Q3(b). Detection limit and unknown by AAS (Fe)
The following data is given for Fe analysis by atomic absorption spectrometry: Fe (mg/L) 0.000, 2.000, 5.000, 15.000, unknown with absorbance 0.040, 0.150, 0.380, 1.020, 0.832. The least square fit of the calibration data is given as y=0.064x+0.032. (i) Can we estimate Fe in a sample with concentration 0.200 mg/L? Explain. (ii) Determine the concentration of Fe in the unknown sample. [Given: The blank sample was measured for 25 times and the standard deviation of the mean value (i.e., 0.040) is 0.008]
Answer in brief
(i) No. LOD =3s/m=3(0.008)/0.064=0.375 mg L⁻¹ and LOQ =10s/m=1.25 mg L⁻¹; 0.200 mg L⁻¹ is below both. (ii) x=(0.832-0.032)/0.064=12.5 mg L⁻¹.
Key points
- Slope m=0.064 L mg⁻¹; s_(blank)=0.008 (absorbance).
- LOD=(3s_b)/(m)=(3(0.008))/(0.064)=0.375 mg L⁻¹; LOQ=(10s_b)/(m)=(0.080)/(0.064)=1.25 mg L⁻¹.
- The predicted signal for 0.200 mg L⁻¹ is 0.064(0.200)+0.032=0.045, only 0.005 above the blank (0.040): less than 3s_b=0.024. The signal cannot be distinguished from the blank, so Fe at 0.200 mg L⁻¹ can be neither detected nor…
The full model answer
Worked out step by step, as it should go in the answer booklet. The solution sheet follows it.
(i) Limits (6 marks)
- Slope m=0.064 L mg⁻¹; s_(blank)=0.008 (absorbance).
- LOD=(3s_b)/(m)=(3(0.008))/(0.064)=0.375 mg L⁻¹; LOQ=(10s_b)/(m)=(0.080)/(0.064)=1.25 mg L⁻¹.
- The predicted signal for 0.200 mg L⁻¹ is 0.064(0.200)+0.032=0.045, only 0.005 above the blank (0.040): less than 3s_b=0.024. The signal cannot be distinguished from the blank, so Fe at 0.200 mg L⁻¹ can be neither detected nor quantified; concentrate the sample or use a more sensitive method (GFAAS, ICP-MS).
(ii) Unknown (4 marks)
x=(y-b)/(m)=(0.832-0.032)/(0.064)=12.5 mg L⁻¹
(inside the calibrated range 0–15 mg L⁻¹; the sample absorbance 0.832 lies between the standards 0.380 and 1.020).
Coverscalibration, LOD and LOQDetection limit and unknown by AAS (Fe)
All parts of Question 3
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