You need to make a 1.00 ppm (aqueous) iron standard to calibrate an analyticalin

Question

You need to make a 1.00 ppm (aqueous) iron standard to calibrate an analyticalinstrument. What mass of “FAS” [Fe(NH4)2(SO4)2·6H2O] is needed to make 1.00 L of this solution? Using commonly available sizes of volumetric flasks and transfer pipets, design a two-step dilution procedure that allows you to startwith a larger mass of FAS—enough to get at least four significant figures in the initialweighing. Specify the concentration and volume of your initial stock solution, the mass of FAS you would have to weigh out to make it, and what glassware you would use todilute it to obtain 1.00 L of 1.00 ppm Fe.

Can someone please explain how you could go about doing this please?

Explanation / Answer

We need 1ppm Fe solution of 1litre

1ppm = 1mg /litre

So, we need total amount of 1mg of Fe in 1litre .This should be preapared in two steps

Let us, arbitrarily, weigh acurately 4.525g of FAS . From this amount of FAS we need to prepare 1ppm of Fe std solution in two steps

First consider the molar mass of Fe which is 392.13 g/ mol

The formula is Fe(NH4)2(SO4)2.6H20

1mol of FAS having 1mol oF Fe

Molar mass of Fe = 55.845g/mol

Therefore, 392.13g of FAS having 55.845g of Fe and the conversion factor of FAS weight to Fe weight is 0.1424

Therefore, Fe present in 4.525 g of FAS is 0.1424 × 4.525 = 0.6444g

Let us dissolve 4.525g of FAS in distilled water (slightly aciify to avoid hydrolysis) and make up it for 500ml in an 500ml standard measuring flask (SMF). Now , this solution having the concentration of 1288.8 mg/ litre(ppm) Fe.

Let us dilute this solution 10 time . For this , if we choose 100 ml SMF ,we pipette out 10ml From the above solution ( 100/10 =10 ) and transfer it into 100 ml SMF and make up it for 100ml with disstilled water.Now , the prepared concentration is 1288.8 /10 = 128.88

Now, we have to dilute this solution to 1ppm ,for this we can use one formula i.e

C1 V1 = C2 V2

Where, C1 and V1 are initial concentration and volume respectively and C2 and V2 are final concentration and volume respectively

C1 = 128.88 ppm , V1 = ?

C2 = 1ppm , V2 =1000ml

128.88 ppm × V1= 1ppm × 1000ml

V1 = (1ppm × 1000ml)/128.88ppm = 7.759ml

So , if we take 7.759ml from the 128.88 ppm solution and transfer it into 1000ml of SMF and make up it for 1000ml with distilled water , we can get 1ppm Fe standard solution finally.

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