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?

Answer 1

concentration of original phosphate standard = 4.27 mM

Concentration of standard 1 = 0.0342 mM

Concentration of standard 2 = 0.0640 mM

Concentration of standard 3 = 0.128 mM

slope of calibration line = 5.63

Step-by-step explanation:

mass of NaH2PO4 = 512.2 mg

moles of NaH2PO4 = (mass of NaH2PO4) / (molar mass of NaH2PO4)

moles of NaH2PO4 = (512.2 mg ) / (119.977 g/mol)

moles of NaH2PO4 = 4.27 mmol

concentration of original solution = (moles of NaH2PO4) / (volume of solution in Liter)

concentration of original solution = (4.27 mmol) / (0.10000 L)

concentration of original solution = 42.7 mM

concentration of original standard solution = (concentration of original solution) * (volume of original solution / volume of standard solution)

concentration of original standard solution = (42.7 mM) * (10.00 mL / 100.00 mL)

concentration of original standard solution = .4.27 mM

concentration of standard 1 = (concentration of original standard solution) * (volume of standard solution / volume of standard 1)

concentration of standard 1 = (4.27 mM) * (0.8 mL / 100.00 mL)

concentration of standard 1 = 0.03415 mM

concentration of standard 2 = (concentration of original standard solution) * (volume of standard solution / volume of standard 2)

concentration of standard 2 = (4.27 mM) * (1.5 mL / 100.00 mL)

concentration of standard 2 = 0.0640 mM

Answer 2

Check the explanation

Step-by-step explanation:

Phosphate is an essential nutrient that controls the rate of microbial creation in a lot of marine and freshwater environments. The real phosphate concentration in waters limited by phosphorus is largely not known because commonly used chemical and radiochemical methods overrate the concentration.

Kindly check the attached image below to see the step by step solution to the above questions.