CHEMISTRY 311 - ASSIGNMENT 2 Hand-in your answers in a neat and organized format to the appropriate number significant

figures, showing chemical equations and calculations where appropriate. Due date: Thursday, Oct. 18th, 2018

1. Propagate the indeterminate readings errors associated with the volumetric analysis of dissolved oxygen using class A analytical laboratory glassware (100.0 +/- 0.08 mL sample, 0.0200 +/- 0.0001 N sodium thiosulfate and 5.00 +/- 0.03 mL of titrant) and compare that to the same using a digital titrator in the field (20.0 +/- 1 mL sample, 0.200 +/- 0.004 N sodium thiosulfate and 100 +/- 2 uL of titrant added). Strategy: Set up the mathematical equation to solve for the concentration, substituting in the values and related absolute uncertainties. Propagate the either absolute or relative errors as appropriate depending on the mathematical operations involved.

2 32 2 322

2

S O S OO

O

N x VNV− −=

Since all mathematical operations are multiplication and division, it is the relative errors that are additive. Solution: Class A glassware

2 2 23

2(0.0200 0.0001 ) (5.00 0.03 ) 0.0001 0.03 0.081.00 10

(100.0 0.08 ) 0.0200 5.00 100.0ON x mLN x N

mL−

⎛ ⎞± ± ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟= = ± + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟± ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠ = 1.000 x 10-3 N +/- 0.0078 (0.78% relative error) Converting this back to absolute error yields, NO2 = (1.000 +/- 0.008) x 10-3 N Any further conversions to different concertation units will result in the same relative error. For instance, 8.00 +/- 0.06 mg/L O2 Digital titrator

2 2 233

2(0.200 0.004 ) (100 2) 10 0.004 2 11.00 10

(20 1 ) 0.200 100 20ON x x mLN x N

mL

−−

⎛ ⎞± ± ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟= = ± + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟± ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

= 1.00 x 10-3 N +/- 0.057 (5.7% relative error) Converting this back to absolute error yields, NO2 = (1.00 +/- 0.06) x 10-3 N Any further conversions to different concertation units will result in the same relative error. For instance, 8.0 +/- 0.5 mg/L O2

2. A Method Detection Limit (MDL) can be estimated according to a procedure described in U.S. EPA Method 300.0 by analyzing at least seven replicates with a concentration close to the estimated method detection limit. The MDL is then calculated as (t) x (SD) where t = Student’s t value for a 99% confidence level with n - 1 degrees of freedom and SD is the sample standard deviation. Check for outliers using the Grubb’s test at the 95% confidence level and estimate the method detection limit for the measurements of a 2.5 ppb NO3

- - N standard solution by ion chromatography. 2.8 1.7 2.4 2.9 2.1 4.0 2.3 2.6 1.8 2.7 Strategy: Test the extreme values (1.7 and 4.0) as possible outliers using Grubb’s test

𝐺!"#! = 𝑥! − 𝑥𝑠

where the critical value (Gtable) is 2.176 at 95% confidence for 10 observations. Using the remaining values, calculate the mean and the standard deviation and student t-value at 99% confidence to calculate MDL. Solution: For the minimum value (1.7), Gcalc = (1.7 – 2.53)/0.657 = 1.26 (retain) For maximum value (4.0), Gcalc = (4.0-2.53)/0.657 = 2.24 (reject) Since 4.0 is rejected, the Method Detection Limit is given by; (3.355) x (0.430 ppb NO3-N) = 1.44 ppb NO3-N where t = 3.355 for df = 8 at 99% (α = 0.05 two tailed) à MDL = 1.4 ppb NO3-N (see attached spreadsheet calculation)

3. Carbon dioxide in the atmosphere was analyzed by bubbling air into a 100.0 mL volume of 5.00 x 10-2 M barium hydroxide ‘trapping solution’ at a flowrate of 0.8 L/min for 90. mins at 12.0 oC and 1.00 atm. The remaining Ba(OH)2 was then back-titrated with 34.32 mL of 0.1600 M HCl (see further the CHEM 311 Lab Manual). a) Determine the concentration of CO2 in the air sample as a mole ratio x 106 (ppmv). b) If all glassware used for the volumetric measurements was class A, what do you suspect is the largest source of indeterminant error in the method described. c) If the air sampled was uncontaminated outdoor air, do you suspect that your answer is exhibiting a bias? Strategy: This is a volumetric analysis by back titration, where the number of moles of CO2 trapped in the solution is given by the number of moles of Ba(OH)2 that have reacted.

CO2(aq) + Ba(OH)2(aq) à BaCO3(s) + H2O

The concentration of CO2 in the atmosphere is then given by the ratio 𝑛!"!𝑛!"#

× 10!

where the nair = PV/RT from the ideal gas law. Solution:

a) #mols Ba(OH)2 reacted = #mols Ba(OH)2 initial - #mols Ba(OH)2 remaining #mols Ba(OH)2 initial = (5.00 x 10-2 mol/L) x (100.0 x 10-3 L) = 5.00 x 10-3 mol #mols Ba(OH)2 remaining = #mols H+ added in titration x (1 mol Ba(OH)2/2 mol H+) = (0.1600 mol/L) x (34.32 x 10-3 L) x 1 mol Ba(OH)2/2 mol H+) = 2.746 x 10-3 mols Ba(OH)2

Therefore, #mols Ba(OH)2 reacted = 2.254 x 10-3 mol = # mols of CO2 trapped

And 𝑛!"# =!.!! !"# (!.! !/!"# × !". !"# )

!.!"#!$ ! !"# !"#!! !!! (!"#.! !)= 3.078 𝑚𝑜𝑙𝑠 𝑜𝑓 𝑎𝑖𝑟 𝑠𝑎𝑚𝑝𝑙𝑒𝑑

So, the concentration of CO2 in the air is given by;

(2.254 x 10-3 mol/3.078) x 106 = 732 ppmv CO2

b) Based on the reported values given in the question the largest contributor to the indeterminant error would be the least precise measured value, which is the flowrate in this case. c) Given that the average outdoor air concentration of CO2 is around 406 ppmv, the determined value from the back titration is exhibiting a significant positive bias of nearly 90%.

4. The following data were collected during a study of the concentration of zinc in samples drawn from several locations in Lake Erie (all concentrations in ppm).

Location Conc Zn2+ at air-water Conc Zn2+ at sediment-water 1 0.430 0.415 2 0.266 0.238 3 0.567 0.390 4 0.531 0.410 5 0.707 0.605 6 0.716 0.609

Determine if there is a significant difference between the concentration of Zn at the air-water interface and the sediment-water interface at the 95% confidence level. Strategy: Use the null hypothesis and the paired t-test to determine if there is a significant difference between surface and depth zinc concentrations in the lake.

𝑑 =𝑡 𝑠!𝑁

where di = deviation between data pairs and sd is the standard deviation in the deviations. Solution: In all locations the concentration of zinc appears higher at the surface than at depth. The deviations are as follows;

Location Difference between surface and depth

1 0.015 2 0.028 3 0.177 4 0.121 5 0.102 6 0.107

Mean deviation = 0.092 ppm Std dev in deviations = 0.061 ppm Calculate a t-value and compare this to ttable at 95% confidence and df=5 (ttable = 2.571);

𝑡!"!" =! !!!

= 3.71 Since tcalc > ttable, the surface water samples can be said to have a higher concentration of Zn at a 95 % confidence level.

5. The following calibration data are provided for a series of external aqueous standards of Cd2+ that had been buffered at a pH of 3.7 and 4.6, respectively. [Cd2+] ppb 15.4 30.4 44.9 59.0 72.7 86.0 Signal (pH = 4.6)

4.8 11.4 18.2 26.6 32.3 37.7

Signal (pH = 3.7)

15.0 42.7 58.5 77.0 101 118

a) Use Excel to derive the best-fit lines using the method of least squares and indicate how much more sensitive this method is at the lower pH. b) Using the Excel Spreadsheet for least squares analysis described in the textbook (Section 4.9), report the standard deviation of the slope and the intercept. c) A single sample buffered at pH 3.7 was analyzed for cadmium yielding a signal of 66.3. Report the concentration of Cd2+ in the sample and the 95% confidence interval. Strategy: Use the data provided to generate two calibration curves and compare slopes to determine which method is more sensitive (ie gives a greater change in signal for a given change in concentration). Then use Equations 4-21 and 4-22 in the text to compute the uncertainty in both the slope and intercept. Finally, use the signal for the unknown to calculate it’s concentration and corresponding uncertainty (sx) from equation below (Equation 4-27 in the text). The 95% confidence limit for a concentration derived from a calibration curve is then given by;

𝐶𝐿 𝑓𝑜𝑟 𝑥 = 𝜇 ± 𝑡 × 𝑠! where sx is given by

𝑠! =𝑠!𝑚

1𝑘+ 1𝑛+

𝑦 − 𝑦 !

𝑚!Σ 𝑥! − 𝑥 !

Solution: See attached calibration curves and Excel spreadsheet. pH 3.7; y = 1.434 x - 5.019 pH 4.6; y = 0.477 x - 2.690 Therefore, the method is 3.0 times more sensitive at pH 3.7 than at pH 4.6. The uncertainty in the slope (sm) = 0.048 ppb-1 The uncertainty in the intercept (sb) = 2.72 The uncertainty in the derived value of x from a single measurement (k = 1) of an unknown with a signal of 66.3 based on the calibration curve at pH 3.7 with six data points (n=6), is 2.1. Hence, the concentration of the unknown is 49.7 +/- 2.1 ppb Cd, where the uncertainty reported is the std deviation. The 95% confidence interval is given by;

t x sx = (2.771) (2.13) = 5.82 ppb Cd Therefore, the concentration of Cd in the unknown sample is 50. +/- 6 ppb with 95% confidence.

y=0.4771x-2.6898R²=0.99714

y=1.4342x-5.0188R²=0.99557

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0

Signal

[Cd2+]nM

Calibra'onCurvesforCd2+Determina'onsatpH3.7and4.6

(Assign2;Q#5)

Least Squares Spreadsheet

number of x y xy x2 d d2

points (n) = 15.40 15.00 231 237.16 -2.0682 4.27736 30.40 42.70 1298.1 924.16 4.1186 16.9626

44.90 58.50 2626.7 2016 -0.8776 0.770259.00 77.00 4543 3481 -2.6001 6.760372.70 101.00 7342.7 5285.3 1.7512 3.066686.00 118.00 10148 7396 -0.3239 0.1049

308.400 412.200 26189 19340 3.19E+01

D= sy= F4=C4-$A$16*B4-$A$1820927.16 2.826 A14=$A$5*E11-B11*B11

m= sm= A16=(D11*$A$5-B11*C11)/$A$141.434 0.048 A18=(E11*C11-D11*B11)/$A$14

b= sb= B14=SQRT(G11/($A$5-2)-5.019 2.717 B16=$B$14*SQRT($A$5/$A$14)

B18=$B$14*SQRT(E11/$A$14)

Finding uncertainty in x with EQN 4-27 (Harris)

measured y= number of replicate values of y measured (k)=66.3 input input 1

derived x= uncertainty in derived x (sx) =49.73 output output 2.13

column sums

6. A combined standard solution was prepared to contain 25.0 uM caffeine and 5.0 uM of an appropriately chosen internal standard. Analysis of this combined standard gave signals for the caffeine and internal standard of 0.155 and 0.233 (arbitrary units), respectively. A 1.00 mL aliquot of internal standard solution was added to a 10.0 mL sample to yield a 10.0 uM solution of the internal standard in the diluted sample. Analysis of the spiked sample yields signals for the analyte and internal standard of 0.274 and 0.198, respectively. a) What considerations go into choosing an appropriate internal standard? b) Is the analytical technique in this case more sensitive to caffeine or the internal standard? c) Report the concentration of caffeine in the sample in units of ppm. Strategy: Use the measured signals for the combined standard to determine the relative response factor (RF).

𝑅𝐹 =

𝐴![𝑋]

𝐴![𝑆]

Then using this RF value, calculate the concentration of the unknown in the diluted sample as follows;

𝑋 = 𝐴! [𝑆]

𝐴! × (𝑅𝐹)

Solution:

a) An internal standard should be chemically similar to the analyte but distinguishable by the method employed. It should also not be naturally present in the sample.

b) The method is more sensitive to the internal standard than the caffeine in this case. We

can see this from the value of RF itself.

𝑅𝐹 =0.155

(25.0 𝑢𝑀)0.233

(5.0 𝑢𝑀)= 0.133

Since RF < 1, the method is more sensitive to the internal standard than the caffeine.

c) The concentration of caffeine is given by;

𝑋 = 0.274 (10.0 𝑢𝑀)0.198 × (0.133)

= 104 𝑢𝑀 𝑐𝑎𝑓𝑓𝑒𝑖𝑛𝑒

Correcting for dilution of original 10.0 mL to 11.0 mL, yields [X]i = [X]f x (11.0/10.0) = 104 x 1.10 = 114 uM Converting this to ppm caffeine, we use the MW= 194.19 g/mol

114 𝑢𝑚𝑜𝑙𝐿

× 194.19 𝑔𝑚𝑜𝑙

× 𝑚𝑜𝑙

10! 𝑢𝑀 ×

10! 𝑚𝑔𝑔

= 22.1𝑚𝑔𝐿= 22.1 𝑝𝑝𝑚

7. The following data were obtained in a determination of chromium in water. The protocol was to run a blank, a 20.00 mL sample, and then the sample with one and two 250. uL standard addition spikes containing each 150. ng/mL of Cr. Three replicates were measured for each sample.

Solution Volume of Cr spike (uL)

Instrument Readings

#1 #2 #3 Blank 0.00 0.005 0.009 0.003 Sample 0.00 0.086 0.088 0.085 Addition 1 250. 0.179 0.191 0.188 Addition 2 500. 0.305 0.310 0.318 a) Assuming that the volume of the spike was negligible, calculate the mean concentration of Cr in ng/mL in the sample. b) Determine the magnitude of the error introduced in part a) and indicate if this is a determinant or indeterminant error. Strategy: In general, for standard additions, we can write;

[X] f

[X] f +[S] f=Ax

Ax+s

where [X]f and [S]f are the concentration of analyte in the unspiked sample and that resulting from the addition of spike, respectively and Ax and Ax+s are the measured signals from the unspiked and spiked samples, respectively. If the unspiked sample was not diluted prior to measurement, then [X]f = [X]i in the numerator and if the addition of spike does not appreciably change the volume of the sample and no other dilutions take place prior to measurement, then [X]f = [X]i in the denominator. We will average the individual instrument readings and subtract the blank from each to give;

Solution Volume of Cr spike (uL)

Mean instrument reading (blank

corrected) Sample 0.00 0.0806 Addition 1 250. 0.1803 Addition 2 500. 0.3053 Solution: a) We will assume that the relatively small volume change resulting from the addition of the spike does not significantly affect the result and can be ignored. Thus, [X]f ≈ [X]i Using data from the Sample and Spiked sample (Addition 1);

[𝑆]! = 250.× 10!! 𝑚𝐿 150.𝑛𝑔/𝑚𝐿

20.00 𝑚𝐿 = 1.875 𝑛𝑔/𝑚𝐿

Hence,

[𝑋]!𝑋 ! + (1.875

𝑛𝑔𝑚𝐿)

= 0.08060.1803 = 0.4470

[X]i = 0.4470 ([X]i + 1.875 ng/mL) = 0.4470 [X]i + 0.8382 ng/mL and [X]i (1 - 0.4470) = 0.8382 ng/mL and [X]i = 0.8382/(1 - 0.4470) = 1.515 ng/mL Cr Similarly, for the Sample and Spiked sample (Addition 2)

[𝑆]! = 500.× 10!! 𝑚𝐿 150.𝑛𝑔/𝑚𝐿

20.00 𝑚𝐿 = 3.750 𝑛𝑔/𝑚𝐿

[𝑋]!𝑋 ! + (3.750

𝑛𝑔𝑚𝐿)

= 0.08060.3053 = 0.2640

[X]i = 0.2640 ([X]i + 3.750 ng/mL) = 0.2640 [X]i + 0.9900 ng/mL and [X]i (1 - 0.2640) = 0.9900 ng/mL and [X]i = 0.9900/(1 - 0.2640) = 1.345 ng/mL Cr Mean (part a) = 1.430 ppm Cr b) We now repeat the calculation taking into account the small dilution that occurs as a result of the spike addition.

[𝑆]! = 250.× 10!! 𝑚𝐿 150.𝑛𝑔/𝑚𝐿

20.25 𝑚𝐿 = 1.852 𝑛𝑔/𝑚𝐿 Hence,

[𝑋]!𝑋 ! + (1.852

𝑛𝑔𝑚𝐿)

= 0.08060.1803 = 0.4470

where [X]f = [X]i (20.00 mL/20.25 mL) = 0.9876 [X]i

[𝑋]!0.9876 𝑋 ! + (1.852

𝑛𝑔𝑚𝐿)

= 0.4470

[X]i = 0.4470 (0.9876 [X]i + 1.852 ng/mL) = 0.4415 [X]i + 0.8278 ng/mL

and [X]i (1 - 0.4415) = 0.8278 ng/mL and [X]i = 0.8278/(1 - 0.4415) = 1.482 ng/mL Cr Comparing this result to that obtained in part a), we observe that neglecting the dilution introduced by the spike leads to a determinant error +0.051 ng/mL Cr Similarly, for the Sample and Spiked sample (Addition 2)

[𝑆]! = 500.× 10!! 𝑚𝐿 150.𝑛𝑔/𝑚𝐿

20.50 𝑚𝐿 = 3.658 𝑛𝑔/𝑚𝐿 Hence,

[𝑋]!𝑋 ! + (3.658

𝑛𝑔𝑚𝐿)

= 0.08060.3053 = 0.2640

where [X]f = [X]i (20.00 mL/20.50 mL) = 0.9756 [X]i

[𝑋]!0.9756 𝑋 ! + (3.658

𝑛𝑔𝑚𝐿)

= 0.2640

[X]i = 0.2640 (0.9756 [X]i + 3.658 ng/mL) = 0.2576 [X]i + 0.9657 ng/mL and [X]i (1 - 0.2576) = 0.9657 ng/mL and [X]i = 0.9657/(1 - 0.2576) = 1.301 ng/mL Cr Comparing this result to that obtained in part a), we observe that neglecting the dilution introduced by the spike leads to a determinant error +0.044 ng/mL Cr Mean (part b) = 1.392 ppm Cr Summary Table Vol of spike added

(uL) Answer part a)

(ppb Cr) Answer part b)

(ppb Cr) % error*

250. 1.515 1.482 +2.2 500. 1.345 1.301 +3.4

* % error = [answer part a) – answer part b)]/answer part b) x 100%

Note: Differences in the results between sample spikes Addition 1 and 2 are indeterminate (random), whereas those between part a) and part b) are determinate (systematic).