Question

A 11.17 mg ore sample containing copper is digested with acid, filtered and transferred to a volumetric flask and diluted to exactly 200.00 mL. A 10.00 mL aliquot is transferred to a 100.00 mL volumetric flask containing zincon reagent and a quantity of a pH 9 buffer solution. It is then diluted to the mark and mixed thoroughly. The measured absorbance of the resulting blue solution at 600 nm was 0.326 which corresponded to a Cu(II) concentration of 0.498 ppm (mg/L). Determine the percent age of copper in the ore sample.

Answer 1

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11.17mg of sample - 200 mL solution A (10mb loom L Solution Now, concentration of Colution B is 0.498 ppm = 0.498 mg/L so, concentration of solution A - ? use av = Cara formula Initial final . Then c = ?? C C = 0.498 ugle (Concentrates (diluted) V = 10mL = 100 mL = C2 V ₂ = 6.498 mg/L) (100m) = 4.98 mg/L løfor and volume of solution of = 200mL concentration of solution A = 4.98 mg/L Here 4.98 mg of ca in 1 L = 1000mL of solution n So g of Cu in of solution 200mL > 4.98ing x 200ml - anong - 0.996 mg of cu 1000mL (This has come from 11.17 ug of) % of Cu = mass of cu : sample mass of J x 100 = 0.996 mg sample all ma mas X LOO 11.17 ng = 8.916 % = 8.92%