The UltraPhen molecule is found to form complexes quickly with Cr^3+, Ni^2+ and
ID: 1058402 • Letter: T
Question
The UltraPhen molecule is found to form complexes quickly with Cr^3+, Ni^2+ and Zn^2+, each with a different wavelength of maximum absorption (410 nm, 535 mu and 610 nm, respectively). This leads to the possibility that solutions containing all 3 ions could be analyzed by simultaneous spectrometric determination. To utilize this application, you dissolve 0.1175 g of a steel sample in acid, add an excess of the UltraPhen, then dilute it to 100.0 mL with deionized water. The following % transmittance values were obtained for the sample solution when measured in a 1.00 cm cuvette at 410 run, 535 mn and 610 mn, respectively: 38.5%, 18.8%, 46.1%; given the following information regarding molar absorptivities, calculate the mass percentages of Cr, Ni and Zn on the steel sample. Molar absorptivity (cm^-1 M^-'1)Explanation / Answer
Ans. Beer-Lambert’s Law, A =e C L - equation 1, where,
A = Absorbance
e = molar absorptivity at specified wavelength
L = path length (in cm)
C = Molar concentration of the solute
Note: A higher value of e means that at particular wavelength (compared to other wavelengths) means that the molecule absorbs light very efficiently at that wavelengths because of electronic transitions to greater extent. Therefore, a wavelength with higher e value shall be preferred because it gives more sensitive (can detect very low concentration of the compound) result.
# Absorbance of a solution can be calculated using given value of transmittance as follow-
Absorbance, A = 2 - log (% transmittance) = 2- log (T)
Part A: Cr3+
Absorbance = 2 – log T = 2- 1.58546 = 0.41453
Now,
C = A/ (eL) = 0.41453 / (1054 M-1cm-1 x 1 cm) = 3.9329 x 10-4 M
That is, 3.9329 x 10-4 moles Cr3+ in 1000 ml (= 1L) solution.
Since, the final volume is made upto 100 mL only, the number of moles of Cr3+ in 100 ml volume = 3.9329 x 10-5. Since 100 ml solution in made from 0.1175 gram steel, total number of moles of the cation in 0.1175 gram sample is equal to 3.9329 x 10-5.
Mass of Cr3+ in 100 ml volume = moles x atomic mass of Cr
= 3.9329 x 10-5 mole x 51.996 g mol-1
= 0.0204496 gram
% of Cr3+ in sample = (Mass of Cr3+ / total mass of steel sample) x 100
= (0.0204496 gram / 0.1175 gram) x 100 = 17.40%
Part B: Ni2+
Absorbance = 2 – log T = 2- 1.27415 = 0.725842
Now,
C = A/ (eL) = 0.725842/ (1263 M-1cm-1 x 1 cm) = 5.746968 x 10-4 M
That is, 5.746968 x 10-4 moles Ni2+ in 1000 ml (= 1L) solution.
Since, the final volume is made upto 100 mL only, the number of moles of this cation in 100 ml volume = 5.746968 x 10-5. Since 100 ml solution in made from 0.1175 gram steel, total number of moles of this cation in 0.1175 gram sample is equal to 5.746968 x 10-5 .
Mass of Ni2+ in 100 ml volume = moles x atomic mass of Cr
= 5.746968 x 10-5 mole x 58.693 g mol-1
= 0.03373 gram
% of Ni2+ in sample = (Mass of Ni2+ / total mass of steel sample) x 100
= (0.03373 gram / 0.1175 gram) x 100 = 28.70%
Part C: Zn2+
Absorbance = 2 – log T = 2- 1.6637009 = 0.336299
Now,
C = A/ (eL) = 0.336299/ (1143 M-1cm-1 x 1 cm) = 0.001455556365 M
That is, 0.001455556365 moles Zn2+ in 1000 ml (= 1L) solution.
Since, the final volume is made upto 100 mL only, the number of moles of this cation in 100 ml volume = 0.0001455556365. Since 100 ml solution in made from 0.1175 gram steel, total number of moles of this cation in 0.1175 gram sample is equal to 0.0001455556365.
Mass of Zn2+ in 100 ml volume = moles x atomic mass of Cr
= 0.0001455556365 mole x 65.380 g mol-1
= 0.0095164275 gram
% of Zn2+ in sample = (Mass of Zn2+ / total mass of steel sample) x 100
= (0.0095164275 gram / 0.1175 gram) x 100 = 8.09 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.