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?
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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