Question #73ac3

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Question #73ac3

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Answer 1 · 2015-01-06T09:18:47

You are indeed dealing with a double replacement precipitation reaction, a reaction in which two soluble ionic compounds react to form an insoluble precipitate.

If you are familiar with the solubility rules (more here: http://www.csudh.edu/oliver/chemdata/solrules.htm), you'll recognize that both $N a O H$ $NaOH$ (being a hydroxide salt of a group I element) and $A g N O_{3}$ $AgNO_3$ (a common soluble salt of silver) are soluble in water. The general reaction looks like this

$N a O H_{(a q)} + A g N O_{3 (a q)} \to A g O H_{(s)} + N a N O_{3 (a q)}$ $NaOH_((aq)) + AgNO_(3(aq)) -> AgOH_((s)) + NaNO_(3(aq))$

However, this reaction will not create siginificant amounts of $A g O H$ $AgOH$ because of the more favorable energetics of this reaction:

$2 A g O H \to A g_{2} O_{(s)} + H_{2} O_{(l)}$ $2AgOH -> Ag_2O_((s)) + H_2O_((l))$

Silver oxide looks like this:

So, your general reaction actually looks like this:

$2 A g N O_{3 (a q)} + 2 N a O H_{(a q)} \to A g_{2} O_{(s)} + 2 N a N O_{3 (a q)} + H_{2} O_{(l)}$ $2AgNO_(3(aq)) + 2NaOH_((aq)) -> Ag_2O_((s)) + 2NaNO_(3(aq)) + H_2O_((l))$

Since we are dealing with ionic compounds, the complete ionic equation looks like this:

$2 N a_{(a q)}^{+} + 2 O H_{(a q)}^{-} + 2 A g_{(a q)}^{+} + 2 N O_{3 (a q)}^{-} \to A g_{2} O_{(s)} + 2 N O_{3 (a q)}^{-} + 2 N a_{(a q)}^{+} + H_{2} O_{(l)}$ $2Na_((aq))^(+) + 2OH_((aq))^(-) + 2Ag_((aq))^(+) + 2NO_(3(aq))^(-) -> Ag_2O_((s)) + 2NO_(3(aq))^(-) + 2Na_((aq))^(+) + H_2O_((l))$

The net ionic equation, which results from eliminating spectator ions, is:

$2 A g_{(a q)}^{+} + 2 O H_{(a q)}^{-} \to A g_{2} O_{(s)} + H_{2} O_{(l)}$ $2Ag_((aq))^(+) + 2OH_((aq))^(-) -> Ag_2O_((s)) + H_2O_((l))$

Here's a video of the reaction:

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Question #73ac3

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