Question #d3b6a
1 Answer
Here's what I got.
Explanation:
Barium nitrate,
"Ba"("NO"_ 3)_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)Ba(NO3)2(aq)→Ba2+(aq)+2NO−3(aq)
"NaF"_ ((aq)) -> "Na"_ ((aq))^(+) + "F"_ ((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"_2BaF2 , an insoluble ionic compound that precipitates out of solution- sodium nitrate,
"NaNO"_3NaNO3 , 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))Ba(NO3)2(aq)+NaF(aq)→BaF2(s)⏐⏐↓+NaNO3(aq)
To balance this chemical equation, add a coefficient of
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)))
