Three unknown solutions were prepared by pipetting 10.00 mL of an unknown nickel
ID: 474478 • Letter: T
Question
Three unknown solutions were prepared by pipetting 10.00 mL of an unknown nickel solution into a 500.00 mL volumetric flask and diluting to the mark. Then three samples were prepared from this dilute unknown solution by pipetting 15.00 mL of the dilute solution into a 100.00 mL flask, to which the reagents were added then it was diluted to the mark. The resulting solutions were analyzed by UV Spectroscopy and the unknown absorbance value were converted to concentration by means of a calibration curve, Table 4-1. Fill in the diagram with appropriate volumes. Determine the concentration of the original nickel solution, in ppm Ni, from each of the three unknown solutions and an average. Included appropriate error calculations. Show complete calculations using Trial 1 data. Present results for all trials and the average with uncertainties.Explanation / Answer
Let the initial concentration of nickel solution = xppm = M1
The volume of initial nickel solution taken = V1mL = 10mL
(i) First dilution is upto 500mL =V2
So the new concentration after first dilution will be = M2
M1V1 = M2V2
M2 = M1 X V1/ V2 = x (10) / 500 = 0.02 x ppm
(ii) in second dilution the above 0.02x ppm solution is further diluted to 100mL (V3) by taking 15mL
So the new concentration = M3 will be
0.02x X 15 = 100 X M3
M3 = 0.003 x ppm
Now the concentrations (M3) for three solutions is given in the table
1) 5.689 +/-0.451
% error = error X 100 / actual value = 0.451 X 100 / 5.689 = 7.93%
0.003x = 5.689
x = 1896.333 ppm
error = 7.93 X 1896.333 / 100 = 150.389
(ii) 5.484 +/- 0.447
% error = error X 100 / actual value = 0.447 X 100 / 5.484 =8.15%
0.003x = 5.484
x = 1828.000 ppm
error = 8.15 X 1828.000 / 100 = 148.982
(iii) 6.004 +/-0.448
% error = error X 100 / actual value = 0.448 X 100 / 6.004 = 7.46%
0.003x = 6.004
x =2001.333 ppm
error = 7.46X 2001.333 / 100 = 149.299
The average value will be
1896.333+1828.000+ 2001.333 / 3 = 1908.555 ppm
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