Pure solid NaH2PO4 is dissolved in distilled water, making 100.00 ml of solution
ID: 1033444 • Letter: P
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) and the 3 other standards.
Explanation / Answer
Ans. Original Solution: Moles of NaH2PO4 taken = Mass/ MW
= 0.5122 g / (119.97701 g/ mol)
= 4.2692 x 10-3 mol
# 1 mol NaH2PO4 yields 1 mol PO43-.
So, the moles of phosphate (PO43-) taken = 4.2692 x 10-3 mol =
Now,
[PO43-] in original solution = 4.2692 x 10-3 mol / 0.100 L
= 4.2692 x 10-2 M
= 42.692 mM
# Solution 2: 10.0 mL of original solution is diluted to 100.0 mL.
Using C1V1 (original solution) = C2V2 (solution 2)
Or, 42.692 mM x 10.0 mL = C2 x 100.0 mL
Hence, C2 = (42.692 mM x 10.0 mL) / 100.0 mL = 4.2692 mM
Therefore,
[PO43-] in solution 2 = 4.2692 mM
# Working standards are prepared from taking certain volume of solution 2 and dilution them to a final volume of 100.0 mL.
# [PO43-] in working std 1, C2 = (4.2692 x 0.8 mL) / 100 mL = 0.0341536 mM
# [PO43-] in working std 2, C2 = (4.2692 x 1.5 mL) / 100 mL = 0.064038 mM
# [PO43-] in working std 1, C2 = (4.2692 x 3.0 mL) / 100 mL = 0.128076 mM
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