Referring to the table in line 2, what is the final concentration of Fe2+ in ppm

Question

Referring to the table in line 2, what is the final concentration of Fe2+ in ppm of the standard prepared in that step? For this question, assume that you added exactly the volumes of all reagents as listed in the table. (remember that the stock solution concentration is 50 ppm).

Suppose that for the standard prepared in line 4, a total of 4.08 mL of the iron stock solution was added to the volumetric flask instead of exactly 4.00 mL. What would the final concentration of iron be in ppm?

Plot the values of the four standards on graph paper with the absorbances on the ordinate and the ppm of iron on the abscissa. Then, using a transparent plastic ruler, draw the best straight line that will fit those four points. It must be noted that in very dilute solutions that line will probably not intersect the origin. Mark the absorbance value of the unknown on that line and drop a perpendicular from that point to the abscissa. You have now established the ppm of iron in your unknown. Empty the volumetric flasks and rinse several times. Repeat the rinsing process with six cuvettes and store them, inverted, in the plastic rack. Table: mL of Reagents added to volumetric flask Fe2+ NH4Ac NH2OH-HCl phen 0.0 1.0 10 2.0 4.0 10 8.0 10 UNKNOWN *Note. The values above are volumes of reagents that are added to make the blank (#1), the four standards

Explanation / Answer

Use the dilution equation: M1*V1 = M2*V2 where M1 = concentration of the stock (concentrated solution), V1 = volume of stock solution taken, M2 = concentration of the dilute solution (solution prepared by mixing all the reagents) and V2 = volume of the dilute solution.

For the solution in line 2, M1 = 50 ppm, V1 = 1.0 mL and V2 = (1.0 + 5 + 5 + 10) mL = 21.0 mL.

Plug in values and write

(1.0 mL)*(50 ppm) = (21.0 mL)*M2

===> M2 = (1.0*50)/(21.0) ppm = 2.381 ppm (ans).

For the solution in line 4, M1 = 50 ppm, V1 = 4.08 mL and V2 = (4.08 + 5 + 5 + 10) mL = 24.08 mL.

Plug in values and write

(4.08 mL)*(50 ppm) = (24.08 mL)*M2

===> M2 = (4.08*50)/(24.08) ppm = 8.472 ppm (ans).

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