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If we have $\ce{A + T -> P}$ where $\ce{A}$ is the analyte, $\ce{T}$ is the titrant, and $\ce{P}$ is the product, photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

When stable complexes formed such as $\ce{Bi}$-EDTA, they will not absorb the radiation (or in small amounts), but when unstable complexes formed such as $\ce{Cu}$-EDTA, they will absorb highly. The same goes also for the acids. Strong acids will absorb much less the electromagnetic radiation than the weak acids. The reason for this is that strong acids are completely dissociated.

Why does that indicate the photometric titration ?

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    $\begingroup$ This statement seems incorrect: "Strong acids will absorb much less the electromagnetic radiation than the weak acids. The reason for this is that strong acids are completely dissociated". Sulphuric acid is a strong acid which is colourless whether dissociated or not. Trinitromethane is a relatively strong acid which is colourless when undissociated, but strongly yellow when dissociated. Furthermore, many pH indicators are weak acids which are strongly coloured both when dissociated or not. I don't see how the proposed relation between colour and acid strength works. $\endgroup$ Commented May 10, 2015 at 22:13

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photometric titration happens only in one condition, that one of the following substances will absorb the electromagnetic radiation.

Almost ;-)

Regardless whether you monitor a complexometric titration by eye or with the help of a spectrometer, it is a good idea to use an indicator, such as Eriochrome Black T. Then, it doesn't matter, if or how strong the EDTA complex of the metal absorbs!

Instead, you monitor the absorption of the indicator, that is liberated from the metal-indicator complex upon addition of EDTA.

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  • $\begingroup$ Yes but i didn't mean classical methods by that $\endgroup$ Commented May 21, 2015 at 13:50
  • $\begingroup$ @NdrinaLimani so what exactly did you mean? I find your question quite unclear as it stands. $\endgroup$
    – MarcoB
    Commented Jun 10, 2015 at 6:54
  • $\begingroup$ Well i think you should read it more carefully, because I re-read it and I dont know any other way to explain it $\endgroup$ Commented Jun 10, 2015 at 20:00

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