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)
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.
2 answers
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 ?