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

To solve this question, it is important to understand the chemical reactions involved when the \(Cu^{2+}\) ion is treated with potassium iodide (KI). Let's break down the reaction step by step.

When \(Cu^{2+}\) ions react with \(KI\), the iodide ion reduces \(Cu^{2+}\) to \(Cu^+\), and iodine is liberated. This leads to the formation of a white precipitate known as cuprous iodide, \(Cu_2I_2\).

The chemical reaction can be written as:

\(2Cu^{2+} + 4I^- \rightarrow Cu_2I_2 \downarrow + I_2\)

The precipitate \(Cu_2I_2\) is a white solid and is denoted by \(X\) in the options provided.

Following this, the liberated iodine \(I_2\) is titrated with sodium thiosulphate, \(Na_2S_2O_3\), which reduces iodine to iodide ions while itself getting oxidized to tetrathionate, \(Na_2S_4O_6\).

The titration reaction is:

\(2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI\)

Thus, \(Y\) is identified as sodium tetrathionate, \(Na_2S_4O_6\).

Based on the above reactions, the correct answer is: \(X = Cu_2I_2\,\,\,Y = Na_2S_4O_6\)

This corresponds to the third option in the list:

• \(X = CuI_2\,\,\,Y = Na_2S_2O_3\)
• \(X = CuI_2\,\,\,Y = Na_2S_4O_6\)
• \(\mathbf{X = Cu_2I_2\,\,\,Y = Na_2S_4O_6}\)
• \(X = Cu_2I_2\,\,\,Y = Na_2S_4O_5\)