Question

The chromium in an aqueous sample was determined by pipetting 10.0 mL of the unknown into each of five 50.0-mL volumetric flasks. Various volumes of a standard containing 12.2 ppm Cr were added to the flasks, following which the solutions were diluted to volume.

(a) Plot the data using a spreadsheet.

(b) Determine an equation for the relationship between absorbance and volume of standard.

(c) Calculate the statistics for the least-squares relationship in (b).

(d) Determine the concentration of Cr in ppm in the sample.

(e) Find the standard deviation of the result in (d).

Short answer

a) The spreadsheet is given below in its respective step.

b) An equation for the relationship between absorbance and volume of standard is y = 0.00881x + 0.2022

c) The standard error in y is 0.001304 and the sum of squares of deviations from the mean for the individual value of x is 1000.

d) The concentration of chromium in ppm in the sample solution is 28.0 ppm.

e) The standard deviation in concentration is 0.258.

Step-by-step solution

Part (a) Step 1: Given Information

A graph should be plotted using the spreadsheet for given information.

Part (a) Step 2: Explanation

The following data is given in the question:

Unknown, mL	Standard, mL	Absorbance
10.0	0.0	0.201
10.0	10.0	0.292
10.0	20.0	0.378
10.0	30.0	0.467
10.0	40.0	0.554

Data for the volume of standard added to 10 mL of unknown sample is taken in x-axis and corresponding absorbance readings is taken in y-axis, when the data is plotted, the following graph is obtained:

Part (b) Step 1: Given Information

The following data is given in the question:

Unknown,mL	Standard, mL	Absorbance
10.0	0.0	0.201
10.0	10.0	0.292
10.0	20.0	0.378
10.0	30.0	0.467
10.0	40.0	0.554

Part (b) Step 2: Explanation

Slope of the graph and y-intercept are calculated using the spreadsheet and the slope is found to be 0.00881 and y intercept is 0.2022.

Therefore, it can be inferred that the equation for the relationship between absorbance and volume of standard can be given by:

y = 0.00881x + 0.2022

Part (c) Step 1: Given Information

The statistics for the least squares relationship for the equation in option b should be determined.

Part (c) Step 2: Explanation

The statistics is calculated by the error analysis. The standard error in ySrand the sum of squares of deviations from the mean for individual value of xSxxis given below:

Slope	0.00881
Intercept	0.2022
Spreadsheet documentation
CellB26=SLOPE(B7:B11,A7:A11)
CELLB27=INTERCEPT(B7:B11,A7:A11)
Equation of line is
Error analysis
Standard error in y	0.0013
N	5
Sxx	1000
Spreadsheet documentation
CellB35=STEYX(B7:B11,A7:A11)
CellB36=COUNT(B7:B11)
CellB37=DEVSQ(A7:A11)

Therefore, it can be inferred that the standard error in y is 0.001304 and the sum of squares of deviations from the mean for the individual value of x is 1000.

Part (d) Step 1: Given Information

The concentration (in ppm) of Chromium in the sample should be determined.

Part (d) Step 2: Explanation

The concentration of chromium in ppm in sample is determined using the x-intercept of the line graph obtained when the absorbance is plotted against the volume of standards and is given as below:

Slope	0.00881
Intercept	0.2022
Spreadsheet documentation
CellB26=SLOPE(B7:B11,A7:A11)
CELLB27=INTERCEPT(B7:B11,A7:A11)
Equation of line is
Error analysis
Standard error in y	0.0013
N	5
Sxx	1000
Spreadsheet documentation
CellB35=STEYX(B7:B11,A7:A11)
CellB36=COUNT(B7:B11)
CellB37=DEVSQ(A7:A11)
x intercept	-22.95119183
Concentration of Cr in sample	28.00045403
Spreadsheet documentation
CellB43= -B27/B26
CellB44= -B43*B4/B3

Therefore, it can be inferred that the concentration of chromium in ppm in the sample solution is 28.0 ppm.

Part (e) Step 1: Given Information

The standard deviation of the concentration of chromium in the sample should be determined.

Part (e) Step 2: Explanation

The average of the volume of standard added,is calculated which is used to calculate the standard deviation in volume and standard deviation in concentration of the sample.

Slope	0.00881
Intercept	0.2022
Spreadsheet documentation
CellB26=SLOPE(B7:B11,A7:A11)
CELLB27=INTERCEPT(B7:B11,A7:A11)
Equation of line is
Error analysis
Standard error in y	0.0013
N	5
Sxx	1000
Spreadsheet documentation
CellB35=STEYX(B7:B11,A7:A11)
CellB36=COUNT(B7:B11)
CellB37=DEVSQ(A7:A11)
x intercept	-22.95119183
Concentration of Cr in sample	28.00045403
Spreadsheet documentation
CellB43= -B27/B26
CellB44= -B43*B4/B3
Average absorbance of standard y bar	0.3784
Standard deviation in volume	0.211628663
Standard deviation in concentration	0.258186969
Spreadsheet documentation
CellB51=AVERAGE(B7:B11)
CellB52=
CellB53=B52*B4/B3

Standard deviation in concentration is calculated as below:

$\mathrm{Standard}\mathrm{deviation}\mathrm{in}\mathrm{concentration}=\mathrm{standard}\mathrm{deviation}\mathrm{in}\mathrm{volume}\times \frac{(\mathrm{Concentration}\mathrm{of}\mathrm{standard})}{(\mathrm{Volume}\mathrm{of}\mathrm{sample})}$

Therefore, it can be inferred that standard deviation in concentration is 0.258.