A 3.026-g portion of a copper(II) salt was dissolved in a 250-mL volumetric flask. A 50.0-mL aliquot was analyzed by adding of KI and titrating the liberated iodine with 23.33mL of $0.04668{\mathrm{MNa}}_2{\mathrm{S}}_2{\mathrm{O}}_3$Find the weight percent of Cu in the salt. Should starch indicator be added to this titration at the beginning or just before the end point?

The Weight Percentage is simply the ratio of a solute's mass to the mass of a solution multiplied by 100. The Weight Percentage is also referred to as the Mass Percentage.

$\mathrm{percent}\mathrm{by}\mathrm{weight}=\frac{\mathrm{gram}\mathrm{of}\mathrm{solute}}{100\mathrm{g}\mathrm{of}\mathrm{solution}}$

In this task we have a type of iodometric titration. lodide reacts with ${\mathrm{Cu}}^{2+}$and triiodide is formed:

$2{\mathrm{Cu}}^{2+}+5{\mathrm{l}}^{-}\rightleftharpoons 3\mathrm{Cil}+{\mathrm{l}}_3^{-}\left(1\right)$

Triiodide is then titrated with thiosulfate and as we can see in equation 16-19, thiosulfate is oxidized to tetrathionate:

${l_3}^{-}+2S_2{O_3}^{2-}\rightleftharpoons 3l^{-}+S_4{O_6}^{2-}\left(2\right)$

To calculate the weight percent of copper, first we need to calculate the number of moles of copper. To do that we can calculate moles of triiodide from thiosulfate and then moles of copper from triiodide.

As we can see in reaction (2), triiodide and thiosulfate react in ratio 1:2 so we can write:

$$\begin{gathered} \mathrm{n}\left({{\mathrm{l}}_3}^{-}\right)=\frac{1}{2}\mathrm{n}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right) \\ \mathrm{n}\left({{\mathrm{l}}_3}^{-}\right)=\frac{1}{2}\mathrm{c}\left({\mathrm{S}}_2{{\mathrm{O}}_3}^{2-}\right).\mathrm{V}\left({\mathrm{S}}_2{{\mathrm{O}}_2}^{2-}\right) \\ \mathrm{n}\left({{\mathrm{l}}_3}^{-}\right)=\frac{1}{2}0.04668\mathrm{mol}/\mathrm{L}.23.33\times {10}^{-3}\mathrm{L} \\ \mathrm{n}\left({{\mathrm{l}}_3}^{-}\right)=5.45\times {10}^{-4}\mathrm{mol} \end{gathered}$$

To recalculate the moles of triiodide to copper, we have to take into account the ratio that copper and triiodide react in and that is so we can write:

$$\begin{gathered} \mathrm{n}\left({\mathrm{Cu}}^{2+}\right)=2.\mathrm{n}\left({\mathrm{l}}_3^{-}\right) \\ \mathrm{n}\left({\mathrm{Cu}}^{2+}\right)=2.5.45\times {10}^{-4}\mathrm{mol} \\ \mathrm{n}\left({\mathrm{Cu}}^{2+}\right)=1.09\times {10}^{-3}\mathrm{mol} \end{gathered}$$

Since we can see in the task that ${\mathrm{Cu}}^{2+}$was dissolved 250mL in but only was took for iodometric titration we have to recalculate the number of moles from 50mL to 250mL

$\frac{250\mathrm{mL}}{\mathrm{x}}=\frac{50\mathrm{mL}}{1.09\times {10}^{-3}\mathrm{mol}}$

By cross multiplying we get:

$$\begin{gathered} \mathrm{x}.50\mathrm{mL}=250\mathrm{mL}.1.09\times {10}^{-3}\mathrm{mol}/:50\mathrm{mL} \\ \mathrm{x}=\frac{250\mathrm{arL}.1.09\times {10}^{-3}\mathrm{mol}}{50\mathrm{mL}} \end{gathered}$$

Moles of copper in 250mL solution

$\mathrm{x}=\mathrm{n}\left({\mathrm{Cu}}^{2+}\right)=5.45\times {10}^{-3}\mathrm{mol}$

Now we can calculate the weight perfect of copper in the salt:

$$\begin{gathered} \mathrm{w}(\mathrm{copper},\mathrm{salt})=\frac{\mathrm{m}(\mathrm{copper})}{\mathrm{m}(\mathrm{salt})}.100\% \\ \mathrm{w}(\mathrm{copper},\mathrm{salt})=\frac{\mathrm{m}(\mathrm{copper}).\mathrm{M}(\mathrm{copper})}{\mathrm{m}(\mathrm{salt})}.100\% \\ \mathrm{w}(\mathrm{copper},\mathrm{salt})=\frac{5.45\times {10}^{-3}.\mathrm{proT}.63.546\mathrm{g}/\mathrm{noT}}{3.026\mathrm{g}}100\% \\ \mathrm{w}(\mathrm{copper},\mathrm{salt})=11.45\% \end{gathered}$$

Since we have a iodometric titration in the task,starch should be added right before the equivalence point so that the triiodide doesn’t bind to it during the reaction.Therefore the weight percent of Cu in the salt is 11.45% .