Question

When $Cu ^{2+}$ ion is treated with $KI$, a white precipitate, $X$ appears in solution The solution is titrated with sodium thiosulphate, the compound $Y$ is formed $X$ and $Y$ respectively are

• A. $X = CuI _2 \,\,\,Y = Na _2 S _2 O _3$
• B. $X = CuI _2 \,\,\, Y = Na _2 S _4 O _6$
• C. $X = Cu _2 I _2 \,\,\,Y = Na _2 S _4 O _6$
• D. $X = Cu _2 I _2 \,\,\, Y = Na _2 S _4 O _5$

Hint:

The reaction between copper ions and KI produces copper(I) iodide as a precipitate, which is then titrated with sodium thiosulphate.

Answer 1

The Correct Option is C

$C{u}^{2+}+2KI⟶\underset{\text{Unstable}}{Cu{I}_2}\downarrow +2K^{+}$
$I^{-}$is strong R.A it reduces $C{u}^{2+}$ to $C{u}^{+}$
$2Cu{I}_2⟶\underset{\text{(White)’X’}}{C{u}_2{I}_2\downarrow }+I_2$
$KI+I_2⟶K^{+}{I}_3^{-}\text{(Brown solution)}$
$I_3^{-}\rightleftharpoons I_2+I^{-}$
$K{I}_3+N{a}_2{S}_2{O}_3\to KI+\underset{(Y)}{N{a}_2{S}_4{O}_6}$

Answer 2

When \( \text{Cu}^{2+} \) reacts with KI, the following reactions occur:

\[ \text{Cu}^{2+} + 2\text{KI} \rightarrow \text{CuI}_2 \downarrow + 2K^+ \]

\[ \text{CuI}_2 \text{(White)} \xrightarrow{\text{Na}_2\text{S}_2\text{O}_3} \text{CuI} + \text{Na}_2\text{S}_4\text{O}_6 \]

Thus, the white precipitate \( X = \text{CuI}_2 \) and the compound formed is \( Y = \text{Na}_2\text{S}_4\text{O}_6 \). 
The reaction with sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)) helps dissolve the precipitate and form a soluble complex, converting copper(I) into copper(I) iodide. 
This results in the formation of a colorless solution, making the reaction useful in quantitative analysis and complexation studies.