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Pro 🔒~18 min

Acid-Base Chemistry & pH

Proton transfer, titration curves, and buffer systems

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).

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Step-by-step

  1. Start with an acid solution and observe the initial pH.
  2. Move the NaOH slider to perform a titration — watch the S-shaped titration curve.
  3. Identify: the initial pH (from Ka), the buffer region (flat slope), the equivalence point (steep jump), and the post-equivalence rise.
  4. Add conjugate base (Pro) to design a buffer solution.

Key formulas

  • pH=−log⁡[H+]Kw=[H+][OH−]=10−14\text{pH} = -\log[\text{H}^+] \quad K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}pH definition and water's ion product at 25°C
  • pH=pKa+log⁡[A−][HA](Henderson-Hasselbalch)\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]} \quad (\text{Henderson-Hasselbalch})Buffer pH — Henderson-Hasselbalch equation
  • Ka=[H+][A−][HA](weak acid dissociation constant)K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \quad (\text{weak acid dissociation constant})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).