Acid-Base Chemistry & pH
Proton transfer, titration curves, and buffer systems
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How it works
Acids donate protons (H⁺); bases accept protons (Brønsted-Lowry definition). Strong acids (HCl, H₂SO₄, HNO₃) dissociate completely: [H⁺] = initial acid concentration. Weak acids partially dissociate; Ka = [H⁺][A⁻]/[HA]. pH = -log[H⁺]. At 25°C, Kw = [H⁺][OH⁻] = 10⁻¹⁴; neutral pH = 7. Titration: adding strong base to acid gradually neutralizes it. The equivalence point is where moles of base = moles of acid (pH = 7 for strong/strong, > 7 for weak acid/strong base). The buffer region (±1 unit from pKa) resists pH changes. Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). Maximum buffer capacity when [A⁻] = [HA] (pH = pKa). Quotable anchors: pH is defined as pH = −log[H⁺], and at 25 °C pure water (Kw = [H⁺][OH⁻] = 10⁻¹⁴) reads exactly pH 7. Same concentration does not mean same pH — at 0.100 M this sim reads pH 1.00 for HCl (complete ionization, every proton released) but pH 2.88 for acetic acid (Ka = 1.8×10⁻⁵, only about 1.3% ionized). At half-equivalence of a weak-acid titration [A⁻] = [HA], so Henderson–Hasselbalch reduces to pH = pKa = 4.74 for acetic acid — in this sim that point is 25.0 mL of 0.100 M NaOH into 50.0 mL of 0.100 M CH₃COOH, and the equivalence point at 50.0 mL reads pH ≈ 8.7, above 7 because the leftover acetate is a weak base. Digit note: this sim solves the full quadratic x² + Ka·x − Ka·C = 0 for the weak-acid pH (sim.js:182), pricing 0.100 M acetic acid at pH 2.875 (displayed 2.88), while the textbook approximation x ≈ √(Ka·C) gives 2.87 — the difference is the solution method, not an error, so a hand-worked 2.87 and the screen's 2.88 are the same answer.
Step-by-step
- Start with an acid solution and observe the initial pH.
- Move the NaOH slider to perform a titration — watch the S-shaped titration curve.
- Identify: the initial pH (from Ka), the buffer region (flat slope), the equivalence point (steep jump), and the post-equivalence rise.
- Add conjugate base (Pro) to design a buffer solution.
Key formulas
- pH definition and water's ion product at 25°C
- Buffer pH — Henderson-Hasselbalch equation
- Ka defines weak acid strength (CH₃COOH: Ka = 1.8×10⁻⁵)
Frequently asked questions
- What is the pH of 0.01 M HCl?
- HCl is a strong acid → fully dissociates. [H⁺] = 0.01 M. pH = -log(0.01) = 2.
- Calculate the pH of 0.1 M acetic acid (Ka = 1.8×10⁻⁵).
- Ka = x²/(0.1-x) ≈ x²/0.1. x = √(0.1×1.8×10⁻⁵) = √(1.8×10⁻⁶) = 1.34×10⁻³ M. pH = -log(1.34×10⁻³) ≈ 2.87.
- Design a buffer with pH = 5.0 using acetic acid (pKa = 4.74). What ratio of [CH₃COO⁻]/[CH₃COOH] is needed?
- PH = pKa + log([A⁻]/[HA]). 5.0 = 4.74 + log(ratio). log(ratio) = 0.26. Ratio = 10^0.26 = 1.82. So [CH₃COO⁻]/[CH₃COOH] ≈ 1.82.
- 50 mL of 0.1 M CH₃COOH is titrated with 0.1 M NaOH. What is the pH at the equivalence point, and why?
- At equivalence, all acid is converted to CH₃COO⁻ (acetate). 0.005 mol CH₃COO⁻ in 100 mL = 0.05 M. Kb(CH₃COO⁻) = Kw/Ka = 10⁻¹⁴/1.8×10⁻⁵ = 5.6×10⁻¹⁰. [OH⁻] ≈ √(0.05×5.6×10⁻¹⁰) = 5.3×10⁻⁶. pOH = 5.3, pH = 8.7 (basic, as expected).