Question

Gold nanoparticles (Figure 17-31) can be titrated with the oxidizing agent TCNQ in the presence of excess of${\mathbf{Br}}_{\mathbf{2}}^{\mathbf{-}}$to oxidize $\mathbf{Au}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{"}\mathbf{to}\mathbf{}{\mathbf{AuBr}}_{\mathbf{2}}^{\mathbf{-}}$in deaerated toluene. Gold atoms in the interior of the particle are Au(0) . Gold atoms bound to ${\mathit{C}}_{\mathit{12}}{\mathit{H}}_{\mathit{25}}\mathit{S}$- (dodecanethiol) ligands on the surface of the particle are Au(I) and are not titrated.

The table gives the absorbance at $856 0.700 mL of $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathbf{MTCNQ}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{05}\mathbf{M}\mathbf{(}{\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{1}}\mathbf{7}{\mathbf{)}}_{\mathbf{4}}{\mathbf{N}}^{\mathbf{+}}{\mathbf{Br}}^{\mathbf{-}}$in toluene is titrated with gold nanoparticles (1.43 g/Lin toluene) from a microsyringe with a Teflon-coated needle. Absorbance in the table has already been corrected for dilution.

(a) Make a graph of absorbance versus volume of titrant and estimate the equivalence point. Calculate the Au(0) in 1.00 g of nanoparticles.

(b) From other analyses of similarly prepared nanoparticles, it is estimated that 25 % of the mass of the particle is dodecanethiol ligand. Calculate mmol of localid="1667559229564" ${\mathbf{C}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{25}}\mathbf{S}$in 1.00 g of nanoparticles.

(c) The Au(I) content of 1.00 g of nanoparticles should be 1.00 mass of Au(0) - mass of ${\mathbf{C}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{25}}\mathbf{S}$. Calculate the micromoles of Au(I) in 1.00 g of nanoparticles and the mole ratio Au(I) :. In principle, this ratio should be 1: 1. The difference is most likely because ${\mathbf{C}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{25}}\mathbf{S}$was not measured for this specific nanoparticle preparation.

Short answer

$\mathrm{a})\mathrm{V}=21.4\mathrm{\mu L};\mathrm{n}(\mathrm{Ag})=2.29\mathrm{mmol}\mathrm{b})\mathrm{n}=1.24\mathrm{mmol}\mathrm{c})\mathrm{n}=1.52\mathrm{mmol};\mathrm{ratio}=1.23;$

Step-by-step solution

Calculate the Au(0)

a) The end point is observed from the graph $\left(\mathrm{V}=21.4\mathrm{\mu L}\right):$

$$\begin{gathered} \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{nano}\mathrm{particle} \\ \mathrm{V}=21.4\mathrm{\mu L} \\ \mathrm{n}\left(\mathrm{Ag}\right)=2.29\mathrm{mmol} \end{gathered}$$

Calculate mmol of C12H25S

b) 1.00g of nanoparticles contains $0.25{\mathrm{gC}}_{12}{\mathrm{H}}_{25}\mathrm{S}(\mathrm{w}=25\mathrm{\%})$. So, the n is:

localid="1667559795352" $$\begin{gathered} \mathrm{n}=\frac{\mathrm{n}}{\mathrm{m}}=\frac{0.25}{201.40\mathrm{g}/\mathrm{mo}l} \\ \mathrm{n}=1.24.{10}^{-3}\mathrm{mol} \\ \mathrm{n}=1.24\mathrm{mmol} \end{gathered}$$

Calculate micromoles of Au(I) 

c)First we calculate the mass of Ag(I) on 1 g of nanoparticles:

$$\begin{gathered} \mathrm{m}\left(\mathrm{Ag}\right)=2.29\cdot {10}^{-3}\mathrm{mol}\cdot 196.97\hspace{0.33em}\mathrm{g}/\mathrm{mol} \\ \mathrm{m}\left(\mathrm{Ag}\right)=0.4511\hspace{0.33em}\mathrm{g} \end{gathered}$$

The mass of Ag(I) is:

role="math" localid="1667559730834" $$\begin{gathered} \mathrm{m}\left(\mathrm{Ag}\right)=1-\mathrm{m}\left(\mathrm{Ag}\right(0\left)\right)-\mathrm{m}\left({\mathrm{C}}_{12}{\mathrm{H}}_{25}\mathrm{S}\right) \\ \mathrm{m}\left(\mathrm{Ag}\right)=(1-0.4511-0.25)\hspace{0.33em}\mathrm{g} \\ \mathrm{m}\left(\mathrm{Ag}\right)=0.2989\hspace{0.33em}\mathrm{g} \end{gathered}$$

The n of Ag(I) is:

$$\begin{gathered} \mathrm{n}=\frac{\mathrm{n}}{\mathrm{M}}=\frac{0.2989g}{196.97\mathrm{g}/\mathrm{mo}l} \\ \mathrm{n}=1.52.{10}^{-3}\mathrm{mol} \\ \mathrm{n}=1.52\mathrm{mmol} \end{gathered}$$

The ratio$\mathrm{Ag}\left(\mathrm{I}\right):{\mathrm{C}}_{12}{\mathrm{H}}_{25}\mathrm{S}$ is:

$\frac{1.52\mathrm{mmol}}{1.24\mathrm{mmol}}=1.23$