${\mathbf{cd}}^{\mathbf{2}\mathbf{+}}$was used as an internal standard in the analysis of ${\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}$by square wave polarography. ${\mathbf{Cd}}^{\mathbf{2}\mathbf{+}}$gives a reduction wave at -0.60 V and ${\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}$gives a reduction wave at –0.40 V. It was first verified that the ratio of peak heights is proportional to the ratio of concentrations over the whole range employed in the experiment. Here are results for known and unknown mixtures:

The unknown mixture was prepared by mixing $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{)}\mathbf{}\mathbf{mL}$of unknown (containing only ${\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}$) plus $\mathbf{10}\mathbf{.}\mathbf{00}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{)}\mathbf{}\mathbf{mL}\mathbf{}$of $\mathbf{3}\mathbf{.}\mathbf{23}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{)}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{}{\mathbf{MCd}}^{\mathbf{2}\mathbf{+}}$and diluting to $\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{(}\mathbf{\pm }\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{)}\mathbf{}\mathbf{mL}$.

(a) Disregarding uncertainties, find $\mathbf{\left[}{\mathbf{Pb}}^{\mathbf{2}\mathbf{+}}\mathbf{\right]}$in the undiluted unknown.

(b) Find the absolute uncertainty for the answer to part (a).

The internal standard equation is $\mathrm{X}={\mathrm{Pb}}^{2+}$and $\mathrm{S}={\mathrm{Cd}}^{2+}$.

The standard mixture response factor is given as

role="math" localid="1667558549294" $$\begin{gathered} \frac{\mathrm{Signal}{}_{\mathrm{X}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{\mathrm{signals}{}_{\mathrm{s}}}{\left[\mathrm{S}\right]} \\ \frac{1.58\mathrm{\mu A}}{[41.8\mathrm{\mu M}]}=\mathrm{F}\left(\frac{1.64\mathrm{\mu A}}{[32.3\mathrm{\mu M}]}\right) \\ \mathrm{F}=0.7445 \end{gathered}$$

Cadmium mixed with an unknown concentration is

$$\begin{gathered} =\left(\frac{10.00}{50.00}\right)\left(3.23\times {10}^{-4}M\right) \\ =6.46\times {10}^{-5}M \end{gathered}$$

The concentration of a lead ion in the diluted unknown and undiluted unknown is calculated as

$$\begin{gathered} \frac{\mathrm{Signal}{}_{\mathrm{X}}}{\left[\mathrm{X}\right]}=\mathrm{F}\frac{\mathrm{signals}{}_{\mathrm{s}}}{\left[\mathrm{S}\right]} \\ \frac{3.00\mu A}{\left[P{b}^{2+}\right]}=0.{744}_5\left(\frac{1.64\mathrm{\mu A}}{[64.6\mu \mathrm{M}]}\right) \\ \left[{\mathrm{Pb}}^{2+}\right]=130.2\mu \mathrm{M} \end{gathered}$$

The strength of undiluted unknown is$\left(\frac{50.00}{25.00}\right)130.2\mu M=2.60\times {10}^{-4}M$

Comparative uncertainty in response factor is

$$\begin{gathered} F=\frac{(1.58\pm 0.03)(32.3\pm 0.1)}{(1.64\pm 0.03)(41.8\pm 0.1)} \\ =0.7445\pm 0.0199(\pm 2.67\mathrm{\%}) \end{gathered}$$

Ambiguity in concentration of lead ion is

$$\begin{gathered} \left[{\mathrm{Pb}}^{2+}\right]=\frac{(3.00\pm 0.03)\frac{(10.00\pm 0.05)}{(50.00\pm 0.05)}\left(3.23(\pm 0.01)\times {10}^4\right)}{(2.00\pm 0.03)(0.7445\pm 0.0199)} \\ \left[{\mathrm{Pb}}^{2+}\right]=2.60(\pm 0.09)\times {10}^{4-}\mathrm{M} \end{gathered}$$