Two Equivalence Points but One Indicator
A student titrates 0.250 g of an unknown solid acid with 0.100 M NaOH. The acid dissolves in 50.0 mL of distilled water. Using a pH meter, the student records a first steep pH rise at 12.5 mL of NaOH (pH 4.2) and a second steep rise at 25.0 mL (pH 8.8). The student then repeats the titration with phenolphthalein, observes the first permanent pink color at 25.0 mL, and calculates the molar mass using the first equivalence volume from the pH curve: 0.250 g / (0.100 mol/L × 0.0125 L) = 200 g/mol. The student concludes: 'The acid is monoprotic because 12.5 mL is the equivalence point, and phenolphthalein confirms it.' The teacher must decide whether the student's conclusion and indicator choice are consistent with the data.
1. Interpret the titration curve. Explain what the two steep rises indicate about the acid and determine the number of acidic protons per molecule.
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Two steep rises mean two equivalence points. The first occurs at 12.5 mL and the second at 25.0 mL, exactly twice the volume. For a diprotic acid H2A, the first equivalence consumes one mole of OH− per mole of acid (H2A → HA−) and the second consumes a second mole (HA− → A2−). The 1:2 volume ratio therefore indicates two acidic protons per molecule. The pH at the first equivalence (4.2) is acidic, consistent with an amphiprotic HA− species; the second equivalence (8.8) is basic, consistent with hydrolysis of A2−.
2. Evaluate the student's molar-mass calculation and conclusion. Is the numerical value 200 g/mol correct? Is the acid monoprotic?
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The numerical molar mass is correct but for a different reason than the student thinks. At the first equivalence, mol NaOH = 0.100 mol/L × 0.0125 L = 0.00125 mol. If the acid is diprotic, this corresponds to one proton per acid molecule, so mol acid = 0.00125 mol. Molar mass = 0.250 g / 0.00125 mol = 200 g/mol. At the second equivalence, mol NaOH = 0.00250 mol, which is two moles OH− per mole acid; mol acid = 0.00250/2 = 0.00125 mol, giving the same 200 g/mol. Thus the value 200 g/mol is consistent with a diprotic acid, but the student's conclusion that the acid is monoprotic is wrong. The equivalent weight is 100 g per acidic proton.
3. The student used phenolphthalein and observed the color change at 25.0 mL. Explain whether phenolphthalein is appropriate for detecting the first or second equivalence point, and recommend a better indicator or method for the first equivalence.
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Phenolphthalein changes over pH 8.2–10.0, which matches the second equivalence at pH 8.8, not the first at pH 4.2. The student's observation of pink at 25.0 mL is therefore evidence for the second equivalence, and it does not confirm the first equivalence volume of 12.5 mL. For the first equivalence, choose an indicator whose transition range brackets pH 4.2, such as methyl orange (3.1–4.4) or bromocresol green (3.8–5.4), or use a pH meter and locate the maximum slope. The indicator must match the steep portion of the curve; using phenolphthalein to stop at the first equivalence would give a late, inaccurate endpoint.