| 1. | The amount of phosphate in urine can be determined as follows:
The proteins are removed, followed by treating 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 (lmax)
at 650 nm.
Over 24 hours, a patient expelled 1122 mL of urine. A 1.00 mL sample was taken and treated as described above. The sample, along with a series of phosphate standards, was run in an absorption experiment, with the following results: |
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| (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.
Sxy = 1.028; Sxx = 5; Syy = 0.2113...; mean(x) = 2.5; mean(y) = 0.538 slope = m = Sxy/Sxx = 0.2056
[PO43-]unk = xunk = (yunk - b)/m = 2.40272... Using the above statistical values for slope, Sxx
and Syy (noting that N = 7),
Therefore, the unknown concentration is 2.40 +/- 0.01 ppm (or 2.403 +/- 0.013) Recall that ppm = mg of solute per g of solution,
and 1 mL of water weighs 1 g at extreme dilution.
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| (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 2.40 mg/g, and the
patient excreted 1122 mL (= 1122 g) of urine,
mass = (2.40 mg/g)*(1122 g) = 2690 mg = 2.69 mg excreted per day |
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| 2. | Quinine is used as an anti-malarial drug. It is also added to
tonic water (incidentally, gin was added to the tonic water to encourage
British troops to take their medicine, hence the origin of the drink gin
and tonic). A stream next to a quinine plant is known to be slightly
contaminated due to a spill. Seven vials of 10.00 mL of stream water
are taken for analysis. Six of them are spiked with a 7.5 ppb (parts
per billion) solution of quinine, and the fluorescence of the resulting
solutions were measured. The following results were obtained:
Determine the concentration of quinine in the stream water. Report a confidence interval based on an analysis of the regression line. |
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| Sol'n | Again, plotting the data and calculating the slope and intercept the
same way as above gives,
Sxy = 125.18; Sxx = 112; mean(x) = 6; mean(y) = 10.50... slope = m = Sxy/Sxx = 1.1176...
Recall that cs = 7.5 and Vu = 10.00
Sr = 0.19138...; x-intercept = -3.3970....
Therefore, the concentration of quinine in the stream is 2.5 +/- 0.1 ppb (or 2.54 +/- 0.12 ppb) |
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| 3. | Acid-base indicators are species that are weak acids. Some are
coloured in both their acidic and basic forms. This sets up the equilibrium,
HIn + H2O <===> In- + H3O+ A small amount of an indicator HIn was dissolved in 1.00 L water.
Absorbances were measured at 450 and 550 nm in 1.00 cm pathlength
cells, to give the following data:
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| Sol'n | To solve for the two metal-containing species' concentrations, you
must set up a set of two equations using Beer's Law:
A450 = eHIn/450bcHIn + eIn-/450bcIn- A550 = eHIn/550bcHIn + eIn-/550bcIn- Solving the two simultaneous equations gives cHIn = 7.00 x 10-5 and cIn- = 3.00 x 10-5 If the autoionisation of water is negligible, then the only source of H+ is the weak acid, and because it is a monoprotic acid, cIn- = cH+ Therefore, Ka = [H+][In-] / [HIn] = (3.00 x 10-5)2 / (7.00 x 10-5) = 1.28 x 10-5. |
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Last updated, February 10, 2003
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MacKinnon