Titration is used in analytical chemistry to determine acid, bases, reductants, oxidants, and other species. Titrations can usually occur in reactions such as redox reactions and acid-base reactions. During the process, two important stages known as endpoint and equivalence point are reached:
The point in titration, usually indicated by a change of color of an indicator, at which a particular reaction is complete, is called end point. It is an experimental value.
The equivalence point represents the state when the equivalent of a reacting species becomes equal to the equivalent of other reacting species. It is a theoretical value.
End point is almost near to equivalence point.
|Point where the indicator changes colour||The point at which the titrant is chemically equivalent to the analyte in the sample|
|Comes after the equivalence point||Comes before the endpoint|
|Weak acids can have only one endpoint||Weak acids can have multiple equivalence point|
There are the substances, the by a sharp colour change, indicate the completion of a chemical reaction.
pH scale used as an indicator
A double indicator titration is a process used in chemistry experiments to determine and analyze the amount and concentration of acids or bases in a solution at two endpoints.
The following points must, therefore, be remembered.
Case I: I endpoint determined using phenolphthalein as an indicator and then methyl orange is used to get II endpoint in continuation.
Case II: Endpoint is determined using phenolphthalein as an indicator. Next time endpoint is determined by taking another (fresh) same volume of the mixture using methyl orange as an indicator.
Meq. Of Acid = meq. Of NaOH = 1/2 meq. Of Na2CO3
For Methyl Orange:Different colors of methyl orange at different pHMeq. Of acid = meq. Of NaOH = meq. Of Na2CO3
Example: 200 mL of a solution of mixture NaOH and Na2CO3 was first titrated with phenolphthalein and N/10 HCl. 17.5 mL of HCl was required for the endpoint. After this methyl orange was added and 2.5 mL of the same HCl was again required for the next endpoint. Find out the mass of NaOH and Na2CO3 in the mixture:
Phenolphthalein as an indicator:
Meq. Of HCl used for 200 mL solution = 17.5 x (1/10) = 1.75
∴ meq. Of NaOH + 1/2 meq. Of Na2CO3 = 1.75
Methyl Orange as indicator: After I end point, methyl orange is added
1/2 meq. Of Na2CO3 = meq. Of HCl used
= 2.5 x (1/10) = 0.25
Meq. Of NaOH = 1.75 – 0.25 = 1.50
∴ (w/40) x 1000 = 1.50
Or WNaOH = 0.06 g per 200 mL
Also, meq. Of Na2CO3 = 0.25 x 2 = 0.50
∴ (w/53) x 1000 = 0.50
= 0.0265 g per 200mL
The technique involving volume measurement during the reaction of gases is called eudiometry.
Example: A mixture of ethane (C2H6) and ethene (C2H4) occupies 40 litres at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O2 to produce CO2 and H2O. Assuming ideal gas behavior, calculate the mole fractions of C2H4and C2H6 in the mixture.
Solution: For a gaseous mixture of C2H6 and C2H4
PV = nRT
∴ Mole fraction of C2H6 = 0.808 / 1.2195 = 0.66
And Mole fraction of C2H4 = 0.34
In percentage: The mass of H2O2 present in 100 mL H2O2 solution is H2O2 concentration in the percentage of H2O2 solution.
In volume: The volume of O2 at STP given by 1 mL H2O2 solution on decomposition is H2O2 concentration of H2O2 in volume.
Direct conversions can be made by using the following relations
(100 + X%) of oleum means ‘X’ g H2O reacts with equivalent amount of free SO3 to give H2SO4.
Example: Calculate the % of free SO3 in oleum (a solution of SO3 in H2SO4) that is labeled 109% H2SO4)
Solution: percentage above 100 represents the mass of H2O that reacts with dissolved SO3 in oleum to give H2SO4, i.e., 9 g H2O reacts with free SO3 to produce H2SO4
18 g H2O reacts with 80 g SO3
∴9 g H2O will react with 40 g SO3
Or % of free SO3 = 40
The reason for choosing CaCO3 as the standard to express hardness, inspite of the fact that CaCO3 is not soluble in water but its molar mass is 100 which makes calculation easy.
The hardness may be removed by either of the reactants on treating with water.