Mn^2+ is analyzed by oxidizing it to MnO^-_4 and measuring the absorbance of the
ID: 515939 • Letter: M
Question
Mn^2+ is analyzed by oxidizing it to MnO^-_4 and measuring the absorbance of the oxidized product, which absorbs strongly at 525 nm. Two solutions are prepared as shown in the table and their absorbance values are measured at 525 nm in a 1.00-cm cuvette. Calculate the concentration of Mn^2+ in the original unknown sample. Beer's law plots are generally done lambda_max, this wavelength (A) the sensitivity is the lowest. (B) the slope of the Beer's law plot is the (C) the absorbance value is independent of pathlength. (D) the absorbance is free from interference absorbers. Which of the following analytes would be for analysis by gas chromatography? (A) proteins in human serum (B) metal cations m wastewater (C) industrial solvents in groundwater (D) amino acids in infant formula mixExplanation / Answer
Ans. #1. Final volume of spiked solution = 10.0 mL
Volume of standard Mn2+ added to spiked-solution = 1.0 mL
Now, using C1V1 = C2V2 - equation 1
C1= Concentration, and V1= volume of initial solution 1 ;(standard Mn2+ soln.)
C2= Concentration, and V2 = Volume of final solution 2 ;( Sample spiked with Mn2+)
Putting the values in equation 1-
0.0010 M X 1.0 mL = C2 x 10.0 mL
Or, C2 = (0.0010 M X 1.0 mL) / 10.0 = 0.00010 M
Thus, increase in concertation of Mn2+ in the spiked sample solution after adding standard Mn2+ solution = 0.00010 M
#2. Absorbance of original un-spiked aliquot = 0.254
Absorbance of spiked sample = 0.355
Increase in concentration of spiked sample = 0.00010 M
Increase in absorbance = Abs of spiked sample - Abs of un-spiked aliquot
= 0.355 – 0.254 = 0.101
Increase in absorbance of spiked sample is solely due to addition of standard Mn2+ solution that increase Mn2+ concertation by 0.00010 M.
Therefore, increase of 0.101 unit of absorbance is equivalent to increase in concertation by 0.00010 M
Thus, 0.101Abs unit = 0.00010 M Mn2++
or, 1 Abs unit = 0.000990 M
[Mn2+] in original sample = Abs of original sample x ([Mn2+] equivalent to 1.0 unit Abs)
= 0.254 x 0.000990 M
= 0.00025146 M
Therefore, [Mn2+] in 10.0 mL original un-spiked aliquot = 0.00025146 M
#3. Note that in the 10.0 mL unknown aliquot, there is 1.0 mL original unknown sample, the rest being water as diluent.
Let the [Mn2+] in 1.0 mL, undiluted original unknown sample be C1.
Putting the values in equation 1-
C1 x 1.0 mL = 0.00025146 M x 10 mL
Or, C1 = 0.0025146 M
Therefore, [Mn2+] in 1.0 mL original unknown sample = 0.0025146 M
= 2.5 mM ; [1 M = 103 mM]
Correct option. C.