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The secondary valence of Pt^(4+) is six. Calculate the number of moles of AgCl participated, when axcess of AgNO_3 solution is added to 2L of 0.1 M PtCl_4 . 4NH_3 solution. |
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Answer» Solution :Secondary valence of `Pt^(4+)` is six it will be satisfied by four `NH_3` and two `Cl^(-)` ions. The remaining two `Cl^(-)` ion satisfy only PRIMARY valence. `PtCl_4 . 4NH_3` can be WRITTEN as `[Pt(NH_3)_4Cl_2]Cl_2`. One mole of this complex GIVES 2 moles of chloride ions in the aqueous solution. NUMBER of moles of complex present in 2L of 0.1 M`PtCl_4 . 4NH_3 = M xx V = 0.1 xx 2 = 0.2`. Number of mole of chloride ions present in 2L of `0.1 M PtCl_4 NH_3` solution `=2 xx 0.2 = 0.4` Number of moles of AgCl precipitated from 2L of `0.1 M PtCl_4 . 4NH_3` is 0.4. |
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