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Topic: Something about the titration by double indicator method  (Read 8010 times)

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Offline shuryanc

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Something about the titration by double indicator method
« on: May 06, 2010, 09:56:01 AM »
The titrate contain a mixture of Na2CO3 and NaOH and titrate with HCl.
The first indicator I use is phenolphthalein and the second is methyl orange.
The thing I want to know is: why can't the order of indicators be reversed? Thank.

Offline AWK

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Re: Something about the titration by double indicator method
« Reply #1 on: May 06, 2010, 10:20:12 AM »
Na2CO3+HCl=NaHCO3+NaCl At equivalence point (pH about 8-9) phenophtalein becomes colourless
NaHCO3+HCl=H2CO3+NaCl At this equivalence point (pH about 7 - note carbonic acid should be decomposed by warming) the second indicator changes its color.

This is just optimized procedure and there is nothing to discuss
« Last Edit: May 06, 2010, 10:44:35 AM by AWK »
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Offline Borek

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ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline shuryanc

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Re: Something about the titration by double indicator method
« Reply #3 on: May 06, 2010, 11:34:58 AM »
So if the order is changed, the first equivalent will need a long time to reach and the second equivalent point will not exist? Any more reasons? Thank(I have given you snack(both))

Offline Pradeep

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Re: Something about the titration by double indicator method
« Reply #4 on: May 06, 2010, 01:04:14 PM »
The active pH range of Methyl orange 3.5-5.5
The active pH range of Phynopthalein 9-11

The initial pH of the Na2CO3 solution is always supposed that above 11.

When U add HCl

Na2CO3 and NaOH will react with HCL. So the pH of the solution will come down. When that pH comes down to below 9 phnopthalein willl be colorless.

when that pH comes down to below 3 methyl orannge will be yello (what ever the changed color)

So color change of the phynopthalein comes first. So u have to use it first and then met Orange.

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« Last Edit: May 06, 2010, 02:37:37 PM by Borek »

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