Titration Curve Plotter

The equivalence point is where the moles of titrant added are stoichiometrically equal to the moles of analyte — the acid and base have completely reacted, in the proportions set by the balanced equation. Mathematically: n_acid × M_acid × V_acid = n_base × M_base × V_base, where n is the stoichiometric coefficient of each species. For 1:1 reactions like HCl + NaOH this collapses to M_acid V_acid = M_base V_base. The pH at the equivalence point is not always 7. Strong acid + strong base → pH = 7. Weak acid + strong base → pH > 7 (the conjugate base hydrolyzes). Strong acid + weak base → pH < 7. Weak acid + weak base → depends on the relative Ka and Kb. Equivalence point vs endpoint: the equivalence point is the theoretical exact point; the endpoint is what the experimenter observes (an indicator color change or the inflection on a pH meter trace). The small gap between the two is the titration error. Worked example: 20.0 mL of HCl titrated to equivalence with 25.0 mL of 0.100 M NaOH gives M(HCl) = (0.100 × 25.0) / 20.0 = 0.125 M.

Use the stoichiometric volumetric equation n1 M1 V1 = n2 M2 V2, where n1 and n2 are the coefficients of the two species in the balanced reaction. For 1:1 reactions the n's cancel and you can use M1 V1 = M2 V2 directly. Procedure: balance the neutralization equation, identify which quantity is unknown, plug in known molarities and volumes, and solve. Keep volumes in consistent units (both mL or both L). Example 1 — 1:1 reaction. How much 0.100 M NaOH neutralizes 25.0 mL of 0.0500 M HCl? 0.0500 × 25.0 = 0.100 × V → V = 12.5 mL. Example 2 — 1:2 reaction (NaOH neutralizing H2SO4). 30.0 mL of NaOH neutralizes 25.0 mL of 0.100 M H2SO4. 2 × M_NaOH × 30.0 = 1 × 0.100 × 25.0 → M_NaOH = 2.5 / 60 = 0.0417 M. For the pH at the equivalence point, use the appropriate hydrolysis: for a weak-acid / strong-base titration, [OH-] ≈ √(Kb_salt × C_salt) where Kb_salt = Kw / Ka_acid; for strong-acid / weak-base, [H+] ≈ √(Ka_salt × C_salt) where Ka_salt = Kw / Kb_base.

The pKa of a weak acid equals the pH at the half-equivalence point of its titration with a strong base. Reasoning: the Henderson-Hasselbalch equation says pH = pKa + log([A-] / [HA]). At half-equivalence exactly half the acid has been deprotonated, so [HA] = [A-], log(1) = 0, and pH = pKa. Procedure: titrate the weak acid with a standardized strong base while recording pH; locate the equivalence point (the steep inflection); the half-equivalence volume is half of that; read the pH at the half-equivalence volume off the curve. That pH is the pKa. Worked example: a weak acid is titrated to equivalence at 20.0 mL of 0.100 M NaOH. Half-equivalence is at 10.0 mL; the pH at that point is 4.75, so pKa = 4.75 and Ka = 10^-4.75 ≈ 1.78 × 10^-5 — consistent with acetic acid. For polyprotic acids, each plateau gives a separate pKa: H3PO4 shows three half-equivalence regions at approximately pKa1 = 2.15, pKa2 = 7.20, pKa3 = 12.35. Bonus: at pH = pKa the buffer is at maximum capacity, which is why this point is also the optimal operating pH for a buffer based on that acid-conjugate-base pair.

It depends on the strengths of the acid and base. Strong acid + strong base: the only species at the equivalence point is a neutral salt, so pH = 7.00 at 25 °C. Weak acid + strong base: the salt's anion is the conjugate base of the weak acid; it hydrolyzes (A- + H2O ⇌ HA + OH-) with Kb = Kw / Ka. Approximate [OH-] ≈ √(Kb × C_salt), then pOH = -log[OH-] and pH = 14 - pOH. Example: 25.0 mL of 0.100 M CH3COOH (pKa = 4.76) titrated with 0.100 M NaOH gives equivalence at 25.0 mL, total volume 50.0 mL, C_salt = 0.0500 M, Kb = 10^-14 / 1.74 × 10^-5 = 5.75 × 10^-10, [OH-] ≈ √(5.75 × 10^-10 × 0.0500) = 5.36 × 10^-6, pOH = 5.27, pH = 8.73. Strong acid + weak base: the salt's cation is the conjugate acid of the weak base; it hydrolyzes (BH+ + H2O ⇌ B + H3O+) with Ka = Kw / Kb. Approximate [H+] ≈ √(Ka × C_salt). Example: 25.0 mL of 0.100 M NH3 (Kb = 1.8 × 10^-5) titrated with 0.100 M HCl gives Ka = 5.56 × 10^-10, [H+] ≈ √(5.56 × 10^-10 × 0.0500) = 5.27 × 10^-6, pH = 5.28. Weak acid + weak base: pH at equivalence ≈ 0.5 × (pKw + pKa - pKb), so the result depends on the relative strengths. The pH may be near 7 only when pKa ≈ pKb. Indicator choice follows the equivalence pH: phenolphthalein (pKa range 8.2-10) for WA+SB; methyl orange (3.1-4.4) or methyl red (4.4-6.2) for SA+WB.

The endpoint is the experimentally observed signal that the titration is complete — most commonly the volume at which the indicator changes color. It is conceptually distinct from the equivalence point: the equivalence point is the stoichiometric exact point determined by the reaction, and the endpoint is the operator's best detection of it. The discrepancy between the two is the indicator (or titration) error. Common acid-base indicators and their transition ranges: methyl orange 3.1-4.4 (red → yellow); methyl red 4.4-6.2 (red → yellow); bromothymol blue 6.0-7.6 (yellow → blue); phenol red 6.4-8.2 (yellow → red); phenolphthalein 8.2-10.0 (colorless → pink). Choose the indicator so its transition range straddles the expected equivalence pH: phenolphthalein for SA+SB (any indicator works, but phenolphthalein is convenient) and for WA+SB; methyl orange or methyl red for SA+WB. WA+WB systems generally need a pH meter rather than a visual indicator. For non-acid-base titrations, endpoint signals differ: a self-indicating reagent (KMnO4 stays purple at the first drop in excess); an external indicator (Mohr's method uses K2CrO4 to give brick-red Ag2CrO4 at the end of a chloride/silver titration); a starch-iodine complex turning blue-black at the end of an iodometric titration; or a sharp pH or potential step on a meter trace.