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Chapter 5

Q. 5.4

A third spectrophotometric method for the quantitative determination of the concentration of Pb^{2+} in blood yields an S_{samp} of 0.193 for a 1.00-mL sample of blood that has been diluted to 5.00 mL. A second 1.00-mL sample is spiked with 1.00 μL of a 1560-ppb Pb^{2+} standard and diluted to 5.00 mL, yielding an S_{spike} of 0.419. Determine the concentration of Pb^{2+} in the original sample of blood.

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Verified Solution

The concentration of Pb^{2+} in the original sample of blood can be determined by making appropriate substitutions into equation 5.7 and solving for C_A. Note that all volumes must be in the same units, thus V_s is converted from 1.00 mL to 1.00 × 10^{–3} mL.

\frac{S_{samp}}{C_A(V_0/V_f)} =\frac{S_{spike}}{C_A(V_0/V_f)+C_S(V_s/V_f)}       (5.7)

\frac{0.193}{C_A(\frac{1.00  ml}{5.00  ml} )} =\frac{0.419}{C_A(\frac{1.00  ml}{5.00  ml}) +1560  ppb\left\lgroup\frac{1.00\times 10^{-3}}{5.00  ml} \right\rgroup }

\frac{0.193}{0.200C_A}=\frac{0.419}{0.200C_A+0.312  ppb }

0.0386C_A+0.0602  ppb=0.0838C_A

0.0452C_A=0.0602  ppb

C_A=1.33  ppb

Thus, the concentration of Pb^{2+} in the original sample of blood is 1.33 ppb.