The titrimetry or titration is a technique of Dosage used in analytical Chimie in order to determine the concentration of a chemical species in solution (or titrates of a solution).

The method of titration the most used is volumetry or volumetric titration. It consists in using a solution of known concentration (called titrating) in order to neutralize a species contained in the unknown solution.

The volumetric titrations most widespread are titrations acid-bases: The operator makes run drop by drop an acid in a basic given volume. Thus the reagents react mol to mol. Base-acid titration is also possible.

The point of neutralization is known thanks to an indicator added in the unknown solution (This indicator changes color at the time of neutralization) or thanks to a variation of the potential or pH (measured by means of a electrode soaking in the unknown solution).

Hardware requirement

A graduated oil-can is in general used when titration is manual or an automatic titrimeter when one wishes to improve the repeatability and the traceability. The volume of the sample is taken by means of a pipette of given volume and is placed in a erlenmeyer. The Burette always contains the titrating solution of reagent which one knows the concentration. The oil-can gives the versed volume of titrating solution and thus the point with equivalence will give us. The solution to be proportioned will be always in a beaker or another clean container, it will be in exactly known Volume.

Titration in aqueous solution

To carry out a titration, one carries out a chemical reaction where the species in solution reacts with another substance. The species to be proportioned is called reactive titrated and the added substance is called reactive titrating

It is possible to carry out a conductimetric titration (for a acido-basic Réaction), a colorimetric Titrage , a titration pH - metric ,…

Titrations pH-metric

This type of titration is realizable only with acids and bases, weak or strong, and under presence of a PH-mètre or an indicating of coloured pH.

Titration of a strong acid by a strong base

In solution, the Acide S forts like hydrochloric acid are completely dissociated and give ions $H\_3O^+\; \backslash ,$. In the same way, the bases S strong dissociate completely to release from the ions $OH^-\; \backslash ,$. The ions $H\_3O^+\; \backslash ,$ and $OH^-\; \backslash ,$ react in the following way: $H\_3O^+\; +\; OH^-\; \backslash \; longrightarrow\; 2:00\; \_2O\; \backslash ,$ (Here the general equation of neutralization)

With volume with equivalence all the ions $H\_3O^+$ and $OH^-$ have reacts, the only reaction which occurs is that of the autoprotolyse of water. The pH is then equal to 7. Equivalence is indicated either by a Indicateur of pH, or indirectly by a curve representing the pH according to versed basic volume. You must for that record the values of the PH-meter for each ml of versed solution. When the pH starts to increase to a significant degree, then pour 0,2 ml of solution between each measurement of pH once the pH will be become again relatively stable take again a measurement of pH all the ml By plotting the curve of the pH according to basic volume poured, have obtains the curve represented as on the cartoon. To find volume with equivalence, you must trace the tangent S at the two points of inflection (where the curve changes direction). the tangents must be parallel S . To trace the Perpendicular on these two lines, then thanks to a compass, to trace the Mediating perpendicular. the mediator must be parallel to the tangents . Volume with equivalence is the place where the mediator and the curve meet, for a pH of 7.

Colorimetric titrations

See also: colorimetric Titration

Conductimetric titrations

A conductimetric titration uses the capacity of the ions to lead the electric current in an aqueous medium, one then measures the Conductance of the solution thanks to an electrode. Like each ion the current leads differently, the conductance varies during proportioning. This concept is directly dependant with the concentration of the ions present. For example : The ion $H\_3O^+$ leads the current best that the ion $OH^-$. If you make react the ions $H\_3O^+$, the conductance of the solution drop because these ions disappear (the concentration of the ions $H\_3O^+$ decreases). Then if you continue titration beyond the point with equivalence, the conductance will go up, because the ions $OH^-$ will be increasingly numerous, (the concentration of the ions $OH^-$ increases). By recording the values indicated by the conductimeter, one can plot the straight line of the conductance according to versed volume. It takes shape two adjacent lines then, in form of V.Le not adjacent indicates volume to equivalence.

Calculations

The goal of a titration is to find the concentration in a given element. There exists for that two possible ways: by the calculation and a promotion table. In both cases it is necessary to know the equation assessment of the reaction.

Mathematical method

With equivalence, there was as many moles of reagent has that of reagent B.

• $C\_A\; \backslash \; times\; V\_A=C\_B\; \backslash \; times\; V\_B\; \backslash ,$

• thus $C\_A\; =\; \backslash \; frac\; \{C\_B\; \backslash \; times\; V\_B\}\; \{V\_A\}$

With:

• $C\_A\; \backslash ,$: concentration of the known solution (titrating)

• $V\_A\; \backslash ,$: volume of titrating cast in ml
• $C\_B\; \backslash ,$: concentration of the unknown solution
• $V\_B\; \backslash ,$: volume of sample used in ml

This formula is general whatever the coefficients (or numbers) stoechiometric.

While reasoning with the concentrations then the stoechiometric coefficients intervene.

Promotion table

The assessment of the reaction is written like this:

At the beginning of titration, (when you did not still pour anything), you have only reagents and any product of reaction. As your reaction proceeds, a quantity X of reagent disappears, whereas at the same time a quantity X of product appears. It is the application of the law of lavoisier. At the end of titration, i.e. when one of your reagents completely disappeared (that being in your erlen) you reached maximum advance, your reaction cannot go further.

At this time there

• $n\; H\_3O^+\; -\; X\; =\; 0\; \backslash ,$
• $n\; OH^-\; -\; X\; =\; 0\; \backslash ,$
• thus $n\; H\_3O^+\; =\; N\; OH^-\; \backslash ,$
• $C\_\; \{H\_3O^+\}\; =\; \backslash \; frac\; \{C\_\; \{OH^-\}\; \backslash \; times\; V\_\; \{OH^-\}\}\; \{V\_\; \{H\_3O^+\}\}$

This table makes it possible to include/understand what this master key during the reaction and not to mislead you with the coefficients. Indeed you have just to defer them in front of X. Normally when you have water in your reaction you must write " in excès" (you have always more water than it of is not necessary so that the reaction proceeds)

Calculation of an excess

At the beginning the graduated solution is in excess. After equivalence, it is the titrating solution which is in excess. For a acido-basic reaction, the acid is limiting to the beginning and for an oxydoreduction, it is the oxidizing one which is limiting before equivalence.

An excess is part of the reagents which does not react during a chemical reaction and which is thus found in the products of the reaction. One can proportion the excess of reagent with a second chemical reaction: it is a back titration.

Recall

• the number of Mole $n\; \backslash ,$ corresponds to the quantity of matter of a chemical species.

• the molar Mass $M\; \backslash ,$ of an element is the quantity of matter which it is necessary to have a mole of this element. Example: $M\_C\; =\; 12g\; \backslash ,$, one needs 12 grams of carbon to have 1 mole.
• the mass $m\; \backslash ,$ is the quantity of weighed matter.

$m\; =\; M\; \backslash \; times\; n$

• the concentration $C\; \backslash ,$ is the quantity of matter contained in one liter of solution. It is expressed in mol/L or g/L.

$C\; =\; N/V$ or $C\; =\; m/V$

Example of calculation

One makes react sodium chloride $NaCl$ with silver nitrate $AgNO\_3$.

One pours 10 ml of NaCl with 0.2 mol/L in a beaker containing 20 ml of $AgNO\_3$ with 0.2 mol/L.

• Mr. Ag =47 g/mol
• MR. NR = 14 g/mol
• MR. O = 16 g/mol

To determine the mass of excess d'$AgNO\_3$.

• $n\; Cl^-\; -\; X\; max\; =\; 0$

• $x\; max\; =\; N\; Cl^-$
• $n\; Ag^+\; -\; xmax\; =\; Ag^+$
• $n\; Ag^+\; -\; N\; Cl^-\; =\; Ag^+$

• $n\text{'}\; Ag^+\; =\; C\; AgNO\_3\; \backslash \; times\; V\; AgNO\_3\; -\; C\; NaCl\; \backslash \; times\; V\; NaCl$

• $m\; AgNO\_3\; =\; (C\; AgNO\_3\; \backslash \; times\; V\; AgNO\_3\; -\; C\; NaCl\; \backslash \; times\; V\; NaCl)\; \backslash \; times\; Mr.\; AgNO\_3\; \backslash ,$

External bonds

• pH-metric Titrations - resolution or simulation with Excel, in English or Portuguese

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