Question #c4cf5

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Question #c4cf5

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Answer 1 · 2015-10-24T08:44:06

The spectator ions are $2 N a^{+}$ $2Na^+$ and $2 N O_{3}^{-}$ $2NO_3^-$

Explanation:

You need to rewrite the equation showing ALL of the ions in their corresponding oxidation states.

Hence, $P b {(N O_{3})}_{2} + 2 N a C l = P b C l_{2} + 2 N a N O_{3}$ $Pb(NO_3)_2 + 2NaCl = PbCl_2 + 2NaNO_3$ should be written as

$P b^{2+} + 2 N O_{3}^{-} + 2 N a^{+} + 2 C l^{-} = P b C l_{2} (s) + 2 N a^{+} + 2 N O_{3}^{-}$ $Pb^"2+" + 2NO_3^"-" + 2Na^"+" + 2Cl^"-" = PbCl_2 (s) + 2Na^"+" + 2NO_3^"-"$

Why write all in its ionic state except the lead (II) chloride? Because it is already in its salt (solid) state - meaning the substance will be difficult to separate since the chemical bond between them is much stronger than those in aqueous states.

Since the sodium and nitrate ions are both spectator ions, you can now simplify the formula as

$P b^{2+} + 2 C l^{-} = P b C l_{2} (s)$ $Pb^"2+" + 2Cl^"-" = PbCl_2 (s)$

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