Three Unlabelled Solutions and a Warm Titration Flask
A laboratory group is given three colorless solutions labelled X, Y and Z. They are told each contains one of: dilute hydrochloric acid, sodium chloride solution, or sodium hydroxide solution, but the labels are missing. The students record: X turns universal indicator red-orange (pH ≈ 2); Y turns it green (pH ≈ 7); Z turns it blue-violet (pH ≈ 12). When a few marble chips (calcium carbonate) are added to each, only X produces bubbles that turn limewater milky. A 25.0 mL sample of X is titrated with Z, using phenolphthalein. The endpoint is reached after 20.0 mL of Z has been added. The teacher asks whether the result proves X is monoprotic and whether the reaction is a neutralization. A classmate suggests that because Z feels slippery and Y is salty, Y must be the sodium chloride. The students also notice the titration flask becomes warm. Use the concentration of Z as 0.100 mol/L.
1. Identify X, Y and Z and justify each using two independent observations.
Read the worked answer
X is the hydrochloric acid: pH ≈ 2 (red-orange indicator) and it is the only solution that reacts with calcium carbonate to give carbon dioxide, which turns limewater milky. The reaction is CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Z is sodium hydroxide: pH ≈ 12 (blue-violet), it feels slippery, and it neutralizes X. Y is sodium chloride: pH ≈ 7 (green), it gives no gas with marble chips, and it does not neutralize X. The classmate’s reasoning is only partly useful: Y’s neutrality and ‘salty’ description are consistent with NaCl, but ‘Z feels slippery’ is a property of bases, not proof that Y is NaCl. The identification of Y rests on the absence of acid–carbonate reaction and on neutral pH, with the other two solutions identified independently.
2. Calculate the amount concentration of X if it is monoprotic, and explain how the result would change if X were diprotic. Include the neutralization equation for HCl and NaOH.
Read the worked answer
n(Z) = cV = 0.100 mol/L × 0.0200 L = 2.00 × 10⁻³ mol. If X is monoprotic, HCl + NaOH → NaCl + H₂O, so n(X) = n(Z) = 2.00 × 10⁻³ mol. Then c(X) = n/V = 2.00 × 10⁻³ / 0.0250 = 0.0800 mol/L. If X were diprotic, one mole of acid would supply two moles of H⁺ and would neutralize two moles of NaOH: H₂A + 2NaOH → Na₂A + 2H₂O. The same titration would then imply n(X) = ½ n(Z) = 1.00 × 10⁻³ mol, so c(X) = 0.0400 mol/L. Therefore the titration data alone do not prove X is monoprotic; the assumed stoichiometry determines the calculated concentration. The student must identify the acid or run another test, such as comparing the amount of gas produced from a known amount of carbonate, before choosing the monoprotic value.
3. The flask warms during titration. Is this evidence for a chemical reaction or just mixing? Explain using bond/energy ideas, and state one control the students should run.
Read the worked answer
The warming is evidence of an exothermic chemical reaction, but it is not conclusive by itself because mixing two solutions can also release or absorb heat through dilution and solvation. Neutralization is strongly exothermic: H⁺(aq) + OH⁻(aq) → H₂O(l) releases energy as new bonds form and ions become hydrated. To separate reaction heat from mixing heat, run a control: add the same volume of Z to pure water, or add Z to a NaCl solution of similar concentration, and measure the temperature change under the same insulation and starting-temperature conditions. Compare the temperature rise in the control with that in the acid–base flask. A significantly larger rise in the neutralization flask supports the reaction interpretation. The students should also stir consistently, use an insulated container, and record temperature against time, because the maximum temperature may occur after mixing is complete.