Question

Question #e6ffb

Answer 1

See explanation.

Explanation:

The idea here is that you must be familiar with the solubility rules for ionic compounds, so make sure that you grab a solubility rules chart

Let's start with the first example

`3"Ba"_ ((aq))^(2+) + 2"PO"_ (4(aq))^(3-) -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr`

The problem wants you to find two soluble ionic compounds that can be mixed in aqueous solution to produce barium phosphate, an insoluble ionic compound that precipitates out of solution.

So, the first ionic compound must dissolve to produce barium cations, `"Ba"^(2+)`, in aqueous solution, so look for an anion that does not form an insoluble compound with barium cations.

One possibility would be the nitrate anion, `"NO"_3^(-)`, meaning that one of the two soluble ionic compounds would be barium nitrate

`"Ba"("NO" _ 3) _ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)`

The second ionic compound must dissolve to produce phosphate anions, `"PO"_4^(3-)`, so look for a cation that does not form an insoluble compound with phosphate anions.

Notice that phosphate anions form soluble ionic compounds with alkali metal cations and the ammonium cation, so one good candidate here would be the sodium cation, `"Na"^(+)`, meaning that the second soluble ionic compound would be trisodium phosphate

`"Na"_ 3 "PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)`

Now, when you mix these two solutions, you will get the complete ionic equation that describes this double replacement reaction -- keep in mind that the net ionic equation must be balanced!

If you start from

`color(purple)(3)"Ba"("NO"_ 3)_ (2(aq)) + color(blue)(2)"Na"_ 3"PO"_ (4(aq)) -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr + 6"NaNO"_ (3(aq))`

you will have

`color(purple)(3) xx ["Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)] + color(blue)(2) xx [3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)] -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr + 6 xx ["Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)]`

which is equivalent to

`3"Ba"_ ((aq))^(2+) + 6"NO"_ (3(aq))^(-) + 6"Na"_ ((aq))^(+) + 2"PO"_ (4(aq))^(3-) -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr + 6"Na"_ ((aq))^(+) + 6"NO"_ (3(aq))^(-)`

To get the net ionic equation, you must eliminate the spectator ions, i.e. the ions that are present on both sides of the chemical equation.

You will have

`3"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(6"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(6"Na"_ ((aq))^(+)))) + 2"PO"_ (4(aq))^(3-) -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr + color(red)(cancel(color(black)(6"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(6"NO"_ (3(aq))^(-))))`

which gets you the starting equation

`3"Ba"_ ((aq))^(2+) + 2"PO"_ (4(aq))^(3-) -> "Ba"_ 3 ("PO"_ 4)_ (2(s)) darr`

So there you have it. I'll leave the second example to you as practice.

Remember, all you have to do is find two soluble ionic compounds that dissolve in water to produce magnesium cations, `"Mg"^(2+)`, and hydroxide anions, `"OH"^(-)`, respectively.