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Question about dissolving Na2CO3.xH2O in water and getting x with a HCl titration

  • 29-05-2007 6:39pm
    #1
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭


    It's an LC experiment. Basically, you dissolve certain mass of hydrated Sodium Carbonate in a specific amount of water(ie. you know the concentration in g/l) and titrate it against HCl to find the concentration of the Na2CO3.xH2O in mol/l. At this point, since you know the concentration in g/l and mol/l you can get the molar mass and put it equal to 106 + 18x to get the value of x.

    My problem with understanding this is that when you dissolve the Na2CO3.xH2O in the water is the xH2O not liberated and therefore is the titration not simply measuring the concentration of Na2CO3 and not the concentration of Na2CO3.xH2O?

    Is it simply assumed for the LC that [Na2CO3] = [Na2CO3.xH2O], or am I wrong in some way?


Comments

  • #2
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Actually, thinking about it now, the concentration of Na2CO3 would be equal to the concentration of Na2CO3.xH2O, wouldn't it? .....

    Still, I'd like someone to clarify, thanks.


  • #3
    DemocAnarchis
    Registered Users Posts: 404 ✭✭DemocAnarchis


    You're basically correct. The complexed H2O will be liberated, but you arent titrating against it, you are titrating against Na2CO3.


  • #4
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    So, to clarify, is it a simplifying assumption or is it correct?

    And is the x just an average number?


  • #5
    DemocAnarchis
    Registered Users Posts: 404 ✭✭DemocAnarchis


    You were making it a little more complicated than nessecary, but yes, it is correct.

    X is not an average number, X should be a whole number, referring to the number of molecules of H2O complexed to the Na2CO3, known as water of hydration. This will be the same for each of the Na2CO3 groups.


  • #6
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    So there's the exact same amount of water molecules bonded to every Na2CO3 group?

    That kinda doesn't make sense to me, but I guess I'll accept it.


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  • #7
    DemocAnarchis
    Registered Users Posts: 404 ✭✭DemocAnarchis


    JC 2K3 wrote:
    So there's the exact same amount of water molecules bonded to every Na2CO3 group?

    That kinda doesn't make sense to me, but I guess I'll accept it.

    Well, think about it like this

    You titrate against HCl to find how many moles of Na2CO3 are present, and calculate moles/litre of Na2CO3. From this the molar mass can be deduced, lets call it Y. The mass of Na2CO3 is 106, and the mass of H2O is 18, so as you said,

    Y = 106 + 18x

    For the purpose at hand, lets say x = 2. What does that mean? That means for every molecule of Na2CO3 present, there are two water molecules present. If the value of x was just an average, you could get a value such as 1.3, 3.6, 6.5 etc. Now, does that make sense? No. The molecular formula is Na2CO3.xH2O, and those values would imply that you could have 1.3 water molecules bound to it. X will always be a whole number, because x is a constant.


  • #8
    JC 2K3
    Closed Accounts Posts: 7,794 ✭✭✭JC 2K3


    Oh, I understand that. What seems odd is that the exact same number of water molecules would be bonded to each molecule of Na2CO3 in the sample. Expecially since the amount of water bonded to it varies from sample to sample....

    Also, in the book it said that not getting a whole number was not uncommon as the crystals lose water over time....


  • #9
    DemocAnarchis
    Registered Users Posts: 404 ✭✭DemocAnarchis


    Well the experiment lead me to believe that you were analysing an unknown, such as Sodium Carbonate Pentahydrate, if it was an impure mix, ie a sample of Na2CO3 was subject to hydration conditions, where the degree of hydration could be different, then of course there would be variation. There could well be a loss of water over time, my statement was based assuming ideal conditions, ie no loss of water to the surroundings.


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