Question

A solution was prepared by weighing 0.269 g of ammonium iron(II) sulfate hexahydrate, (NH4)2Fe(SO4)2*6H2O, and diluting...

A solution was prepared by weighing 0.269 g of ammonium iron(II) sulfate hexahydrate, (NH4)2Fe(SO4)2*6H2O, and diluting to 500.00 mL in a volumetric flask. A 5.00-mL sample of this solution was transferred to a 250-mL volumetric flask and diluted to the mark with water. What is the concentration of sulfate ions in the solution?

Homework Answers

Answer #1

molarity of (NH4)2Fe(SO4)2*6H2O(M1) = w/M*1000/V in ml

                   = (0.269/392.14)*(1000/500)

                 M1 = 0.00137 M

concentration of diluted solution(M2) = ?

V1 = 5 ml

V2 = 250 ml

FROM DILUTION FORMULA

M1V1 = M2V2

(0.00137*5) = M2*250

M2 = 2.74*10^-5 M

now, 1 mol (NH4)2Fe(SO4)2*6H2O = 2 mol SO4^2-

concentration of sulfate ions in the solution = 2*2.74*10^-5 = 5.48*10^-5 M

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