Calcium fluorapatite$\mathbf{\left(}{\mathbf{Ca}}_{\mathbf{10}}\mathbf{\right(}{\mathbf{PO}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{6}}{\mathbf{F}}_{\mathbf{2}}\mathbf{,}\mathbf{FM}\mathbf{1008}\mathbf{.}\mathbf{6}\mathbf{)}$laser crystals were doped with chromium to improve their efficiency. It was suspected that the chromium could be in the$\mathbf{+}\mathbf{4}$oxidation state.

1. To measure the total oxidizing power of chromium in the material, a crystal was dissolved in $\mathbf{2}\mathbf{.}\mathbf{9}\mathit{M}\mathit{H}\mathit{C}\mathit{L}{\mathit{O}}_{\mathbf{4}}$ at $\mathbf{100}\mathbf{°}\mathit{C}$, cooled to $\mathbf{20}\mathbf{°}\mathit{C}$ , and titrated with standard $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ , using Pt and Ag - AgCl electrodes to find the end point. Chromium above the 3 + state should oxidize an equivalent amount of $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ in this step. That is,$\mathit{C}{\mathit{r}}^{\mathbf{4}\mathbf{+}}$would consume one $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ , and $\mathit{C}{\mathit{r}}^{\mathbf{6}\mathbf{+}}$in $\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}$ would consume three $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ :

role="math" localid="1664873864085" $$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}{\mathit{r}}^{\mathbf{4}\mathbf{+}}\mathbf{+}\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\to }\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}} \\ \frac{\mathbf{1}}{\mathbf{2}}\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{3}\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\to }\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{3}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}} \end{gathered}$$

2. In a second step, the total chromium content was measured by dissolving a crystal in $\mathbf{2}\mathbf{.}\mathbf{9}\mathit{M}\mathit{H}\mathit{C}\mathit{L}{\mathit{O}}_{\mathbf{4}}$ at and cooling to $\mathbf{20}\mathbf{°}\mathit{C}$ . Excess and were then added to oxidize all chromium to $\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}$ . Unreacted ${\mathit{S}}_{\mathbf{2}}{{\mathit{O}}_{\mathbf{8}}}^{\mathbf{-}\mathbf{2}}$was destroyed by boiling, and the remaining solution was titrated with standard $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ . In this step, each Crin the original unknown reacts with three $\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}$ .

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}{\mathit{r}}^{\mathbf{x}\mathbf{+}}\mathbf{+}\stackrel{{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{8}}^{\mathbf{2}\mathbf{-}}}{\mathbf{\to }}\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}} \\ \frac{\mathbf{1}}{\mathbf{2}}\mathit{C}{\mathit{r}}_{\mathbf{2}}{\mathit{O}}_{\mathbf{7}}^{\mathbf{2}\mathbf{-}}\mathbf{+}\mathbf{3}\mathit{F}{\mathit{e}}^{\mathbf{2}\mathbf{+}}\mathbf{\to }\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{3}\mathit{F}{\mathit{e}}^{\mathbf{3}\mathbf{+}} \end{gathered}$$

In Step 1,0.4375g of laser crystal required 0.498mL of (prepared by dissolving in ). In step , of crystal required of the same solution. Find the average oxidation number of in the crystal and find the total micrograms ofpre gram of crystal.

A redox titration happens when the analyte and the titrant undergo an oxidation–reduction process. The endpoint is frequently detected using an indicator, much as it is in acid–base titrations. Oxalic acid titrated against potassium permanganate in acid medium is an example of redox titration.

In Step 1, chromium above +3 state was used to oxidize $F{e}^{2+}$. By looking at the first reaction in step 1 , we can see that the ratio of moles of $C{r}^{4+}$and $F{e}^{4+}$is 1:1

$$\begin{gathered} c\left(F{e}^{2+}\right)=2.786mM=2.786\times {10}^{-3}M \\ V\left(F{e}^{2+}\right)=0.498mL=0.498\times {10}^{-3}L \\ m\left(crystal\right)=0.4375g \end{gathered}$$

Number of moles of $F{e}^{2+}$used in step is:

$$\begin{gathered} n\left(F{e}^{2+}\right)=c\left(F{e}^{2+}\right)\times V\left(F{e}^{2+}\right) \\ n\left(F{e}^{2+}\right)=0.498\times {10}^{-3}L\times 2.786\times {10}^{-3}M \\ n\left(F{e}^{2+}\right)=1.387\times {10}^{-6}mol \end{gathered}$$

Hence the number of moles of $F{e}^{2+}$is $1.387\times {10}^{-6}mol$.

In Step 2 , the total chromium content was determined by using the same solution. The only difference is the ratio of moles. Each chromium reacts with three $F{e}^{2+}$.

$$\begin{gathered} c\left(F{e}^{2+}\right)=2.786mM=2.786\times {10}^{-3}M \\ V\left(F{e}^{2+}\right)=0.703mL=0.703\times {10}^{-3}L \\ m\left(crystal\right)=0.1566g \end{gathered}$$

The average oxidation number is:

Average oxidation number= role="math" localid="1664877813554" $$\begin{gathered} 3+\frac{n(Crabove3+)}{n(totalCr)} \\ 3+\frac{3.17\times {10}^{-6}mol}{4.17\times {10}^{-6}mol} \end{gathered}$$

= 3+0.76

=3.76

Hence the average oxidation number is

Total micrograms of chromium per gram of crystal can be calculated by using total moles and molar mass of chromium:

$$\begin{gathered} m(totalCrincrystal)=n(totalCr)\cdot M\left(Cr\right) \\ m(totalCrincrystal)=4.17\times {10}^{-6}mol\cdot 51.996g/mot \\ m(totalCrincrystal)=2.17\times {10}^{-4}g=217\mu g \end{gathered}$$

Hence the total amount of chromium is $217\mu g$.