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    zeroerror's Avatar
    zeroerror Posts: 2, Reputation: 1
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    #1

    Jul 3, 2009, 06:54 PM
    Determination of sulphate by gravimetric analysis
    Hi, when they asked to add dilute solution of barium chloride to a hot solution of sodium sulphate slightly acidified with concentrated HCL. Is there a difference if I add BACL to HCL first then add NASO4? Or if I add NASO4 and HCL then BACL.

    I'm not sure if I'm correct but if you add NASO4 to HCL first, it causes the dissociation of ions in solution so SO42- is free to react with BACL upon addin it later?
    Perito's Avatar
    Perito Posts: 3,139, Reputation: 150
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    #2

    Jul 3, 2009, 07:07 PM
    When asked to add dilute solution of barium chloride to a hot solution of sodium sulphate slightly acidified with concentrated HCl. Is there a difference if I add BaCl to HCl first then add Na2SO4? Or, should I add Na2SO4 and HCl then BaCl?

    I'm not sure if I'm correct but if you add Na2SO4 to HCl first, it causes the dissociation of ions in solution so SO4^(2-), is free to react with BaCl upon adding it later?
    I would add the Na2SO4 last since that's what will cause the precipitation. You want the solution to be complete and slightly acidic before trying to precipitate anything.

    You're not correct that adding Na2SO4 to HCl will cause the dissociation of ions in solution. The ions will dissociate regardless. The only advantage to adding the HCl first is to ensure that you have slightly acidic solution to begin with. This will prevent any carbonates from precipitating. Basic or neutral solutions will absorb carbon dioxide from the air and form carbonates. The acid tends to force the CO2 out of solution. You don't want any carbonates to precipitate so you acidify the solution first.

    Here's a link to some questions about this analysis that I answered some time ago.

    Link to "Gravimetric Analysis URGENT!"
    zeroerror's Avatar
    zeroerror Posts: 2, Reputation: 1
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    #3

    Jul 3, 2009, 07:28 PM
    oh okay. Thanks thanks! =]
    anafiz's Avatar
    anafiz Posts: 1, Reputation: 1
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    #4

    Mar 24, 2011, 07:07 AM
    Show that the volume 0.25 M BaCl2 solutions needed is 17ml

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