5. A student used the method of standard additions (volume of the diluted unknow

Question

5. A student used the method of standard additions (volume of the diluted unknown was 10.0 mL) and fluorescence spectroscopy to measure the amount of quinine (CH2aN,02, MW 324.43 g/mol) in tonic water. A 50.0 ppm quinine sulfate ([GoHaNz012-H2SO4·2H20, Mw = 782.97 g/mol) solution was used as the standard. a. A linear regression analysis was performed on the data obtained from the experiment and produced the equation y-1532.9 x + 180. Calculate the amount in ppm of quinine sulfate in the diluted sample. b. The diluted sample of quinine sulfate was prepared by placing 500 mL of the unknown in a 250.0 mL volumetric flask and adding distilled water to the mark. Calculate the amount in ppm of quinine sulfate in the undiluted sample c. Assuming that your original (undiluted) sample contained 153 ppm (mg/L) of quinine sulfate, calculate the molarity of quinine in the original sample

Explanation / Answer

a) The concentration of quinine in the diluted sample is obtained by letting y = 0 in the regression equation. Put y = 0 and obtain

0 = 1532.9x + 180

====> -180 = 1532.9x

====> x = -180/1532.9 = -0.1174

Since the concentration of the unknown cannot be negative, hence, ignore the minus sign and report the concentration of quinine sulfate in the diluted sample as 0.1174 ppm (ans).

b) The unknown sample of quinine sulfate was prepared by placing 5.00 mL of the unknown in a 250.0 mL volumetric flask and making up to the mark with distilled water. The dilution factor is (250.0 mL)/(5.00 mL) = 50.0.

The concentration of quinine sulfate in the undiluted sample = (0.1174 ppm)*(dilution factor) = (0.1174 ppm)*(50.0) = 5.87 ppm (ans).

c) The original sample of quinine sulfate contained 153 mg/L of quinine sulfate.

The molar mass of quinine sulfate = 782.97 g/mol.

The molar concentration of quinine sulfate corresponding to 153 mg/L = (153 mg/L)*(1 g/1000 mg)*(1 mole/782.97 g/mol)*(1 M/1 mol/L) = 1.9541*10-4 M 1.95*10-4 M (ans).

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