Question

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.

Answer 1

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)

Answer 2

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 ?