Q5(b)(ii). Phosphate in urine by standard addition
2.00 mL of the urine sample A was treated with reagent to generate colour with phosphate, and the resulting content was diluted to 100 mL. A second 2.00 mL sample B of the same urine was added to exactly 5.0 mL of phosphate solution (0.0300 mg/mL), which was treated in the same way as the sample A. The absorbance of the solution from A = 0.428, and that from B = 0.538. Calculate the concentration of phosphate (mg/mL) in the urine sample A.
Answer in brief
Phosphate in 2.00 mL of urine = 0.584 mg, so the urine contains 0.292 mg mL⁻¹ of phosphate.
Part of VKC Basic
Read the key points and the full model answer with VKC Basic
VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversstandard additionphosphate determinationspectrophotometryBeer's lawabsorbance
All parts of Question 5
- 5(a) Confidence interval of a mean at the 99 % level5 marks
- 5(b)(i) Wavelength of the sodium resonance line5 marks
- 5(b)(ii) Phosphate in urine by standard addition5 marks
- 5(c)(i) Ultimate analysis of a fuel; nitrogen by the Kjeldahl method10 marks
- 5(c)(ii) Fe₃O₄ in an ore by gravimetry5 marks