A 50.00 mL sample containing an analyte gives a signal of 11.5 (arbitrary units)
ID: 1072530 • Letter: A
Question
A 50.00 mL sample containing an analyte gives a signal of 11.5 (arbitrary units). A second 50 mL aliquot of the sample, which is spiked with 1.00 mL of a 10.0-ppm standard solution of the analyte, gives a signal of 23.1. What is the analytes concentration in the original sample? A 50.00 mL sample containing an analyte gives a signal of 11.5 (arbitrary units). A second 50 mL aliquot of the sample, which is spiked with 1.00 mL of a 10.0-ppm standard solution of the analyte, gives a signal of 23.1. What is the analytes concentration in the original sample?Explanation / Answer
Ans. Given,
Signal of original analyte solution = 11.5 unit
Signal of the spiked solution = 23.1 unit
Increase in signal = (23.1 – 11.5) unit = 11.6 unit
Calculate: Concentration of standard only in the spiked solution –
Using, C1V1 = C2V2 -- equation 1
where
C1= Concentration of initial solution 1, V1= volume of initial solution 1 ;( 1.0 mL standard)
C2= Concentration of final solution 2, V2= volume of final solution 2 ; (50.0 mL aliquot)
Or, 10.0 ppm x 1.0 mL = C2 x 50.0 mL
Or, C2 = (10.0 ppm x 1.0 mL) /50.0 mL = 0.2 ppm
So, Concentration of standard only in the spiked solution = 0.2 ppm
Now, increase is signal is due to addition of the standard.
Thus,
11.6 unit signal is equivalent to 0.2 ppm analyte
Or, 1 unit - - (0.2/ 11.6) ppm
= 0.017241379310 ppm
Thus, 1 unit signal represents a concentration of 0.017241379310 ppm.
Calculate: Concentration of original analyte:
Concentration of original analyte= signal x concentration per unit signal
= 11.5 unit x 0.017241379310 ppm per unit signal
= 0.198275862 ppm
Thus, concentration of original analyte solution = 0.198275862 ppm
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