Standard solution: A. KH2PO4 (FM: 136.09): 81.37mg dissolved in 500.0 mL H2O. B.

Question

Standard solution:

A. KH2PO4 (FM: 136.09): 81.37mg dissolved in 500.0 mL H2O.

B. Na2MoO4 * 2H20: 1.25 g in 50 mL of 5 M H2SO4

C. H3NH3: 0.15 g in 100 mL H20

Proedure:

Place a sample (unknown or standard phosphate solution, A) in a 5 mL volumetric flask and add 0.500 mL of B and 0.200 mL of C. Dilute to almost 5 mL with water and heat at 100C for 10 mins to form a blue product (H3PO4(MoO3)12, 12-molybdophosphoric acid). Cool the flask to room temperature, dilute to the mark with water, mix well, and measure absorbance at 830 nm in a 1.00cm cell.

A. when 0.140 mL of solution A was analyzed, an absorbane of 0.829 was recorded. A blank carried through the same procedure gave an aborbance of 0.017. Find the molar absorptivity of blue product?

B. A solution of the phosphate-containing iron-storage protein ferritin was analzyed. Unknown containing 1.35 mg ferritin was digested in a total volume of 1.00 mL to release PO4 from the protein. Then 0.300mL of this solution was analyzed and gave an absorbance of 0.836. A blank carried through the procedure gave an absorbance of 0.038. Find the wt% phosphorus in the ferritin?

Explanation / Answer

(A) concentration of B = 1.25 g/205.92 g/mol x 0.05 L = 0.12 M

12 moles of Na2MoO4 gives 1 mole of H3PO4(MoO3)12

moles of B in solution = 0.12 M x 0.5 ml = 0.06 mmol

concentration of H3PO4(MoO3)12 = 0.06 mmol/12 x 5 = 0.001 M

Absrobance = molar absorptivity x conc. x path length

molar absorptivity of compound = (0.829 - 0.017)/0.001 x 1 = 812 M-1.cm-1

(B) concentration of PO4 in solution = absorbance/molar absorptivity x path length

molar absorptivity taken from literature = 3000 M-1.cm-1

concentration = (0.836 - 0.038)/3000 x 1 = 2.66 x 10^-4 M

concentration of P in sample = 2.66 x 10^-4 M/3 = 8.86 x 10^-5 M

Mass of P in sample = 8.86 x 10^-5 x 1ml x 31 g/mol = 0.275 mg

wt% of P in sample = 0.275 x 100/1.35 = 20.37%

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