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## 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.

## 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.