Class 11 Chemistry Notes Stocihiometry -Double indicator acid-base titration

Double indicator acid-base titration:

In the acid-base titration the equivalence point is known with the help of indicator which changes its color at the end point. In the titration of polyacidic base or polybasic acid there are more than one end point. One indicator is not able to give color change at every end point. So to find out each end point we have to use more than one indicator. For example in the titration of Na₂CO₃ against HCl there are two end points.

Na₂CO₃ + HCl → NaHCO₃ + NaCl

NaHCO₃ + HCl → H₂CO₃ + NaCl

When we use phenolphthalein in the above titration it changes its color at first end point when NaHCO₃ is formed and with it we cannot know second end point. Similarly with methyl orange it changes its color at second end point only and we cannot know first end point. It is because all indicator changes color on the basis of pH of medium. So in titration of against acid phenolphthalein cannot be used.


A sample contains NaOH, Na₂CO₃& NaHCO₃. This solution is titrated with HCl or H₂SO₄

(i) When phenolphthalein (HPh) is used as indicator the reaction are given below.

Na₂CO₃ + HCl → NaHCO₃ + NaCl

This is the half neutralization of Na₂CO₃

NaHCO₃ + HCl → X

NaOH + HCl → NaCl + H₂O

Hence

Meq. Of HCl = Meq. Of NaOH + (1/2) Meq. Of Na₂CO₃

(ii) When methyl orange (MeOH) is used as indicator

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

This is the complete neutralization

NaHCO₃ + HCl → NaCl + H₂O + CO₂

NaOH + HCl → NaCl + H₂O

in this case

Meq. Of HCl = Meq. Of NaOH + Meq. Of Na₂CO₃ + Meq. Of NaHCO₃

(iii) If MeOH is added after the first end point obtained from HPh

NaHCO₃ + NaHCO₃ + HCl → NaCl + H₂O + CO₂

Produced original

Meq. Of HCl = Meq. Of NaHCO₃ (produced) + Meq. Of NaHCO₃ (original)

Note:
When we carry out dilution of solution, Meq eq. Milli mole or mole of substance does not change because they represent amount of substance, however molar concentration may change.
Ex.1 A solution contains a mixture of Na₂CO₃ and NaOH. Using phenolphthalein as indicator 25 ml. of mixture required 19.5 ml of 0.995 N HCl for the end point. With methyl orange. 25 ml of solution required 25 ml. of the same. HCl for the end point. Calculate grams per liter of each substance in the mixture.
Sol.

Let a & b are the Meq. Of Na₂CO₃ and NaOH respectively

When phenolphthalein in is used as indicator

(1/2) Meq. Of Na₂CO₃ + Meq. of NaOH = Meq. of HCl

(a/2) + b = 19.5 × 0.995

(a/2) + b = 19.4 …(1)

When MeOH is used as indicator

Meq. Of Na₂CO₃ + Meq. Of NaOH = Meq. Of HCl

a + b = 25 × 0.995

a + b = 24.875 …(2)

From (1) & (2)

(a/2) = 5.475 Þ a = 10.95

B = 24.875 – 10.95 = 13.925

Wt of Na₂CO₃/lit = (10.95/25) × 10^-3 × (106/3) × 1000 = 23.2 gm/lit

Wt of NaOH/lit = b × 10^-3 × 84/25 × 1000 = 22.28 gm/lit Ans.

Ex.2 2.5 g of a mixture containing CaCO₃, Ca(HCO₃)₂ and NaCl was dissolved in 100 mL water and its 10 mL portion required 10 mL 0.05 M H₂SO₄ solution to reach the phenolphthalein end point. An another 10 mL portion of the same stock solution required 32.35 mL of the same acid solution to reach the methyl orange end point. Determine mass percentage of CaCO₃ and Ca (HCO₃)₃in the original mixture.
Sol.

Let a and b are the Meq. of CaCO₃ and Ca(HCO₃)₂ respectively.

When HPh is used in indicater then only CaCO₃ will react with H₂SO₄

Meq. of H₂SO₄ = (1/2) Meq. of CaCO₃

Meq. of H₂SO₄ = ((10 × 0.05 × 100 × 2)/10) = 10

Meq. of CaCO₃ = 20

When MeOH is used as indicator

Then Meq. or CaCO₃ + Meq. of Ca(HCO₃)₂ = Meq. of H₂SO₄

a + b = ((32.35 × 0.05 × 2)/10) × 100

a + b = 32.35

b = 12.5

wt. of CaCO₃ = a × 10^-3 × 100/2

= 1 gm

wt. of Ca(HCO₃)₂ = b × 10^-3 × 162/2 = 1 gm

% of CaCO₃ = (1/2.5) × 100 = 40%

% of Ca(HCO₃)₂ = (1/2.5) × 100 = 40%

Ex.3. 6.4 g of a pure monobasic organic acid is burnt completely in excess of oxygen and CO₂ evolved is absorbed completely in one liter of an aqueous solution of NaOH. A 10 mL portion of this solution required 14.5 mL of a normal HCl solution to reach the phenolphthalein end point. Another 10 mL portion of the same solution required 18 mL of the same HCl solution to reach the methyl orange end point. If the organic acid contains 25% oxygen by weight, deduce the empirical formula of this acid and strength of original NaOH solution
Sol.

Let 100 ml of solution contains a Meq. of NaOH & b Meq. of Na₂CO₃

When HPh is used as indicator then

Meq. of NaOH + 1/2 Meq. of Na2CO3=Meq. of HCl

Ex.4 A solution contains Na₂CO₃ and NaHCO₃. 10 mL of the solution requires 2.5 mL of for neutralisation using phenolphthalein as an indicator. Methyl organge is added when a further 2.5 mL of 0.2 M H₂SO₄ was required. Calculate the amount of Na₂CO₃ and NaHCO₃ in one litre of the solution.
Sol.

Share this article with your friends.

• Facebook
• Twitter
• Google+
• Pinterest
• StumbleUpon
• Dig
• LinkedIn