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 concentration
of analyte in the original sample?

Answer 1

Not 100% sure about this, but this is my approach:

A=εbC

where ε is the molar absorptivity constant (since both samples contain the same analyte, assume ε is constant for both), b is the path length (assuming you're measuring both samples using the same size cell, assume b is the same for both samples), and C is concentration

letting the first 50 mL sample have absorbances A1 and C1, and the spiked sample having absorbances A2 and C2

C1 = x ppm = x mg / (50/1000 L)

meaning the mass x = C1 * (50/1000 L)

now, looking at solution 2

C2 = (the mass of a 50 mL aliquot + the mass of the 1.0 mL spike) / (50 + 1 mL)

or converting to the more appropriate ppm (mg/L) units

C2 = (C1 * (50/1000 L) + (10.0mg/L) * (1/1000)L) / (51/1000 L) which gives us C2 in terms of C1, in ppm

going back to the original Beer's equation:

A1=εbC1
A2=εbC2

dividing these gives us

A1/A2 = C1/C2

you have C2 in terms of C1, so you can plug in the C2 equation, and the absorbances, and solve for C1