The Double Indicator Method

Titration of a Carbonate

The figure below shows the pH changes during the titration of sodium carbonate with hydrochloric acid. 3

 – 3

A few drops of thymol blue indicator blue indicator are added to 25 cm  of 0.10 mol dm  sodium carbonate  – 3 solution and the mixture is titrated with 0.20 mol dm  hydrochloric  hydrochloric acid until it becomes bec omes 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 HC l (aq) can be represented in two stages: (a) and (b)

(a)

Na2CO3(aq) + HCl(aq)

(b) NaHCO3(aq) + HCl(aq)  Na2CO3(aq) + 2HCl(aq)

NaHCO3(aq) + NaCl(aq) NaCl (aq) + CO2(g) + H2O(l)

 Indicator to exhibit completion of reaction thymol blue

methyl orange

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

If thymol blue is blue is used as the indicator, it changes colour from  blue (basic) to (basic) to pale  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) yellow  (basic) to  to orange(end orange(end  point) (red is acidic) when acidic) when reaction (b) is completed, i.e. when sodium carbonate is completely neutralized to give CO 2 and H2O. From the stoichiometries of the two reactions, it can be seen that 3 if volume of HC l required for reaction (a) = x cm , 3 then volume of HC l required for reaction (b) =  x cm , and volume of HC l required when using phenolphthalein as indicator =  x cm3 3 volume of HC l required when using methyl orange as indicator = 2 x cm 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 b y 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 HC l used with thymol blue as indicator = x mol Further amount of HC l used when methyl orange as indicator =  y mol Amount of HC l used for the complete titration neutralisation of Na 2CO3 = 2 y mol Amount of Na2CO3 reacted = (½)(amount of HC l which reacted with Na 2CO3) = (½)(2 y) = y mol Amount of HC l 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 3

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

NaCl (aq) + CO 2(g) + H2O(l)

Amount of HC l which reacted with NaHCO3 = = Hence amount of NaHCO3 reacted = Amount of HC l reacted = Since one mole of NaHCO3 comes from one mole of Na 2CO3, amount of Na 2CO3 reacted = amount of NaHCO 3 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 HC l which reacted with both Na 2CO3 and NaOH = = Amount of HC l reacted with Na2CO3 to give = Amount of HC l which reacted with  NaHCO3  NaHCO3 = Amount of HC l which reacted with NaOH = = Concentration of NaOH in the solution =

Example 2: 3

A 25.0 cm  portion of a solution containing sodium carbonate and sodium hydrogen carbonate 3 -3 needed 22.50 cm  of a solution of hydrochloric acid of concentration 0.100 mol dm  to pale green 3 of thymol blue. On addition of methyl orange, a further 28.50 cm  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 HC l which reacted with Na 2CO3 to form NaHCO3 = Hence amount of Na 2CO3 reacted = amount of HC l reacted =  concentration of Na2CO3 in the solution =

∴

With methyl orange as indicator,  NaHCO3(aq) + HCl(aq)

NaCl (aq) + CO 2(g) + H2O(l)

Amount of HC l which reacted with all the NaHCO 3 = = Amount of HC l which reacted with NaHCO 3 from Na2CO3 = Hence amount of HC l used to react with NaHCO 3 initially present = = concentration of NaHCO3 in the given solution = =

∴