When Cu2+ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium…

Step 1: Reaction of $\ce{Cu^{2+}}$ with $\ce{KI}$

When $\ce{Cu^{2+}}$ (cupric) ions are treated with an excess of Potassium Iodide ($\ce{KI}$), $\ce{Cu^{2+}}$ oxidizes the iodide ions ($\ce{I-}$) into elemental iodine ($\ce{I2}$ or brown $\ce{I3-}$), and the copper is simultaneously reduced to $\ce{Cu+}$ to form an insoluble white precipitate of copper(I) iodide ($\ce{Cu2I2}$ or $\ce{CuI}$). $\ce{2Cu^{2+} + 4I- -> Cu2I2 v + I2}$ (Note: $\ce{Cu2I2}$ is often written natively as $2\ce{CuI}$ taking white/off-white appearance visually masked by iodine, but classical texts prefer formulating cuprous halides as dimers like $\ce{Cu2I2}$). Therefore, X is $\ce{Cu2I2}$. Highly unstable Cupric iodide ($\ce{CuI2}$) does not exist as a final product here.

Step 2: Titration with Sodium Thiosulphate

The liberated iodine ($\ce{I2}$) in the solution can be quantified by titrating against a standard solution of sodium thiosulphate ($\ce{Na2S2O3}$), notoriously used in iodometric titrations. Iodine uniquely oxidizes the thiosulphate specifically into a tetrathionate ($\ce{S4O6^{2-}}$) ion: $\ce{I2 + 2Na2S2O3 -> Na2S4O6 + 2NaI}$ Therefore, the compound Y formed during titration is sodium tetrathionate, $\ce{Na2S4O6}$.

Step 3: Conclusion

X is $\ce{Cu2I2}$ and Y is $\ce{Na2S4O6}$.

Key Points to Remember:

• Cupric iodide, $\ce{CuI2}$, is highly unstable and rapidly decomposes into white cuprous iodide ($\ce{Cu2I2}$) and iodine gas.
• This is the cornerstone reaction of iodometric titration for copper estimation.
• Thiosulphate ($\ce{S2O3^{2-}}$) reacts with mild oxidizer iodine ($\ce{I2}$) primarily to form tetrathionate ($\ce{S4O6^{2-}}$).