Question

A$\mathbf{0}\mathbf{.}\mathbf{326}\mathbf{}\mathbf{8}\mathbf{-}\mathbf{g}$unknown containing $\mathbf{Pb}{\left({\mathrm{CH}}_3{\mathrm{CHOHCO}}_2\right)}_{\mathbf{2}}$(leadLactate, FM 385.3) plus inert material was electrolyzed to produce $\mathbf{0}\mathbf{.}\mathbf{1111}\mathbf{g}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{PbO}}_{\mathbf{2}}\mathbf{}\left(\mathrm{FM}239.2\right)$. Was the PbO₂deposited at the anode or at the cathode? Find the weight percent of lead lactate in theUnknown.

Short answer

PbO₂was deposited at the anode. The weight percent of lead lactate is $\mathrm{w}\left(\mathrm{lead}\mathrm{lactate}\right)=54.76\%$.

Step-by-step solution

Electrogravimetric Analysis

• Separation along with the quantification of metals are mainly observed in electrogravimetricanalysis.
• Electrolysis of the analyte solution is done during this procedure.

Determine the lead lactate

The reaction is

$\mathrm{Pb}{\left({\mathrm{CH}}_3\mathrm{CHOHCO}\right)}_2+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{PbO}}_2+2\mathrm{lactate}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

${\mathrm{Pb}}^{2+}$Is oxidized to ${\mathrm{Pb}}^{4+}$at the anode

First, we need to calculate n of ${\mathrm{PbO}}_2$

role="math" localid="1663650526992" $$\begin{gathered} \mathrm{n}\left({\mathrm{PbO}}_2\right)=\frac{\mathrm{m}\left({\mathrm{PbO}}_2\right)}{\mathrm{M}\left({\mathrm{PbO}}_2\right)}=\frac{0.1111\mathrm{g}}{239.2\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}\left({\mathrm{PbO}}_2\right)=4.64\times {10}^{-4}\mathrm{mol} \\ \mathrm{n}\left({\mathrm{PbO}}_2\right)=\mathrm{n}\left(\mathrm{lead}\mathrm{lactate}\right),\mathrm{the}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{lead}\mathrm{lactate}\mathrm{is} \\ \mathrm{m}\left(\mathrm{lead}\mathrm{lactate}\right)=\mathrm{n}\left(\mathrm{lead}\mathrm{lactate}\right)·\mathrm{M}\left(\mathrm{lead}\mathrm{lactate}\right) \\ \mathrm{m}\left(\mathrm{lead}\mathrm{lactate}\right)=4.64\times {10}^{-4}\mathrm{mol}\times 385.3\mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left(\mathrm{lead}\mathrm{lactate}\right)=0.17896\mathrm{g} \end{gathered}$$

So, the w lead lactate is:

$$\begin{gathered} \mathrm{w}\left(\mathrm{lead}\mathrm{lactate}\right)=\frac{\mathrm{m}\left(\mathrm{lead}\mathrm{lactate}\right)}{\mathrm{m}} \\ \mathrm{w}\left(\mathrm{lead}\mathrm{lactate}\right)=\frac{0.17896\mathrm{g}}{0.3268\mathrm{g}} \\ \mathrm{w}\left(\mathrm{lead}\mathrm{lactate}\right)=0.5476=54.76\% \end{gathered}$$

The lead lactate is $54.76\%$.