12. Inorganic phosphorous is spectrophotochemically measured in blood serum by f

Question

12. Inorganic phosphorous is spectrophotochemically measured in blood serum by formation of a reduced heteropoly acid (heteropoly blue) and comparison of its absorbance with a standard treated by the same procedure. Ifa 1.00 mL blood serum sample gives an absorbance of 0.217, while a 2.00 mL aliquot of a standard containing 91.2 mg KH2PO4 per liter gives an absorbance of 0.285, calculate the milligram percent of P (milligrams per 100.0 mL) in the blood sample. Note: both aliquots are diluted to the same volume prior to absorbance measurements.

Explanation / Answer

Ans. Part A: Given, [KH2PO4] = 91.2 mg/ L

Moles of KH2PO4 in standard solution = Mass / molar mass

                                                = 0.0912 g / (136.085542 g / mol)

                                                = 6.701667 x 10-4 mol

1 mol KH2PO4 consists of 1 mol P.

So,

Moles of P in 1.0 L standard solution = 6.701667 x 10-4 mol

Maas of P in 1.0 L standard solution = Moles x Molar mass

                                                = (6.701667 x 10-4 mol) x (30.973762 g/ mol)     

                                                = 0.020757 g

                                                = 20.758 mg                                   

Therefore, [P] in standard solution = 20.758 mg/ L

# Part B: Beer-Lambert’s Law, A = e C L             - equation 1,              

where,

                                                A = Absorbance

                                                e = molar extinction coefficient (M-1cm-1)

                                                p = path length (in cm)

                                                C = concentration

Putting the values for standard solution in equation 1-

            0.285 = e x (20.758 mg/ L) x p

            Or, e = 0.285 / (20.758 mg p L-1) = 0.01373 mg-1 p-1 L

# Part C. Using “e” to calculate [P] in blood sample

            0.217 = (0.01373 mg-1 p-1 L) x C x p

            Or, C = 0.217 / (0.01373 mg-1 L)

            Or, C = 15.805 mg/ L

Therefore, [P] in blood = 15.805 mg/ L = 15.805 mg / 1000 mL

# mg % P in blood = (15.805 mg / 1000 mL) x 100 mL

                                    = 1.58 %

Note: Beer-Lambert law is based in concertation of solution but not on the volume of aliquot. So, volume of aliquots shall not be accounted unless stated otherwise.

The path length is assumed to be constant at “p”. Since the question does not mentions change in path length, it’s assumed to be constant.

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