| If sodium carbonate is titrated with hydrochloric acid using methyl orange as the
indicator, the titre is twice what it is if phenolphthalein is used as the indicator. The
reason is that the two indicators do not change colour at the same pH.
Methyl orange changes at about pH 3 – 4; this corresponds to the complete neutralisation of sodium carbonate with two moles of hydrochloric acid: Na2CO3 + 2HCl à 2NaCl + H2O + CO2. On the other hand phenolphthalein changes at pH 8.5 – 10; this colour change corresponds to one mole of hydrochloric acid reacting with sodium carbonate: Na2CO3 + HCl à NaHCO3 + NaCl. The titration curve for sodium carbonate has two vertical portions, corresponding to the two reactions given above. Sodium hydroxide has only one, and gives the same titre with both indicators. The result is that, if a mixture of sodium carbonate and sodium hydroxide is titrated with hydrochloric acid using phenolphthalein, the volume corresponds to the sodium hydroxide plus half the sodium carbonate. Addition of methyl orange to this mixture and further titration gives the volume of acid corresponding to half the sodium carbonate.
Method: The mixed alkali solution used for analysis should be about 0.5 mol dm-3 in sodium hydroxide and 0.25 mol dm-3 in sodium carbonate. 1 Pipette 25.0 cm3 of the mixed alkali solution into a 250 cm3 graduated flask, make to the mark with pure water, and mix thoroughly. 2 Pipette 25.0 cm3 of this diluted solution into a 250 cm3 conical flask, add 3 – 4 drops of phenolphthalein indicator, and titrate with standard 0.100 mol dm-3 hydrochloric acid until the pink colour is just discharged. Note the volume required. 2 Add to the same mixture 3 – 4 drops of methyl orange indicator (or screened methyl orange if preferred) and continue titration until the methyl orange changes from yellow to red. 3 Repeat to give three consistent sets of titres.
Results:
Mean titre (phenolphthalein)/cm3: Mean titre (methyl orange)/cm3:
Calculation: If the first titre (phenolphthalein), x cm3, represents the hydroxide plus half the carbonate, and the difference between the phenolphthalein and methyl orange titres, y cm3, represents half the carbonate, then the hydroxide is equivalent to (x – y) cm3 of acid, and the carbonate to 2y cm3 of acid. Find the concentration of each of the compounds in mol dm-3. |