A titration curve from the full equilibrium: which indicator finds the equivalence point? changedclaude-science/example-gallery/chem-titration
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99## Prompt
1010
11Copy the prompt and paste it into a new session in [Claude Science](/docs/claude-science/overview). By default, Claude asks with a [permission card](/docs/claude-science/core-concepts#permission-cards) before it runs code or installs packages.
11Copy the prompt and paste it into a new session in [Claude Science](/docs/claude-science/overview.md). By default, Claude asks with a [permission card](/docs/claude-science/core-concepts.md#permission-cards) before it runs code or installs packages.
1212
1313```text Prompt wrap theme={null}
1414I'm titrating 25.0 mL of 0.100 M acetic acid (pKa 4.76) with 0.100 M NaOH. Compute the pH after each addition of base, from 0 to 50 mL, by solving the full equilibrium (mass balance, charge balance, Kw = 1.0 × 10⁻¹⁴) without the usual shortcuts (no Henderson–Hasselbalch, no 'x is small'), using concentrations, at 25 °C. Plot the titration curve and mark the starting point, the half-equivalence point and the equivalence point. Show that the acid alone sets the start, pH = pKa half-way, and acetate acting as a weak base sets the equivalence point.
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