An unknown sample of Cu^2+ gave an absorbance of 0.262 in an atomic absorption a

Question

An unknown sample of Cu^2+ gave an absorbance of 0.262 in an atomic absorption analysis. Then 1.35 mL of solution containing 100.0 ppm (= mu g/mL) Cu^2+ was mixed with 93.0 mL of unknown, and the mixture was diluted to 100.0 mL in a volumetric flask. The absorbance of the new solution was 0.503. 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.) 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.) Find [Cu^2+]| in the unknown.

Explanation / Answer

a) Denoting the initial concentration as [cu2+] write expression for the final concentration cu2+] after dilution

[Cu2+]f =[Cu2+]i x Vi /Vf

= =[Cu2+]i x ( 93 /100 )

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=[s ]f = [S]

Vi sp / Vfsp       = (100 ppm ) (1.35 mL/ 100mL) =1.35 ppm [Cu2+] ---------------answer

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Ax proportional                 [Cu2+]           initial

A x+s proportional [Cu2+]           after spike

=[Cu2+] after spike =[Cu2+]i x Vi /Vf          +[Cu2+]i x   Vi sp / Vfsp     

=0.95 [Cu2+]   +1.35 ppm [Cu2+]

Ax / Ax +s   = [Cu2+]i / 0.95 [Cu2+]i   +1.35 ppm= 0.2262 /0.500=0.524

[Cu2+]i =0.497 [Cu2+]i + 0.524 ppm

0.502 [Cu2+]i =0.524 ppm

[Cu2+]i =0.524 ppm /0.502= 1.0 4 ppm --------answr

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