Pure solid NaH2PO4 is dissolved in distilled water, making 100.00 ml of solution

Question

Pure solid NaH2PO4 is dissolved in distilled water, making 100.00 ml of solution. 10.00ml of this solution is diluted to 100.00 ml to prepare the original phosphate standard solution. Three working standard solutions are made from this by pipetting 0.8ml, 1.5ml and 3.0 ml of the original standard solution into 100.00 ml volumetric flasks. Acid and molybdate reagent are added and the solutions are diluted to 100.00 ml. You may assume that all these absorbance measurements have already been corrected for any blank absorbance. The absorbance of each is measured in the spectrophotometer. Mass of NaH2PO4 (mg) 512.2 Absorbance Standard 1 (0.8 ml) 0.1922 Standard 2 (1.5 ml) 0.3604 Standard 3 (3.0 ml) 0.7209 Calculate the following Concentration of original phosphate standard (mM) , Concentration of Standard 1(mM), Concentration of Standard 2(mM), Concentration of Standard 3(mM), Slope of calibration line?

Explanation / Answer

Concentration NaH2PO4 stock = 512.2 mg in 100 ml = (0.5122 g/119.98 g/mole) / 0.1 L = 0.04269 M

First dilution 10ml to 100 ml, hence the concentration shall be calculated by

M1V1 = M2V2, M2 = M1V1/V2 = 10 X 0.04269 / 100 = 0.004269 M

Concentration od Standard-1: (0.8 ml to 100 ml)

If calculated as above

M2 = 0.8 x 0.004269 / 100 = 3.4152 x 10-5 M

Concentration od Standard-2: (1.5 ml to 100 ml)

M2 = 1.5 x 0.004269 / 100 = 6.4035 x 10-5 M

Concentration od Standard-3: (3.0 ml to 100 ml)

M2 = 3.0 x 0.004269 / 100 = 1.2807 x 10-4 M

Hence Three data points are

(3.4152 x 10-5, 0.1922), (6.4035 x 10-5, 0.3604), (1.2807 x 10-4, 0.7209)

Slope = 5629.43 = 5.629 x 103

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