Solution #1 absorbance= 0.09 #2 = 0.13 #3 = 0.25 #4 =0.36 #5 = 0.45 (Also attach

Question

Solution #1 absorbance= 0.09 #2 = 0.13 #3 = 0.25 #4 =0.36 #5 = 0.45 (Also attached a table of absorbances) Data -Using the calibration curve provided by your instructor, determine the concentration of FeSCN in each ofyour solutions #1-5. For each solution, find the initial amount of Fe3+ and SCN (in mol). The calibration curve is provided below: Absorbance 0.061 0.104 0.217 0.280 0.350 [FeSCN2+1 (10-5 M) 7.2 9.6 12 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 y 0.0278x + 0.0154 R2 0.9991 0 24 6 8 10 12 14 [FeSCN2 1/105M

Explanation / Answer

The equation of the plot of absorption versus concentration of the solution is given as

Absorbance (A) = 0.0278*[FeSCN2+] + 0.0154

[FeSCN2+] value can be written as [FeSCN2+]*10-5 M

If A = 0.09

i.e. 0.09 = 0.0278*[FeSCN2+] + 0.0154

i.e. [FeSCN2+] = (0.09-0.0154)/0.0278 = 2.7*10-5 M

If A = 0.13

i.e. 0.13 = 0.0278*[FeSCN2+] + 0.0154

i.e. [FeSCN2+] = (0.13-0.0154)/0.0278 = 4.1*10-5 M

If A = 0.25

i.e. 0.25 = 0.0278*[FeSCN2+] + 0.0154

i.e. [FeSCN2+] = (0.25-0.0154)/0.0278 = 8.4*10-5 M

If A = 0.36

i.e. 0.36 = 0.0278*[FeSCN2+] + 0.0154

i.e. [FeSCN2+] = (0.36-0.0154)/0.0278 = 12.4*10-5 M

If A = 0.45

i.e. 0.45 = 0.0278*[FeSCN2+] + 0.0154

i.e. [FeSCN2+] = (0.45-0.0154)/0.0278 = 15.6*10-5 M

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