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March 26, 2017

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You weigh out 196.07mg ferrous ammounium sulfate (FW 392.14) to prepare 250mL of solution which is 0.002M with respect to that compound. The salt is quantitatively transferred into a 250mL volumetric flask and 8mL of 3M H2SO4 is added and then diluted to the mark (call this stock Fe solution). 10mL of the stock solution is pipeted into a 100mL volumetric flask, adding 4mL of 3M H2SO4 and diluted to the mark (call this original Fe solution).

10mL of the original fe solution is pipeted into a 50mL volumetric flask (call this standard Fe solution). What is the concentration of the standard Fe solution (mg/L)?

It should be somewhere around 2, but I can't seem to calculate it correctly.

  • Chemistry - ,

    You weigh out 196.07mg ferrous ammounium sulfate (FW 392.14) to prepare 250mL of solution which is 0.002M with respect to that compound.
    The salt is quantitatively transferred into a 250mL volumetric flask and 8mL of 3M H2SO4 is added and then diluted to the mark (call this stock Fe solution).


    So this 250 ml solution is

    784.28 mg L^-1 (as FAS hydrate)
    0.00200 M
    112 mg L^-1 as Fe




    10mL of the stock solution is pipeted into a 100mL volumetric flask, adding 4mL of 3M H2SO4 and diluted to the mark (call this original Fe solution)

    Here we have dilutes by a factor of 10

    So this 100 ml solution is

    78.428 mg L^-1 (as FAS hydrate)
    0.000200 M
    11.2 mg L^-1 as Fe

    10mL of the original fe solution is pipeted into a 50mL volumetric flask (call this standard Fe solution).

    I presume this is made to the mark? In which case there is a further dilution by a factor of 5.

    15.69 mg L^-1 (as FAS hydrate)
    0.0000400 M
    2.24 mg L^-1 as Fe

    So perhaps the question is looking for the Fe concentration in mg L^-1?

    (You should check my figures as I have rounded in places)

  • Chemistry - ,

    Um, exactly how did you calculate the 112 mg L^-1 as Fe, 11.2 mg L^-1 as Fe and 2.24 mg L^-1 as Fe ?

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