An unknown sample of Cu^2+ gave an absorbance of 0.264 in an atomic sorption ana

Question

An unknown sample of Cu^2+ gave an absorbance of 0.264 in an atomic sorption analysis. Then 1.20 mL f solution containing 100.0 ppm (= mu g/mL) Cu^2+ was mixed with 92.0 mL of unknown, and the mixture was diluted to 100.0 mL in a volumetric flask. The absorbance of the new solution was 0.505. (a) Denoting the initial unknown concentration as [Cu^2+]_i, write an expression for the final concentration, [Cu^2+]_f, after dilution. Units of concentration are ppm. (Use the following as necessary: V_0, V_s, and V.) [Cu^2+]_f = [Cu^2+]_i (b) In a similar manner, write the final concentration of added standard Cu^2+, designated as [S]_f. (Use the following as necessary: V_0, V_s, and V.) (c) Find [Cu^2+]_i in the unknown.

Explanation / Answer

Here, Ix =0.264 , Ixs =0.505 , [S]i = 100

(a) Xf = Xi . 92/100

or [Cu2+]f =[Cu2+]i x 92/100

(b) [S]f = [S]i x 1.20 /100

(c) Ix / Xi = Ixs / Xf + Sf

=> 0.264 / Xi =0.505 / Xi * 0.92 +1.20

=>0.505 Xi =0.243 Xi +0.3168

=> 0.262 Xi =0.3168

=> Xi =1.21 ppm

Thus, [Cu2+]i =1.21 ppm

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