Question

A Cu(II) Standard solution has been diluted as following table concentration and react with NH₃ solution to form a colored complex as Cu(NH₃)Cl₂. Each of the standard has been measured the % Transmission in a 1.00 cm cell at 590 nm :

Concentration of Cu(II) in ppm	% T A510
10	63.1
20	47.9
30	37.2
40	30.9
50	25.7
60	19.5
70	31.6
80	12.0

a. You took a Copper wire sample, weight 1 gram and dissolved to 1 L using volumetric flash with HCl, then take 10 mL fo the solution and complex with Ammonia and dissolved in 250 mL volumetric flash using deionized water. This sample is measured the % T and the result is 35.6. What is the % Cu in your wire sample ?

b. calculate the ε.

Answer 1

20 20 60 1 concentration of u (1) ppmco YT Absorbance = 2-logy.T 63.1 0:20 47.9 0.32 37.2 0.43 30:9 0:51 25.7 0.59 19.5 0.71 70 31.6 0.50 12.0 0.92 Unknown . 35.6 0:45 @ According to Beer's Law =) Absorbance = e.cl From curve, concentration corresponding to Absorbance = 0:45 i's 37.5 ppm 250 ml dilute solution contains = 37.5x250 : 1000 -complise = 9,375 mg cult » » I so, 10 ml ce2+ solution contains 9.375 mg cu 2+ 1:1000 ml Cu2+ o * s9 nge 9. (9-315x1000) = 937.5 mg cost (91375x100 2+ in = 0.9375g cult ::.%. Cue in the sample =(0-9375 x100)% = 93.75% (ans) © slope = txt Lo 0.012 ppñ= txt cu or € = 0.012 ppă cu' (ans)

Plot of Absorbance Vs concentration X-Axis concentration : Ismall unit = 0.5ppm Y- Anu's → Absorbance: I small unel = 0.005 Slobe 3485X0.005 X 15 4= 0.012 bpm 4 ppio OD = (75X05)= 37.5 ppm 0-5 H00 saco 70 go 2030 D40 e (ppm)