3. An unknown sample of cu gave an absorbance of o.262 in an atomic absorption w

Question

3. An unknown sample of cu gave an absorbance of o.262 in an atomic absorption with analysis. Then 1.00 ml of solution containing 100.0 ppm ug/mLI cu was mixed 95.0 of unknown, and the mixture was diluted to 100.0 mL in a volumetric flask. The cu in the unknown. (10 points) a a 1.00 SS02. Soaa 4. Cobalt was used as an internal standard to analyze a sample of titanium with atomic absorption spectroscopy. A mixture was prepared by combining a 6.00 mL Ti solution of unknown concentration with 1.50 mL of a 12.8 ug/ml solution of Co. The atomic absorbances were measured as 0.134 and 0.226 for Ti and Co, respectively. As a reference, a standard mixture containing 190 Hg Co/mL and 1.87 ug Ti/mL was prepared and measured to have a signal-to-signal ratio of 2.69 Ti: 1.00 Co. Determine the concentration (moles/t) of titanium in the original unknown solution. (10 points)

Explanation / Answer

3. Concentration of standard in diluted solution = 100 ppm x 1 ml/100 ml = 1 ppm

absorbance for unknown = 0.262

absorbance for unknown + standard = 0.5

So,

concentration of unknown in diluted solution = 0.262 x 1 ppm/0.5 = 0.524 ppm

Concentration of Cu2+ in unknown solution = 0.524 ppm x 100/95 = 0.5515 ppm

4. Responde factor (F) = (Area of Co/conc. of Co)std/(area of Ti/conc of Ti) std

                                     = (1/1.9)/(2.69/1.87)

                                     = 0.526/1.438

                                     = 0.366

For unknown Ti solution,

concentration of Co = 1.5 ml of 12.8 ug/ml = 12.8 x 1.5/7.5 = 2.56 ug

So,

0.366 = (0.226/2.56)/(0.134/conc. of unknow Ti solution)

concentration of diluted unknown Ti solution = 0.555 ug

So, concentration of Ti in the original unknown solution = 0.555 x 7.5/6 = 0.694 ug

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