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## Q. 5.5

A fourth spectrophotometric method for the quantitative determination of the concentration of $Pb²^+$ in blood yields an $S_{samp}$ of 0.712 for a 5.00-mL sample of blood. After spiking the blood sample with 5.00 mL of a 1560-ppb $Pb²^+$ standard, an $S_{spike}$ of 1.546 is measured. Determine the concentration of $Pb²^+$ in the original sample of blood.

## Verified Solution

The concentration of $Pb²^+$ in the original sample of blood can be determined by making appropriate substitutions into equation 5.9 and solving for $C_A$.

$\frac{s_{samp}}{C_A} =\frac{S_{spike}}{C_A[V_0/(V_0+V_s)]+C_S[V_s/(V_0+V_s)]}$      (5.9)

$\frac{0.712}{C_A} =\frac{1.546}{C_A\left[\frac{5.00 ml}{\left(5.00 ml+5.00\times 10^{-3} ml\right) }\right] +1560 ppb\left[\frac{ 5.00 \times 10^{-3 ml}}{(5.00 ml + 5.00\times 10^{-3} ml)}\right] }$

$\frac{0.712}{C_A} =\frac{1.546}{0.9990C_A+1.558 ppb}$

$0.7113C_A+1.109 ppb=1.546C_A$

$C_A$ = 1.33  ppb

Thus, the concentration of $Pb²^+$ in the original sample of blood is 1.33 ppb.