Question

2. a. was The analysis of the complex formed between Cu(I) and 1,10-phenanthroline conducted using molecular spectroscopic method. The results showed that this complex has a molar absorptivity (8) of 7000 L cm'mol and maximum wavelength absorption (1) of 435 nm. Calculate the absorbance of a 6.77 x 10-6 M solution of the complex when measured in a 1.00 cm cell at wavelength of 435 nm. b. External calibration method is applied to determine the concentration of Na metal in unknown sample by flame atomic emission spectroscopy (AES). The results obtained are shown in TABLE 03. YAB2063 TABLE Q3 : Results of AES Solution Na emission intensity 0.2 ppm Na 0.22 0.5 ppm Na 0.53 2.0 ppm Na 2.05 5.0 ppm Na 5.00 Unknown sample 0.88 I. Plot a calibration curve and deduce a linear regression equation for this calibration curve. ii. Determine the concentration of Na metal in unknown sample using a linear regression equation in part b(i).

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Answer 1

Ans 2a. Given is the molar absorptivity of the complex, ε = 7000 L cm-- mol-. and Amax = 435 nm The molarity of the complex = 6.77 x 10-SM and the cell length = 1.00 cm Using the formula; A= E. c. 1 -(1) Putting values in (1), A = 7000 L cm - mol2 x 6.77 x 10-5Mx 1.00 cm A = 7000 L cm - mol? x 6.77 x 10-5 mol L ? x 1.00 cm A = 0.4739 is the answer Ans 2b. TABLE Q3 : Results of AES Solution Na emission intensity 0.2 ppm Na 0.22 0.5 ppm Na 0.53 2.0 ppm Na 2.05 5.0 ppm Na 5.00 Unknown sample 0.88 (1) The calibration curve for this is plotted using Origin software with linear fitting, shown below: 5.5 5.0 4.5 4.0 3.5 3.0 emission 2: 2.0 1.5 Equation Weight Residual Sum of Squares Pearson's Adj. R-Square y = a +bx No Weighting 926594E-4 1.0 0.99997 0.9999 0.5 Value 0.03478 0.99492 Intercept Slope Standard Error 0.01531 0.00566 B 0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 concentration Here, the slope is 0.99492 and the intercept is 0.03478.

So, the linear regression equation will be, le = 0.03478 +0.99492 x conc.of Na (ppm) (Ie = emission intensity) (ii) Now, for the unknown sample, the Ie = 0.88 Thus from (2), 0.88 = 0.03478 +0.99492 x conc.of Na (ppm) = 0.84954 ppm conc.of Na=0.88 -0.03478 0.99492 Hence is the answer.