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Potentiometric titration _mathematical endpoint__rev

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					R.A. van Iterson Drenthe College Emmen Holland for www.standardbase.com

Mathematical determination of the endpoint in potentiometric titration.
Titration is an analytical technique which allows the quantitative determination of a specific substance (analyte)
dissolved in a sample. It is based on a complete chemical reaction between the analyte and a reagent (titrant) of
known concentration which is added to the sample. A well-known example is the titration of acetic acid in vinegar
with sodium hydroxide:
                                         -        +
NaOH+ HCl -> H2O + NaCl          or   OH + H3O -> 2 H2O

The titrant is added until the reaction is complete. In order to be suitable for a
determination, the end of the titration reaction has to be easily observable. This means
that the reaction has to be monitored (indicated) by appropriate techniques, e.g.
potentiometry (potential measurement with a sensor) or with colour indicators. The
measurement of the dispensed titrant volume allows the calculation of the analyte
content based on the stoichiometry of the chemical reaction. The reaction involved in
a titration must be fast, complete and observable.

When performing a standard titration on a solution of hydrochloric acid, the endpoint
will be clear to see. When adding an indicator such as phelolftalein to a transparant solution, the endpoint is easy to
see. When we use ‘real’samples from industry, most solutions are not clear. This will interfere with the visual
endpoint.
Now potentiometric analyses can be used to let a pH-meter and electrode ‘see’the endpoint.

This special electrochemical titration uses an indication electrode to ‘see’ (measure) the pH of the solution.
With a burette you can add small portions of titrant near the endpoint.
To find the endpoint-pH value it is common to make a quick titration so we can roughly judge where the endpoint
                        lies. The basic idea of potentiometric titrations is to go beyond the endpoint and then
                        calculate the endpoint from the measurements. For acidbase titration we use a glass
                        electrode as indication and a reference in order to have a stable reference value. Before
                        starting a titration the electrodes and ph meter must be calibrated with buffers at room
                        temperature. Normally pH=7 and pH=9 are used (pH=10 is not recommended as higher pH
                        solutions absorb carbon dioxide more easily).

                         The same technique, potentiometry, can be used for redox titrations. In this case the use of a
Pt-electrode (just a Pt-wire as indication electrode is required. The reference can be any common reference electrode
such as Ag/Ag-sulfate. The use of calomel containing Hg-salts is not recommended.

Practice
For redox-titrations the electrode system cannot be calibrated with pH-buffers. Normally when the titration is ready
to start, the potential on the meter is set to 0.00 V. The solution should be swirled with a magnetic stirring rod during
the titration. Usually we need just one quick titration to see where the endpoint occurs.
Then the real titration can start by adding titrant with a special burette. At first, increments of 0.5-1 mL will do. Near
the endpoint, small steps: 0.1 for small increments and 0.2-0.5 for larger increments.

Where is the endpoint ?
To calculate the endpoint volume there are several methods.
1       For sharp larger jumps in curves that are symmetrical we can use
        1a        mathematical method. (second derivative) or
        1b        graphical method with parallel lines.
2       For asymmetric curves we can use a graphical method known as the circle-method.

In the case of redox titration of iron(II) with potassium permanganate method 1a is recommended.
R.A. van Iterson Drenthe College Emmen Holland for www.standardbase.com

Mathematical second derivative
When we calculate the 1st derivative its graph would resemble the slope of athe line at any measurement. The curve
                                    would size until the endpoint is reached and then fall back after the endpoint.
                                    The second derivative would show a rising positive hyperbolic curve towards
                                    the endpoint and a larger negative value just after the endpoint rising to zero.
                                    The endpoint can be found at the point where the value goes from positive to
                                    negative. When using the same steps the following simplified method is used.

                                    I’m afraid we can’t follow this




                                  .
Titrant (mL)                 E(mV )                        (difference)               )
12.20                        512
                                                           +63
12.30                        575                                                         +95
                                                           +158
12.40                        733                                                         -111
                                                           +47
12.50                        780                                                         -15
                                                           +32
12.60                        812

The endpoint must lie between 12.40 mL and 12.50 mL since that’s where the value of the second derivative
changes sign. The formula for the calculated endpoint is:

                                         95
V(endpoint) i= 12.30 mL + 0.1 ml x             = 12.346 mL (or 12.35 mL)
                                      95  111
To determine any endpoint, 5 data points are needed.
For other values we can replace the 0.1 by the actual increments
When increments are not used, and various adding has taken place, we should divide any calculated difference by
the increment (so each value in the last two columns should be divided by 0.1) When using the same increments as
in our example this is not necessary, since all values would be divided by the same 0.1.

				
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