Answer to Question #158150 in General Chemistry for Mo

Question

One common way to determine phosphorous in urine is to treat the sample after removing the protein with molybdenum (VI) and then reducing the resulting 12-molybdophosphate complex with ascorbic acid to give an intense blue-coloured species called molybdenum blue. The absorbance of molybdenum blue can be measured at 650 nm. A 24-hour urine sample was collected, and the patient produced 1122 mL in 24 hours. A 1.00 mL aliquot of the sample was treated with Mo (VI) and ascorbic acid and diluted to a volume of 50.00 mL A calibration curve was prepared by treating 1.00 mL aliquot of phosphate standard solution in the same manner as the urine sample. The absorbances of the standards and the urine sample were obtained at 650 nm and the following results were obtained.

Concentration (ppm) Absorbance at 650 nm

1.00

0.230

2.00

0.436

3.00 0.638

4.00 0.848

Urine sample 0.518

1.1 Identify the absorption line of molybdenum blue.

(1 Mark)

1.2 In which region of the electromagnetic spectrum is the analysis performed? Justify.

(2 marks)

1.3 The energy associate with a photon is given by the formula below.

E = ℎ𝐶/ʎ, where E = Energy of photon, h = Plank constant (6.62 x 10-34 J.s), c = speed of light (3.0 x

108 m/s) and ʎ = wavelength of the photon.

Calculate in joule the energy of a photon at the analyte line.

(3 marks)

1.4 Construct a calibration curve for the determination of phosphorous in urine.

(6 marks)

1.5 Determine the slope

(2 Marks)

1.6 Determine the intercept

(2 Marks)

1.7 Determine the concentration of phosphorous in the original urine sample.

(3 Marks)

1.8 Convert the unknown concentration calculated into % (m/v)

(3 marks)

1.9 Calculate the mass in grams of phosphorous was eliminated per day by the patient

(5 Marks)

1.10 Calculate, in mM, the concentration of phosphorous in the urine.

(3 marks)

[Sub T

Answer 1

1.1 - the absorbance increases linearly with concentration

1.2 - visible region of the electromagnetic spectrum is the analysis performed .because

Visible light covers the range of wavelengths from 400 – 750 nm and here wavelength = 650nm = 6.5×10^-7

1.3 - E = hC/$\lambda$

E =( 6.62×10^-34)(3×10⁸) / (6.5×10^-7)

E = 3.05 ×10^-19 Joule

1.4 -

1.5 - by two points (1,0.230)(2,0.436)

m = (0.436-0.230)/(2-1)

m = 0.206

1.6 - The x-intercept is where a line crosses the x-axis, and the y-intercept is the point where the line crosses the y-axis. Thinking about intercepts helps us graph linear equations. y = mx + b

1.7 - M = no.of moles of solute/ volume of solvent

no. of moles = 1mol

Volume = 50ml

M = 1/50 = 0.02 M concentration

1.8 - 1g phosphorus and 50ml

%(m/v) = (1/50)×100 = 2%

1.9 - (Beer’s law holds)

$\frac{C_{unk}}{C_{std}}$ = $\frac{A_{unk}}{A_{std}}$

Using the standard with the closest absorbance reading gives

C_unk = (3.00 ppm)(0.0848)/(0.0638) = 3.98 ppm P in urine

$\frac{3.98×10^-⁶gP}{g .dil.sol.}$ ×$\frac{1g.dil.sol}{1ml.dil.sol.}$ × 50ml dil sol

= 1.99×10^-4 gP in 1ml urine

$\frac{1.99×10^-⁴gP}{1ml .urine}$ ×$\frac{1270ml.urine}{day}$

= 0.2527g P /day

1.10 - 0.02 M into nm

1 Molar [M] = 1 000 000 000 Nanomolar [nM]

0.02M = 0.02×1000000000

20,000,000 nm