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How does K2[PtCl4] gets ionized when dis...

How does `K_2[PtCl_4]` gets ionized when dissolved in water? Will it form precipitate when `AgNO_3` solution is added to it? Give a reason for your answer.

Text Solution

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When `K_(2)[PtCl_(4)]` is dissolved in water then it ionizes as follows :
`K_(2)PtCl_(4)rarr2K^(+)+[PtCl_(4)]^(2-)`
and when `AgNO_(3)` is added to `K_(2)[PtCl_(4)]` then it will not form the white precipate with `AgNO_(3)` solution since `Cl^(-)` ions are not free instead of it grey precipitate is obtained.
`K_(2)[PtCl_(4)]+2AgNO_(3)rarr2KCl(aq)+underset("Grey precipitate")(PtCl_(2)(AgNO_(3))_(2))(s)`
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