Question #d3b6a

1 Answer
Jan 10, 2017

Here's what I got.


Barium nitrate, #"Ba"("NO"_3)_2#, and sodium fluoride, #"NaF"#, are soluble ionic compounds, which means that they exist as ions in aqueoues solutions.

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

#"NaF"_ ((aq)) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-)#

When two aqueous solutions of barium nitrate and sodium fluoride are mixed, two ionic compounds are formed

  • barium fluoride, #"BaF"_2#, an insoluble ionic compound that precipitates out of solution
  • sodium nitrate, #"NaNO"_3#, a soluble ionic compound that exists as ions in the resulting solution

You can write the unbalanced chemical equation that describes this double replacement reaction like this

#"Ba"("NO"_ 3)_ (2(aq)) + "NaF"_ ((aq)) -> "BaF"_ (2(s)) darr + "NaNO"_ (3 (aq))#

To balance this chemical equation, add a coefficient of #2# to sodium fluoride

#color(darkgreen)(ul(color(black)("Ba"("NO"_ 3)_ (2(aq)) + 2"NaF"_ ((aq)) -> "BaF"_ (2(s)) darr + "NaNO"_ (3 (aq)))))#

Now, you can start from the balanced chemical equation and write the complete ionic equation

#"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2 xx ["Na"_ ((aq))^(+) + "F"_ ((aq))^(-)] -> "BaF"_ (2(s)) darr + 2 xx ["Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)]#

This is equivalent to

#"Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-) + 2"Na"_ ((aq))^(+) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr + 2"Na"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-)#

In order to get the net ionic equation, you must remove the spectator ions, i.e. the ions that are present on both sides of the equation.

#"Ba"_ ((aq))^(2+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))#

This will get you

#color(darkgreen)(ul(color(black)("Ba"_ ((aq))^(2+) + 2"F"_ ((aq))^(-) -> "BaF"_ (2(s)) darr)))#