The Double Indicator Method 1f4m50

Titration of a Carbonate The figure below shows the pH changes during the titration of sodium carbonate with hydrochloric acid. A few drops of thymol blue indicator are added to 25 cm3 of 0.10 mol dm – 3 sodium carbonate solution and the mixture is titrated with 0.20 mol dm – 3 hydrochloric acid until it becomes colourless. Some methyl orange is now added and the titration is continued until the solution again changes colour. The diagram shows how the pH changes during the titration and also shows the pH range of some indicators.

thymol blue

bromothymol blue bromophenol blue methyl orange

The reaction of Na2CO3(aq) and HCl (aq) can be represented in two stages: (a) and (b)

(a) Na2CO3(aq) + HCl(aq)

NaHCO3(aq) + NaCl(aq)

(b) NaHCO3(aq) + HCl(aq)

NaCl (aq) + CO2(g) + H2O(l)

Na2CO3(aq) + 2HCl(aq)

Indicator to exhibit completion of reaction thymol blue methyl orange

2NaCl(aq) + CO2(g) + H2O(l)

If thymol blue is used as the indicator, it changes colour from blue (basic) to pale green/greenyellow (end point) (yellow at acidic) when reaction (a) is completed, i.e. when sodium carbonate is only half-neutralised. If methyl orange is used as the indicator, it changes colour from yellow (basic) to orange(end point) (red is acidic) when reaction (b) is completed, i.e. when sodium carbonate is completely neutralized to give CO2 and H2O. From the stoichiometries of the two reactions, it can be seen that if volume of HCl required for reaction (a) = x cm3, then volume of HCl required for reaction (b) = x cm3, and volume of HCl required when using phenolphthalein as indicator = x cm3 volume of HCl required when using methyl orange as indicator = 2x cm3 Use can be made of this two stage titration to estimate the sodium carbonate and sodium hydrogen carbonate.

The Double Indicator Method A mixture of sodium hydroxide and sodium carbonate can be analysed by the double indicator method. The reaction of sodium hydroxide and sodium carbonate with hydrochloric acid are represented by the following equations: Reaction 1 2(a) 2(b)

NaOH + HCl (x – ymol) Na2CO3 + HCl (y mol) NaHCO3 + HCl (y mol)

NaCl + H2O

Indicator to show the completion of reaction any indicator

NaHCO3 + NaCl NaCl + CO2 + H2O

thymol blue methyl orange

If hydrochloric acid is added to a mixture of sodium hydroxide and sodium carbonate and using thymol blue as the indicator is discharged when reactions 1 and 2(a) are completed. If methyl orange is then added and the titration continued, a further amount of hydrochloric acid is required to bring about a colour change. This amount of acid is required for reaction 2(b). Amount of HCl used with thymol blue as indicator = x mol Further amount of HCl used when methyl orange as indicator = y mol Amount of HCl used for the complete titration neutralisation of Na2CO3 = 2y mol Amount of Na2CO3 reacted = (½)(amount of HCl which reacted with Na2CO3) = (½)(2y) = y mol Amount of HCl which reacted with NaOH = amount of NaOH present = x – y mol The concentrations of sodium hydroxide and sodium carbonate in the solution can then be calculated from the data obtained.

Example 1 25.0 cm3 of a solution containing sodium hydroxide and sodium carbonate was titrated against 0.100 mol dm-3 hydrochloric acid, using thymol blue as indicator. After 30.00 cm3 of acid has been used, the indicator was pale green. Methyl orange was added and a further 12.50 cm3 of hydrochloric acid were needed to turn the indicator orange. Calculate the concentrations of sodium hydroxide and sodium carbonate in the solution. Solution With methyl orange as indicator , NaHCO3(aq) + HCl(aq)

NaCl (aq) + CO2(g) + H2O(l)

Amount of HCl which reacted with NaHCO3 = = Hence amount of NaHCO3 reacted = Amount of HCl reacted = Since one mole of NaHCO3 comes from one mole of Na2CO3, amount of Na2CO3 reacted = amount of NaHCO3 reacted = Hence concentration of Na2CO3 in the solution = = With thymol blue as indicator, NaOH (aq) + HCl (aq) Na2CO3 (aq) + HCl (aq)

NaCl (aq) + H2O (l) NaHCO3 (aq) + NaCl (aq)

Amount of HCl which reacted with both Na2CO3 and NaOH = = Amount of HCl reacted with Na2CO3 to give = Amount of HCl which reacted with NaHCO3 NaHCO3 = Amount of HCl which reacted with NaOH = = Concentration of NaOH in the solution =

Example 2: A 25.0 cm3 portion of a solution containing sodium carbonate and sodium hydrogen carbonate needed 22.50 cm3 of a solution of hydrochloric acid of concentration 0.100 mol dm-3 to pale green of thymol blue. On addition of methyl orange, a further 28.50 cm3 of the acid was needed to turn this indicator to its neutral colour. Calculate the concentrations of sodium carbonate and sodium hydrogen carbonate in the solution. Solution With thymol blue as indicator, Na2CO3(aq) + HCl(aq)

NaHCO3(aq) + NaCl(aq)

Amount of HCl which reacted with Na2CO3 to form NaHCO3 = Hence amount of Na2CO3 reacted = amount of HCl reacted = ∴ concentration of Na2CO3 in the solution = With methyl orange as indicator,

NaHCO3(aq) + HCl(aq)

NaCl (aq) + CO2(g) + H2O(l)

Amount of HCl which reacted with all the NaHCO3 = = Amount of HCl which reacted with NaHCO3 from Na2CO3 = Hence amount of HCl used to react with NaHCO3 initially present = = ∴concentration of NaHCO3 in the given solution = =