Chemistry - Salt Analysis Question with Solution | TestHub

The reaction between ${\mathrm{Cu}}^{2+}$ and $\mathrm{KI}$ will takes place as:

${\mathrm{Cu}}^{2+}+2\mathrm{KI}⟶\underset{\mathrm{Unstable}}{{\mathrm{CuI}}_2\downarrow }+2 {\mathrm{K}}^{+}$

${\mathrm{I}}^{-}$ is a strong reducing agent, it reduces ${\mathrm{Cu}}^{2+}$ to ${\mathrm{Cu}}^{+}$ and a white precipitate of ${\mathrm{Cu}}_2{\mathrm{I}}_2$ is formed.

$2{\mathrm{CuI}}_2⟶\underset{\left(\mathrm{White}\right)\text{'}\mathrm{X}\text{'}}{{\mathrm{Cu}}_2{\mathrm{I}}_2\downarrow }+{\mathrm{I}}_2$

Sodium thiosulphate is used as reducing agent to titrate iodine, with iodide being the product of the reaction.

$2{\mathrm{Na}}_2 {\mathrm{S}}_2{\mathrm{O}}_3 + {\mathrm{I}}_2 \to 2\mathrm{NaI}+ \underset{\left(\mathrm{Y}\right)}{{\mathrm{Na}}_2 {\mathrm{S}}_4{\mathrm{O}}_6}$

So, the compound $\mathrm{X} \text{and} \mathrm{Y}$ are ${\mathrm{Cu}}_2{\mathrm{I}}_2 \text{and} {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$ respectively.