CHM 235 Quantitative Analysis
Spring 2007
Dr. S.A. Skrabal
SOLUTIONS TO PROBLEM SET 4
Calibration curves
6 February 2007
1. Titanium in rock samples can be determined by absorption spectrophotometry after reacting it with hydrogen
peroxide in acidic solution. A regression analysis (using Excel®) of the absorbances of standard solutions
containing 0-60.0 mg/L of Ti yielded the following parameters, with uncertainties expressed as standard
deviations: slope = 1.02413 x 10-2  3.21 x 10-4; y-intercept = 1.099 x 10-3  1.10 x 10-3; standard deviation of y
values (sy)= 2.633 x 10-3. An unknown sample was treated and measured in the same manner as the standards.
The absorbance of the unknown sample was measured at 0.145. (A) Express the regression parameters (m, sm,
b, sb, sy) with the correct number of significant figures. (B) What is the concentration of Ti (in mg/L) of the
unknown sample? (Note: No uncertainty analysis is required.)
(A)
m and sm: 1.024 ( 0.032) x 10-2 Note: This is same as 1.024 x 10-2 ( 0.032 x 10-2)
b and sb: 1.0 (1.1) x 10-3
Same as 1.0 x 10-3 ( 1.1 x 10-3)
sy: 2.6 x 10-3
abs b
(B) conc 
m

0.145 1.0 x 10 3
1.02 4 x 10  2
 14.0 6 mg / L or 14.1 mg / L
2. A calibration curve for dissolved iron was obtained by measuring absorbances in a spectrophotometric
analysis, using Fe standards in a range from 0 to 3.5 µM. The data were analyzed using Excel®, and the
following parameters of the regression were calculated:
slope: 0.1533 ± 0.0083
y-intercept: 0.0027 ± 0.0010
std. dev. of y values: 0.0034
An unknown sample is analyzed, yielding an absorbance of 0.312. Assume the blank is zero.
Calculate (A) the Fe concentration of the unknown sample (in µM) and (B) the relative and absolute uncertainty of
this concentration.
(A) conc 
(B)
abs ( s y )  b ( sb )
m ( s m )

0.312 ( 0.003 4 )  0.002 7 ( 0.0010 )
0.1533 ( 0.008 3 )
enum  (0.0034 ) 2  (0.0010 ) 2  0.0035
%enum 
0.0035
(100)  1.1%
(0.312  0.002 7 )
%edenom 
0.0083
(100)  5.4 %
0.1533
%eoverall  (1.1%) 2  (5.4 %) 2  5.5 %
eoverall = (5.5%)(2.017 µM) = (0.055)(2.017 µM) = 0.11
Answer (with correct sf): 2.01  0.11 µM and 2.01 µM  5.5%
or 2.0  0.1 µM and 2.0 µM  6 %
 2.017  eoverall M
3. A standard curve (using enzyme standards containing 0 to 20 pg) to determine the concentration of an enzyme
in biological samples gave the following results, where the uncertainties are expressed as standard deviations:
slope = 0.04068 + 0.00032; y-intercept = 0.0090 + 0.0012; standard deviation of y values (sy)= 0.0069. The
absorbance of an unknown sample is measured at 0.222. What is the enzyme content (in pg) of this sample, and
what is the uncertainty (absolute and relative) associated with this result?
pg enzyme 
abs ( s y ) b ( sb )
m ( s m )
0.222 ( 0.006 9 )  0.009 0 ( 0.0012 )

0.0406 8 ( 0.0003 2 )
 5.235  eoverall pg
enum  (0.0069 ) 2  (0.0012 ) 2  0.007 0
%enum 
0.007 0
(100)  3.2 %
(0.222  0.009 0 )
%edenom 
0.00032
(100)  0.7 8 %
0.04068
%eoverall  (3.2 %) 2  (0.78 %) 2  3.2 %
eoverall = (3.2%)(5.235 pg) = (0.032)(5.235 pg) = 0.16 pg
Answer (with correct sf): 5.23  0.16 pg and 5.23 pg  3.2%
or 5.2  0.2 pg and 5.2 pg  3 %
4. From the following data of phosphorus standards and absorbances from a colorimetric determination, (A)
prepare a calibration curve; (B) find the line of best fit; (C) find the concentration of phosphorus in the unknown
urine sample. You can use Excel® or a graphing calculator for this. No error analysis is necessary.
P (ppm)
1.00
2.00
3.00
4.00
Urine sample
Absorbance
0.204
0.411
0.614
0.821
0.455
(A) and (B) See graph below, including line of best fit, prepared using EXCEL.
(C) Rearrange y = mx + b to solve for x (concentration):
abs  b 0.455  1.00 x 10 3
Conc 

 2.210
m
2.054 x 10 1
or 2.21 ppm
Absorbance
Phosphorus standard curve
1 y = 0.20540x - 0.00100
0.8
R2 = 0.99998
0.6
0.4
0.2
0
0
1
2
[P] (ppm)
3
4