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

1) Concentration of original phosphate standard (mM)  

2) Concentration of Standard 1(mM)

3) Concentration of Standard 2(mM)

4) Concentration of Standard 3(mM)

5) Slope of calibration line

Answer 1

ans)

for standard 1

Given,

Mass of NaH₂PO₄ = 512.2 mg per 1000 mL

=  51.22 mg per 100 mL

number of moles = mass / molar mass of NaH₂PO₄

= 51.22 mg / 120000 mg/mmol

= 0.00043 mmol

Now, molarity or concentration = number of moles / volume

= 0.00043 mmol / 0.8 mL

= 0.00054 mM

3)

for standard 2

Given,

Mass of NaH₂PO₄ = 512.2 mg per 1000 mL

=  51.22 mg per 100 mL

number of moles = mass / molar mass of NaH₂PO₄

= 51.22 mg / 120000 mg/mmol

= 0.00043 mmol

Now, molarity or concentration = number of moles / volume

= 0.00043 mmol / 1.5 mL

= 0.00029 mM

4)

for standard 3

Given,

Mass of NaH₂PO₄ = 512.2 mg per 1000 mL

=  51.22 mg per 100 mL

number of moles = mass / molar mass of NaH₂PO₄

= 51.22 mg / 120000 mg/mmol

= 0.00043 mmol

Now, molarity or concentration = number of moles / volume

= 0.00043 mmol / 3mL

= 0.00014 mM