I an currently studying acid-base titrations. There is a concept of double titration. Can someone provide me an idea on what actually it is? Also what resources can I refer to?


Double titration also often refers to Gilson's titration method (also known as "Gilman color tests" or "Gilman double titration") for determining the concentration of organometallic compounds in hydrocarbon solvents. At the first stage the total concentration of the base in the solution is determined. At the second stage the residual amount of alkali after the organometallic compound reacts with the halide is established, allowing to calculate the concentration of metalloorganic compound. [1]

For the double titrations, a 3 ml aliquot of the solution to be analyzed was withdrawn by pipette using a suction bulb and added to 10 ml of ether. The solutionwas hydrolyzed with 10 ml of distilled water and titrated with standard acid, using phenolphthalein as indicator, to give the total present. A second 3 ml aliquot was withdrawn and added 10 of ether, containing 1 ml of the halide, under an atmosphere of nitrogen. The solution was swirled gently and allowed to stand for 2 min, after which it was hydrolyzed with distilled water and titrated immediately with standard acid, again using phenolphthalein as indicator.

In titrating the latter solution the end point is easily overstepped, since the aqueous layer decolorizes before the ether layer. Vigorous shaking near the end point is recommended. There is the possibility of interference from hydrolysis of the halide, so that the titration should be carried out within a reasonable length of time.

An updated cumulative summary for the method can be found in Wakefield's Organolithium Methods [2, pp. 17–18]:

For detecting the presence of an organolithium compound (for example for checking whether any remains at the finish of a reaction) Gilman's Colour Test 1 is convenient:

$$ \begin{align} \ce{(4-Me2NC6H4)2C=O + RLi &-> (4-Me2NC6H4)2C(R)OLi}\\ \ce{ &->[(i) H2O][(ii) I2, AcOH] [(4-Me2NC6H4)2CR]+ I-} \end{align} $$

A sample (0.5—1 ml) of the test solution is added to an equal volume of a I solution of Michler's ketone, 4.4'-bis(dimethylamino)benzophenone, in dry benzene. A few drops of water are added, and the mixture is shaken.* A few drops of an 0.2% solution of iodine in glacial acetic acid are added, resulting in the formation of a blue or green colour. Gaidis proposed two modifications, to distinguish between alkyl- and aryllithium compounds (15).

(a) Following hydrolysis, the solution is buffered to pH 9 by the addition of 20% aqueous catechol (1—2 ml). A few drops Of iodine in benzene are added and the solution is shaken. A green to blue colour in the organic layer indicates the presence of an aryllithium compound. The mixture is then acidified with acetic acid to reveal the presence Of an alkyllithium compound.

(b) Following hydrolysis the solution is acidified with acetic acid (5—15 drops). 20% aqueous sodium bisulphite is added; aryl- but not alkyllithium compounds give a green colour.

Gilman's Colour Test 2 (16) which also distinguishes alkyl- from aryllithium compounds, depends on the rapid metal—halogen exchange of the former with 4-bromo-N,N-dimethylaniline:

$$ \begin{align} \ce{4-BrC6H4NMe2 + RLi &<=>> 4-Li2C6H4NMe2 + RBr}\\ \ce{4-Li2C6H4NMe2 &->[(i) Ph2CO][(ii) H3O+] [Ph2CC6H4NMe2]+} \end{align} $$

The solution to be tested (0.5-1 ml) is added to an equal volume of a solution of 4-bromo-N,N-dimethylaniline in dry benzene. A 15% solution of benzophenone in dry benzene (1 ml) is added. After a few seconds, water is added and then the mixture is acidified with concentrated hydrochloric acid. A positive result is indicated by a red colour in the aqueous layer.

* Aerial oxidation sometimes leads to formation of the blue-green at this stage.


  1. Gilman, H.; Cartledge, F. K.; Sim, S.-Y. The Analysis of Organic-Substituted Group IVB Lithium Compounds. Journal of Organometallic Chemistry 1963, 1 (1), 8–14. DOI: 10.1016/S0022-328X(00)80039-X.
  2. Wakefield, B. J. Organolithium Methods; Best synthetic methods; Academic Press: London; San Diego, 1988. ISBN 978-0-12-730940-8.
  • 5
    $\begingroup$ Andselisk, You have given a modern version of the meaning of double titration. In Indian textbooks/test preparation books, a century old meaning is still used. The answer by Wasnik reflects that meaning. That is also historically correct. $\endgroup$
    – M. Farooq
    Jul 20 '20 at 3:21

Double titration is a method of determining the amount of substance present in the form of a solution along with another solution. It can be used in the case of a mixture of bases involving $\ce{NaOH}$, $\ce{Na2CO3}$ or $\ce{NaHCO3}$, $\ce{Na2CO3}$. These bases are titrated against strong acids with proper indicators. Generally a single indicator is not sufficient so two indicators are used.


Firstly Indicator 1 (generally phenolphthalein) is added to the solution and the solution is titrated against a strong acid of known normality which on complete neutralization gives endpoint 1. Then Indicator 2 (generally Methyl Orange) is added to the remaining solution and is titrated against a strong acid which gives endpoint 2.

For example: In case of titration of $\ce{NaOH}$, $\ce{Na2CO3}$ at the first endpoint only half of $\ce{Na2CO3}$ will be converted to $\ce{NaHCO3}$ At the second endpoint remaining half gram equivalent of $\ce{Na2CO3}$ and $\ce{NaHCO3}$ is neutralized. Thus by equating the gram equivalents acid 1 with half equivalent of $\ce{Na2CO3}$ and half equivalent of $\ce{Na2CO3}$ plus equivalent of $\ce{NaHCO3}$ to equivalent of acid 2 we can get the amount of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ in the solution.

  • 2
    $\begingroup$ Another option is titration with the pH meter. $\endgroup$
    – Poutnik
    Apr 26 '19 at 15:24
  • $\begingroup$ I once faced the following problem in my textbook: "When 10 g mixture of Na2CO3 and NaOH was titrated against HCI by using phenolphthalein, followed by adding methyl orange after first equivalence point, 7.8 g of HCI was consumed in total. Find the mass of Na2CO3 in the mixture." Its solution involved the use of equivalence that rather confused me. Actually it turns out the problem simply asks to find the mass of Na2CO3 in a 10g NaOH-Na2CO3 mixture that requires 7.8 g HCl to get neutralized. And this can solved using elementary mole concept. $\endgroup$
    – Shub
    Sep 28 '21 at 6:14
  • $\begingroup$ This is a problem with two unknowns. Needs two equations. 1) The initial mixture contains $a$ mol $\ce{NaOH}$. and $b$ mol $\ce{Na2CO3}$. The mass of this mixture is $40 a + 106 b = 10$. This is first equation. 2) This mixture needs $7.8$ g $\ce{HCl}$ to be titrated, or $7.8/36.5 = 0.214$ mol $\ce{HCl}$ The second equation is : $a + 2b = 0.214$. So now, go on ! $\endgroup$
    – Maurice
    Dec 5 '21 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.