasselbalch equation?
pH = pKa + log([A⁻]/[HA])
Explain amphoteric behavior and give an example.
Substances that act as acid or base; example: H₂O, HCO₃⁻.
If a solution has pOH = 12.7, what is the [OH⁻]?
[OH⁻] = 10⁻¹².⁷ ≈ 2.0 × 10⁻¹³ M
Explain how adding common ion affects weak acid equilibrium.
Shifts equilibrium left; suppresses ionization (Le Châtelier).
Calculate volume of 0.250 M NaOH needed to neutralize 25.0 mL of 0.150 M HCl.
M₁V₁ = M₂V₂ → V = (0.150 × 25.0) / 0.250 = 15.0 mL
A buffer has 0.40 M NH₃ and 0.25 M NH₄⁺; pKa = 9.25. What is pH?
pH = 9.25 + log(0.40/0.25) = 9.25 + 0.20 = 9.45