Question #31ade

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Question #31ade

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Answer 1 · 2015-05-22T15:32:41

You can add a solution of potassium iodide, $K I$ $KI$ .

The balanced chemical equation for this double replacement reaction would look like this

$P b {(N O_{3})}_{2 (a q)} + 2 K I_{(aq]} \to 2 K N O_{3 (a q)} + P b I_{2 (s)} ⏐ ⏐ ↓$ $Pb(NO_3)_(2(aq)) + 2KI_text((aq]) -> 2KNO_(3(aq)) + PbI_(2(s)) darr$

Lead (II) iodide is a yellow precipitate that will be formed in the reaction.

![http://pixgood.com/http://lead-iodide-precipitate.html](https://useruploads.socratic.org/yEgfkXjQua2tYzVnGlJW_Fphoto-29919106A-2RM.jpg)

The ney ionic equation, which you get if the spectator ions are removed, looks like this

$P b_{(a q)}^{2 +} + 2 I_{(a q)}^{-} \to P b I_{2 (s)} ⏐ ⏐ ↓$ $Pb_((aq))^(2+) + 2I_((aq))^(-) -> PbI_(2(s)) darr$

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Question #31ade

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