# Regression Analysis

This chapter is part of Pharmaceutical Chemistry (PHR 1601).

A stock solution of Ferric ammonium sulfate Fe(NH4)(SO4)2 (Fe = 56, N = 14, S = 32, O =16) was prepared by dissolving 1.7214 grams of Fe(NH4)(SO4)2.12H2O accurately weighed in distilled water and made up to 20 mL in a volumetric flask.

5 mL of the stock solution was diluted to 500 mLs in a volumetric flask and labelled Solution X.

A series of Standard solutions were prepared as described below. Samples from each solution were reacted with mercaptoacetic acid in buffered ammonium citrate and the intensity of purple colour so formed measured using a  visible spectrophotometer.

10 mLs of solution X diluted to 100mL                                     1.2

20 mLs of solution X diluted to 250mL                                     0.94

12 mLs of solution X diluted to 200mL                                     0.7

10 mLs of solution X diluted to 200mL                                     0.6

5 mLs of solution X diluted to 250mL                                        0.26

Calculate the concentration in mg/l of iron in the stock solution, solution labelled X and in each of the standard solutions used for calibration.

Using regression analysis determine the best-fit equations for the calibration line.

What is the concentration of iron in a sample of water if when treated in exactly the same manner as the standard solutions used for calibration the absorbance was found to be 0.42?

Mass of Fe(NH4)(SO4)2.12H2O

1 mole of Fe(NH4)(SO4)2.12H2O = 56 + 14 + 4 + 64 + 128 + 12*18= 482

Concentration of Fe(NH4)(SO4)2.12H2O in stock solution

1 mole = 482g

? = 1.7214g 0.00357 moles

0.00357 moles = 20mL

? = 1000mL 0.179M

Concentration of Fe(NH4)(SO4)2.12H2O in solution X

0.179 moles = 1000mL

? = 5mL 0.000893 moles

0.000893 = 500mL

? = 1000mL 0.00179 Molar

Standard Solution

As the methodology for the concentration of the standard solution is the same as the examples shown above, only the concentration will be given for each of these solutions.

All these concentrations may also be found using C1V1=C2V2

10 mLs of solution X diluted to 100mL                     0.00179*10=C2*100         0.000179M

20 mLs of solution X diluted to 250mL                     0.00179*20=C2*250         0.000143M

12 mLs of solution X diluted to 200mL                     0.00179*12=C2*200         0.0001074M

10 mLs of solution X diluted to 200mL                     0.00179*10=C2*200         0.0000895M

5 mLs of solution X diluted to 250mL                        0.00179*5=C2*250           0.0000358M

Finding the parameters for the intercept and gradient equations   This would give the equation as y = 6531.5x + 0.0154.

Finding the concentration when the absorbance is 0.42. The concentration would then be found as 0.00006195 mol-3.