Question 1 Zn(II) and the ligand L form a product that absorbs strongly at 600 n
ID: 546828 • Letter: Q
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Question 1 Zn(II) and the ligand L form a product that absorbs strongly at 600 nm. As long as the molar concentration of L exceeds that of Zn(II) by a factor of five, the absorbance is dependent only on the cation concentration. Neither Zn(II) nor L absorbs at 600 nm. A solution that is 1.60x104 M in Zn(II and 1.00x10-3 M in L has an absorbance of 0.464 in a 1.00 cm cell at 600 nm. Calculate (a) the percentage transmittance of this solution. (b) the percent transmittance of this solution in a 2.50 cm cell. (c) the molar absorptivity of the complex.Explanation / Answer
(a) The absorbance is given as A = 0.464; use the relation A = 2 – log (%T) to find the percentage transmittance (%T = percentage transmittance).
0.464 = 2 – log (%T)
====> log (%T) = 2 – 0.464 = 1.536
====> %T = antilog(1.536) = 34.35579 34.356
The percentage transmittance of the solution is 34.356% (ans).
(b) Determine the ratio of the concentration of L and Zn(II) as [L]/[Zn(II)] = (1.00*10-3 M)/(1.60*10-4 M) = 6.25 > 5 ([..] denote molar concentrations); therefore, we can assume the absorbance of the solution to be dependent only on the concentration of Zn(II). Use Beer’s law:
A = *[Zn(II)*l where = molar absorptivity of the complex and l = path length of the solution = 1.00 cm. Plug in values.
0.464 = *(1.60*10-4 M)*(1.00 cm)
====> = (0.464)/(1.60*10-4 M).(1.00 cm) = 2900 M-1cm-1.
The molar absorptivity of the complex is constant for a particular complex. The concentration of Zn(II) is kept fixed at 1.60*10-4 M; however, the path length is now l = 2.50 cm. Plug in values to get the absorbance A’.
A’ = (2900 M-1cm-1)*(1.60*10-4 M)*(2.50 cm) = 1.16.
Again, log(%T) = 2 – A’ = 2 – 1.16 = 0.84
===>%T = antilog(0.84) = 6.9183 6.918
The percentage transmittance of the solution in a 2.50 cm cell is 6.918% (ans).
(c) The molar absorptivity of the complex has been obtained in part (b) as 2900 M-1cm-1 (ans).
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