Titration with a water-rinsed burette
A student titrates 10.0 mL of hydrochloric acid of unknown concentration with 0.10 mol/L sodium hydroxide solution. Phenolphthalein is added to the acid in a conical flask. The burette is rinsed with distilled water but not with the sodium hydroxide solution before filling. Three trials give endpoints at 11.8 mL, 12.0 mL, and 12.2 mL. The student uses the average, 12.0 mL. A classmate notices that the third trial turned pale pink after only a brief swirl and then faded. The teacher asks the student to calculate the acid concentration, to judge whether the water rinse makes the result too high or too low, and to explain the endpoint in terms of ions. The reaction is HCl + NaOH → NaCl + H2O.
1. Calculate the concentration of the hydrochloric acid using the 12.0 mL average. Show the mole relationship.
Read the worked answer
The amount of sodium hydroxide used is n = cV = 0.10 mol/L × 0.0120 L = 1.20 × 10^-3 mol. From HCl + NaOH → NaCl + H2O, hydrochloric acid and sodium hydroxide react in a 1:1 mole ratio. Therefore the acid contains 1.20 × 10^-3 mol in 10.0 mL, or 0.0100 L. Its concentration is c = n/V = 1.20 × 10^-3 mol / 0.0100 L = 0.120 mol/L. The calculation assumes the sodium hydroxide concentration is exactly 0.10 mol/L and that 12.0 mL is the true endpoint volume.
2. Does the water rinse make the calculated acid concentration too high, too low, or unchanged? Explain.
Read the worked answer
The calculated acid concentration would be too high. Rinsing the burette with distilled water leaves water in the burette, and this dilutes the sodium hydroxide solution when it is added. The diluted sodium hydroxide has a concentration lower than 0.10 mol/L, so a larger volume is needed to supply the same number of moles of OH^- ions. The student, however, calculates the moles of NaOH using the assumed concentration 0.10 mol/L and the larger volume. That overcounts the moles of NaOH and therefore overcounts the moles of HCl. The final acid concentration is reported higher than the true value. Rinsing with the sodium hydroxide solution itself would avoid this error.
3. Write the ionic equation for the neutralization and explain why phenolphthalein changes color near the endpoint. Why is the pH not necessarily exactly 7 at the first pale pink?
Read the worked answer
The ionic equation is H^+ + OH^- → H2O. Sodium and chloride ions are spectator ions and are not included. Phenolphthalein is colorless in acidic solution and pink in basic solution. Near the endpoint, almost all the H^+ has been neutralized, and a slight excess of OH^- makes the solution basic, so the indicator turns pale pink. The first pale pink does not have to appear at pH exactly 7; it appears when the indicator's color-change range is reached. For a strong acid and strong base, the equivalence point is at pH 7, but a very small excess of sodium hydroxide can push the pH above 7. The fading in the third trial suggests the endpoint was judged too early, before the color persisted.