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.

Double indicator acid-base titration


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
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)


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.


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.


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


Double indicator acid-base titration

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          

Double indicator acid-base titration

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.


Double indicator acid-base titration

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