Question

Solution A: KH2PO4 with (FM 136.09) : 81.37 mg dissolved in 500.0mL H20.

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

Solution 3: H3NNH3 ^2+ and SO4 ^2- (ionic) : 0.15 g in 100 mL H20

Procedure: place sample of standard phosphate solution A in a 5mL volumetric flask and add .500mL of B and .200mL of C. Dilute to almost 5 mL with water and heat at 100 degrees C for 10 min to form a blue product (H3PO4(MoO3)12). Cool the flask to room temp, dilute to the mark with water, mix, and measure absorbance at 830nm in 1.00cm cell.

a) When .140 ml of solution A was analyzed, an absorbance of .829 was recored. A blank carried through the same procedure gave an absorbance of .017. Find the molar absorbtivity of the blue product.

b) A solution of the phosphate containing iron storage protein ferritin was analyzed. Unknown containing 1.35 mg ferritin was digested in a total volume of 1.00mL to release PO4^3- from the protein. Then .300mL of this solution was analyzed and gave an absorbance of .836. A blank carried through the procedure gave an absorbance of .038. FInd weight % phosphorus in the ferritin.

Answer 1

Solution.

a) The molar concentration of the solution A is 81.37*2*10^-3/136.09 = 1.196*10^-3 M;

The solution A was diluted, and its final concentration is 1.196*10^-3 *(0.140/5) = 3.35*10^-5 M; after the reaction completed, the concentration of the blue product is the same.

The molar absorbtivity of the blue product can be found using a formula:

A-A₀=klc; l = 1cm;

k = (A-A₀)/c = (0.829-0.017)/(3.35*10^-5) = 2.42*10⁴.

b) The concentration of the blue product should be found first:

c = (A-A₀)/k = (0.836-0.038)/(2.42*10⁴) = 3.30*10^-5 M.

The amount corresponding to Phosphorus in ferritin is n= cV = (3.30*10^-5 M)(0.001 L) = 3.30*10^-8 moles. The mass of Phosphorus is (3.30*10^-8 mol)(30.97 g/mol) = 1.067*10^-9 g or 1.067*10^-6 mg.

The weight % phosphorus in the ferritin is: