Assignment 2 - Solutions

1. NOTE: There were several version of question 1.  The solution to one set of data is given.

The amount of phosphate in urine can be determined as follows:  The proteins are removed, followed by treatment with Mo(VI), followed by reduction to a complex called 12-molybdophosphate, which is intensely coloured when ascorbic acid is added.  The name of this species is “Molybdenum Blue,” with a maximum absorbance (λmax) at 650 nm.

Over 24 hours, a patient expelled 1228 mL of urine.  A 40.00 mL sample was taken and treated as described above.  The sample was divided into eight 5.00 mL samples; seven of the samples were spiked with a 5.00 ppm phosphate solution.  The samples were all topped up to 100.00 mL and the UV-visible absorption experiment gave the following results:

spiking
absorbance at 650 nm
1.00 ppm phosphate
2.00 
3.00 
4.00
5.00
6.00
7.00
0.497
0.656
0.785
0.962
1.095
1.249
1.399
urine sample (no spike)
0.357
(a) Determine the concentration of phosphate in the patient's urine.  Based on the regression analysis, give a confidence range for the number.  Give this number in ppm and also in molarity (mol/L).  You can assume that the density of this dilute solution is the same as that of water.
 
Sol'n Either using a spreadsheet or the formulae from the text, solve for slope and intercept of the best-fit line for the calibration curve.  You should include the point 0, 0.357 (i.e. no added spike gives the reading 0.357).

Sxy = 6.274; Sxx = 42; Syy = 0.93769...;  mean(x) = 3.5;  mean(y) = 0.875 

slope = m = Sxy/Sxx = 0.14938...
intercept = b = mean(y) - m*mean(x) = 0.35216....

Recall that cs = 5.00 and Vu = 5.00
For a spiking experiment, cu = b*cs / (m*Vu) = 2.3575.... (in ppm since the spike value is in ppm)

Sr = 0.0088873; x-intercept = -2.3575....
                              - note: this is only the same as cu because cs and Vu have the same numerical value.

Therefore, using Su formula for a spiking experiment, Su = 0.05774...

Therefore, the concentration of phosphate in the urine is 2.36 +/- 0.06 ppm

In mol/L, recall that ppm = mg of solute in kg of solvent.  If the densities are the same for solution and solvent (as it is in this case), then ppm = mg of solute per kg of solution, which is easier to convert.
Recalling that 1 g H2O = 1 mL of H2O, and that phosphate has a molar mass of 95 g/mol
2.36 ppm = 2.36 mg/kg = 2.36 mg/L = (0.00236 g/L)/(95 g/mol) = 2.48 x 10-5 mol/L
 
(b) How much phosphate does the patient lose through urination in a day?  Give the answer in grams (remember, phosphate is PO43-)
 
Sol'n If the concentration is 2.36 μg/g, and the patient excreted 1228 mL (= 1228 g) of urine,
mass = (2.36 μg/g)*(1228 g) = 2898 μg = 2.90 mg excreted per day
 
2. NOTE: There were several version of question 2.  The solution to one set of data is given.

Ethylenediaminetetraacetate (EDTA) forms complexes with many metals, including cobalt(II), copper(II), and nickel(II), which are often found together at mine sites.  The molar absorptivity of these metal complexes at various wavelengths are as follows:
 

metal
 ε463 (L mol-1 cm-1)
ε732 (L mol-1 cm-1)
ε379 (L mol-1 cm-1)
Co2+
Cu2+
Ni2+
15.8
2.32
1.79
2.11
95.2
3.03
3.11
7.73
13.5

A solution known only to contain Co2+ and Cu2+ is analysed for concentration of the two metals.  The total absorbance at 732 nm is 0.453, and at 463 nm it is 0.107 (through a 1.00 nm cell).  What is the concentration of Co2+ and Cu2+ in the solution?
 
Sol'n To solve for the two metal-containing species' concentrations, you must set up a set of two equations using Beer's Law: 
A732 = εCo732bcCo + ε732CubcCu
A463 = εCo463bcCo + e463CubcCu
Solving the two simultaneous equations gives cHIn = 4.62 x 10-3 M and  cIn- = 6.09 x 10-3 M.
 
3. Predict the frequency of vibration (in wavenumbers, cm-1) for CO and NO.  Are the species resolvable by IR spectroscopy?  Is there a component of air that is likely to interfere with the measurement of either of these species in an infrared (IR) absorption experiment?  Assume air is made up of N2, O2, Ar, H2O, and CO2.

bond
stretching force constant (k)
C≡O
N≡O
18.7 x 105 dyne/cm
15.5 x 105 dyne/cm
Hints: dyne = 10-5 N = g cm s-2
          use mass of a single atom (not g/mol!)
 
Sol'n In grams, the mass of single atoms are MWt/NA.
mass of 14N = (14.00 g/mol) / (6.022 x 1023 mol-1) = 2.325 x 10-23 g
mass of 12C = 1.993 x 10-23 g
mass of 16O = 2.656 x 10-23 g
Therefore, the reduced mass of CO is μ = 1/(1.993 x 10-23 g) + 1/(2.656 x 10-23 g) = 1.1385 x 10-23 g.
                                  - and for NO is μ = 1.2399 x 10-23 g.

ΔE = hν = (h/2π)*(k/μ)^0.5
     - cancelling h will give a value of ν in s-1 (be sure your units work out!)
νCO = 1/(2π)*[(1.87 x 106 g cm s-2 / cm) / (1.1385 x 10-23 g)]^0.5 = 6.4501 x 1013 s-1 (Hz)
νNO = 5.6273 x 1013 s-1.

Using c = λν,
λCO = (2.9979 x 108 m/s) / (6.4501 x 1013 s-1) = 4.622 x 10-6 m = 4.622 x 10-4 cm.
λNO = 5.298 x 10-6 m = 5.298 x 10-4 cm.

Therefore, in wavenumbers (1/λ),
frequency of CO vibration = 1/4.622 x 10-4 cm = 2163 cm-1.
frequency of NO vibration = 1887 cm-1.

I did not give you frequencies for O2, N2, and Ar because they do not have IR vibrations.  H2O has a vibration at ~3900 cm-1, CO2 has a vibration at 2750.  Neither of these will interfere with the CO or NO stretch.  The two peaks are well-separated and can be resolved from each other.


 

Last updated, February 4, 2005
Page maintained by Craig MacKinnon