End Point, Equivalence Point & Indicators | Physical Chemistry PDF Download

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:

End Point 

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.

Equivalence Point

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.

Difference Between Endpoint and Equivalence Point

• Although the endpoint is normally regarded as the equivalence point, they are not the same. 
• But since there is only a slight difference between an equivalent point and an endpoint, it can be considered the same for laboratory purposes. 
• The main difference between an equivalence point and an endpoint is that the former marks the end of the reaction whereas the latter is a point where the indicator changes colour. 

Indicators

There are the substances, the by a sharp colour change, indicate the completion of a chemical reaction.

pH scale used as an indicator

• Indicators for titration of acids and bases are usually weak organic acids or bases, yielding ions of a different color from the unionized molecule, e.g., litmus is red with acids and blue with alkalies, a change in color indicates that neutralization is complete.
• On the basis of different titrations, indicators are classified into several types, viz.,
  (a) neutralization indicators (acid-base or pH indicators,
  (b) oxidation-reduction indicators,
  (c) adsorption indicators,
  (d) mixed indicators,
  (e) screened indicators,
  (f) metal indicators and
  (g) universal indicators

Double Indicators Titrations for Two End Points

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.

• For the titration of alkali mixtures (e.g., NaOH + Na₂CO₃) or (Na₂CO₃ + NaHCO₃), two indicators phenolphthalein and methyl orange are used. 
• Phenolphthalein is a weak organic acid and gives an endpoint between pH 8 to 10, while methyl orange, a weak base itself indicates an endpoint sharply between pH 3.1 to 4.4. 

The following points must, therefore, be remembered.

•  Phenolphthalein is not a good indicator for weak alkali titrations.
•  Methyl orange is not a good indicator for weak acid titrations.
• In the case of strong acid vs. Strong base titrations using phenolphthalein or methyl orange as an indicator, the equivalent point suggests that meq. Of acid = meq. Of alkali.
• In the case of strong acid vs. Strong base such as Na₂CO₃ (which acts as a strong base in I step of dissociation and weak base in II step dissociation), using phenolphthalein as indicator, the m_eq. Of acid are titrated only up to NaHCO₃ state.
  Na₂CO₃ +  H⁺ → NaHCO₃ + Na⁺
  i.e., meq. Of acid = 1/2 meq. Of Na₂CO₃
  (if Eq. Mass of Na₂CO₃ = M/2)
  Or
  meq. Of acid = meq. Of Na₂CO₃
  (if Eq. Mass of Na₂CO₃ = M/1)
  However, titration with methyl orange as indicator, the m_eq. Of acid corresponds of total meq. Of alkali present at that time in mixture.
• One should be careful in calculating m_eq. Of alkalies, if methyl orange is used as an indicator in titration, i.e., for fresh mixture or in continuation of phenolphthalein, e.g., Na₂CO₃ + NaOH mixture v.s HCl

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.

For Phenolphthalein:
M_eq. Of Acid = m_eq. Of NaOH = 1/2 m_eq. Of Na₂CO₃

For Methyl Orange:Different colors of methyl orange at different pHM_eq. Of acid = m_eq. Of NaOH = m_eq. 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:

Solution: 
Phenolphthalein as an indicator:
M_eq. Of HCl used for 200 mL solution = 17.5 x (1/10) = 1.75
∴ m_eq. Of NaOH + 1/2 m_eq. Of Na₂CO₃ = 1.75
Methyl Orange as indicator: After I end point, methyl orange is added
1/2 m_eq. Of Na₂CO₃ = m_eq. Of HCl used

= 2.5 x (1/10) = 0.25

By Eqs.
M_eq. Of NaOH = 1.75 – 0.25 = 1.50
∴ (w/40) x 1000 = 1.50
Or W_NaOH = 0.06 g per 200 mL
Also, m_eq. Of Na₂CO₃ = 0.25 x 2 = 0.50
∴ (w/53) x 1000 = 0.50
 = 0.0265 g per 200mL

Eudiometry

The technique involving volume measurement during the reaction of gases is called eudiometry. 

• Since volume of gas, V ∝ number of mole at constant P, T and thus, volume ratio of gases can be directly used in place of mole ratio to analyse the given data.

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

To Represent the Concentration of H₂O₂ Solution:

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

• % strength 17/56 x  volume strength
• Volume strength = 5.6 x Normality
• Volume strength = 11.2 x Molarity

To Represent the Concentration of Oleum:

(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

To Determine Hardness of Water:

• Water, which gives foams easily with soap is called soft water and if not then hard water.
• The hardness of water is due to the presence of bicarbonates, chlorides, and sulphates of Ca and Mg.
• The hardness is temporary due to bicarbonates and permanent due to chlorides and sulphates of Ca and Mg
• The extent of hardness is known as degree of hardness defined usually as the number of parts by mass of CaCO₃  present per million parts by mass of water:
• Hardness is expressed in ppm i.e., 1 ppm = 1 part of CaCO₃ in 106 part of hard water.

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.

Water Softeners

The hardness may be removed by either of the reactants on treating with water.