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    Chemistry

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    vijay
    This document discusses double or poly titrations using two indicators. Double indicators are used to titrate mixtures containing two bases. There are two types of double titrations: 1) separate experiments adding each indicator or 2) adding the second indicator after the first reaction is complete. Phenolphthalein is an acid indicator used in acidic media while methyl orange is a basic indicator used in acidic or basic media. Three examples of titration reactions are given using various mixtures of NaOH, NaHCO3 and Na2CO3 with the amounts reacted by each indicator calculated. Three problems are presented where the masses of NaOH or Na2CO3 in a mixture are calculated based on the titration results.

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    Chemistry

    Uploaded by

    vijay
    88 views2 pages
    This document discusses double or poly titrations using two indicators. Double indicators are used to titrate mixtures containing two bases. There are two types of double titrations: 1) separate experiments adding each indicator or 2) adding the second indicator after the first reaction is complete. Phenolphthalein is an acid indicator used in acidic media while methyl orange is a basic indicator used in acidic or basic media. Three examples of titration reactions are given using various mixtures of NaOH, NaHCO3 and Na2CO3 with the amounts reacted by each indicator calculated. Three problems are presented where the masses of NaOH or Na2CO3 in a mixture are calculated based on the titration results.

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    This document discusses double or poly titrations using two indicators. Double indicators are used to titrate mixtures containing two bases. There are two types of double titrations: 1) separate experiments adding each indicator or 2) adding the second indicator after the first reaction is complete. Phenolphthalein is an acid indicator used in acidic media while methyl orange is a basic indicator used in acidic or basic media. Three examples of titration reactions are given using various mixtures of NaOH, NaHCO3 and Na2CO3 with the amounts reacted by each indicator calculated. Three problems are presented where the masses of NaOH or Na2CO3 in a mixture are calculated based on the titration results.

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    © All Rights Reserved
    Download as DOC, PDF, TXT or read online from Scribd
    Download as doc, pdf, or txt
    88 views2 pages

    Chemistry

    Uploaded by

    vijay
    This document discusses double or poly titrations using two indicators. Double indicators are used to titrate mixtures containing two bases. There are two types of double titrations: 1) separate experiments adding each indicator or 2) adding the second indicator after the first reaction is complete. Phenolphthalein is an acid indicator used in acidic media while methyl orange is a basic indicator used in acidic or basic media. Three examples of titration reactions are given using various mixtures of NaOH, NaHCO3 and Na2CO3 with the amounts reacted by each indicator calculated. Three problems are presented where the masses of NaOH or Na2CO3 in a mixture are calculated based on the titration results.

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    DOUBLE INDICATOTRS

    If two or more indicators are used to titrate the mixture of solutions, then it is called
    double or poly titrations, here the indicators are called double or poly indicators.
    Ex: Phenolpthalein( HPh) is a weak organic acid, it is used in only acidic media
    If HPh is used as indicators with various reagents the following reactions can take place.
    1. Reaction with NaOH : complete reaction can take place
    NaOH + HCl or H2SO4 NaCl or Na2SO4 + H2O
    2.Reaction with NaHCO3 : No reaction can take place due to high common ion effect
    NaHCO3 + HCl or H2SO4 No Reaction
    3. Reaction with Na2CO3 : 1st step reaction only can take place or half-neutralisation can
    Take place
    Na2CO3 + HCl or H2SO4 NaHCO3 + NaCl or NaHSO4
    NaHCO3 + HCl or H2SO4 No reaction.
    Ex: Methyl orange(MeOH) : weak organic base, it can be used in acidic as well as basic
    Media.
    1. Reaction with NaOH : complete reaction can take place
    NaOH + HCl or H2SO4 NaCl or Na2SO4+ H2O
    2.Reaction with NaHCO3 : complete reaction can take place
    NaHCO3 + HCl or H2SO4 NaCl or Na2SO4+ H2O + CO2
    3. Reaction with Na2CO3 : Complete reaction can take place
    Na2CO3 + HCl or H2SO4 NaCl or Na2SO4+ H2O + CO2
    Double titrations are two types
    1. In two separate experiments addition of HPh and MeOH
    MIXTURE

    HPh (reacts with)

    1. NaOH ( x meq)+ NaHCO3 (y-meq)


    2. NaOH ( x meq)+ Na2CO3 (z-meq)

    3.

    NaHCO3 ( y meq)+ Na2CO3 (z-meq)

    MeOH (reacts with)

    x-meq

    (x + y) meq

    z
    x+
    2

    (x + z) meq

    z
    2

    (y + z) meq

    2. 2nd type of titration: first addition of HPh after the completion of reaction, for the same beaker adding
    MeOH, unreacted part reacts with Acids in presence of MeOH
    MIXTURE

    HPh (reacts with)

    1. NaOH ( x meq)+ NaHCO3 (y-meq)


    2. NaOH ( x meq)+ Na2CO3 (z-meq)

    3.

    NaHCO3 ( y meq)+ Na2CO3 (z-meq)

    MeOH (reacts with)

    x-meq

    y- meq

    z
    x+
    2

    z
    - meq
    2

    z
    2

    (y +

    z
    ) meq
    2

    Problems:
    1. A mixture of NaOH and Na2CO3 required 15ml of N/20 HCl in presence of HPh as a
    indicator. But the same amount of solution using MeOH 25ml of N/20 HCl is required.
    Find the mass of NaOH.
    Ans: it is a 1st type of titration.
    Using HPh : NaOH ( x meq)+ Na2CO3 ( y/2 -meq)
    x + y/2 = 15 x N/20 ----- 1
    Using MeOH: NaOH ( x meq)+ Na2CO3 ( y-meq )
    x + y = 25 x N/20 ------2
    -3
    From 1 and 2 x =0.25meq. == 0.25 x 10 eq
    W
    W
    number of equivalents

    0.25 x103 102 g


    E.Wt
    40
    2. 10ml of solution containing NaOH and NaHCO3 required 2.5ml of 0.1M H2SO4 in
    presence of HPh as a indicator. MeOH is added for the same container after the
    completion of HPh reaction. 2.5ml of 0.2M H2SO4 is required. Find the mass of NaOH.
    nd
    Ans: it is a 2 type of titration.
    Using HPh : NaOH ( x meq) only consumed
    x = 2.5 x 1x0.1M
    Using MeOH: already NaOH is consumed remaing is NaHCO3 ( y-meq )
    y = 2.5 x 1x0.2
    -3
    x =0.25meq. == 0.25 x 10 eq
    W
    W
    number of equivalents

    0.25 x103 102 g /10ml


    E.Wt
    40
    3. 10ml of solution containing Na2CO3 and NaHCO3 required 2.5ml of 0.1M H2SO4 in
    presence of HPh as a indicator. MeOH is added for the same container after the
    completion of HPh reaction. 2.5ml of 0.2M H2SO4 is required. Find the mass of Na2CO3..
    nd
    Ans: it is a 2 type of titration : == 84 x 10-3 g/10ml.

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