Question

After dissolving $0.20$ g sample of the alloy in acid, an excess of $KI$ is added and the $C{u}^{2+}$ and $I^{-}$ ions undergo the reaction:
$C{u}^{2+}\left(aq\right)+I^{-}\left(aq\right)\to CuI\left(s\right)+I_3^{-}\left(aq\right)$
$I_3^{-}\left(aq\right)+S_2{O}_3^{2-}\left(aq\right)\to S_4{O}_6^{2-}\left(aq\right)+3I^{-}\left(aq\right)$
If $20$ ml of $0.1M$ $N{a}_2{S}_2{O}_3$ is required for the titration to the equivalent point, what is the weight percent of $Cu$ in the alloy?

• A. $6.35\%$
• B. $31.75\%$
• C. $63.5\%$
• D. $95.25\%$

Answer 1

Answer: (C)

$63.5\%$

Balanced equations are as follows:

$C{u}^{2+}\left(aq\right)+4I^{-}\left(aq\right)\to CuI\left(s\right)+I_3^{-}\left(aq\right)$

$I_3^{-}\left(aq\right)+2S_2{O}_3^{2-}\left(aq\right)\to S_4{O}_6^{2-}\left(aq\right)+3I^{-}\left(aq\right)$

From the equation, it can be seen that:

Number of moles of $I_3^{-}=2\times$ number of moles of $S_2{O}_3^{2-}=$ number of moles of $C{u}^{2+}=2$ millimoles.

So, mass of Cu in alloy $=2\times \frac{63.4}{1000}=0.127$ g.

$\%Cu$ $=0.127\times \frac{100}{0.20}=63.5\%$.