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Principles of Instrumental Analysis, 6th ed. Chapter 9
 
 7
9-20. Setting up two equations in two unknowns 
 0.599 = 0.450m + b 
 0.396 = 0.250m + b 
 Subtracting the second equation from the first gives 
 0.203 = 0.200 m 
 or the slope m = 0.203/0.200 = 1.015 
 Then using this value in the first equation gives 
 b = 0.599 – 0.450 × 1.015 = 0.14225 
 The unknown Pb concentration is then x = (0.444 – 0.14225)/1.015 = 0.297 ppm Pb 
9-21. In the spreadsheet below we first calculate the equation for the line in cells B10 and B11. 
 y = 0.920 x + 3.180 
 The mean readings for the samples and blanks are calculated in cells D7:G7, and the 
mean Na concentrations in cell D8:G8. These values are uncorrected for the blank. The 
blank corrections are made in Cells B14:D14. These are then converted to %Na2O in 
cells B15:D15 by the following equation 
 2
2 6
conc. Na O in μg/mL 100.0 mL %Na O = 100% 
1.0 g 10 μg/g
×
×
×
 
 The error analysis shows the calculations of the 4 standard deviations by the equations in 
Appendix 1, Section a1-D2. Since the final result is obtained by subtracting the blank 
reading, the standard deviations must be calculated from the difference in cells E24:G24 
by 2
d A = + s s s2
bl where sA is the standard deviation for the concentration of A and sbl is 
the standard deviation of the blank. These are then converted to absolute and relative 
standard deviations of %Na2O in cells E25:G26.