Question #929f6

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Question #929f6

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Answer 1 · 2016-12-20T00:03:34

The $C o^{2 +} and P O_{4}^{3 -}$ $Co^(2+)andPO_4^(3-)$ ions will form a precipitate.

Explanation:

So we can leave out the $C l^{-} and N a^{+}$ $Cl^(-) andNa^+$ ions, as they are only 'spectators', i.e. they will be left unchanged in the solution.

Since $C o$ $Co$ has a charge of +2 and $P O_{4}$ $PO_4$ has a charge of -3, we will have to take 3 of the first and 2 of the other to make charges equal (but opposite):

$3 C o^{2 +} (a q) + 2 P O_{4}^{3 -} (a q) \to C o_{3} {(P O_{4})}_{2} (s)$ $3Co^(2+)(aq)+2PO_4^(3-)(aq)->Co_3(PO_4)_2(s)$

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Question #929f6

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