A third spectrophotometric method for the quantitative determination of the concentration of Pb^2+ in blood yields an Ssamp 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,

Question

A third spectrophotometric method for the quantitative determination of the concentration of [latex]Pb^{2+}[/latex] in blood yields an [latex]S_{samp}[/latex] 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 [latex]Pb^{2+}[/latex] standard and diluted to 5.00 mL, yielding an [latex]S_{spike}[/latex] of 0.419. Determine the concentration of [latex]Pb^{2+}[/latex] in the original sample of blood.

Accepted Answer

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

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

[latex]\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 imes 10^{-3}}{5.00 ml} ight group } [/latex]

[latex]\frac{0.193}{0.200C_A}=\frac{0.419}{0.200C_A+0.312 ppb } [/latex]

[latex]0.0386C_A+0.0602 ppb=0.0838C_A[/latex]

[latex]0.0452C_A=0.0602 ppb[/latex]

[latex]C_A=1.33 ppb[/latex]

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

Question 5.4: A third spectrophotometric method for the quantitative deter...

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