Question

How do you write the molecular, complete ionic, and net ionic equation for lithium sulfide reacting with calcium hydroxide? Assume that at the concentrations chosen, calcium hydroxide is soluble.

Answer 1

Well, you would do all three in sequence anyway, if you wanted to break down the steps... eventually, when you would have gotten better with this over time, you would have skipped straight to the net ionic equation had you been given the opportunity.

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This is a double-replacement reaction, i.e. the cations switch or the anions switch (your pick). Charge is conserved, and the resultant compound is made with that in mind.

This would be the molecular equation:

`barul(|stackrel(" ")(" ""Li"_color(red)(2)"S"(aq) + "Ca"("OH")_color(red)(2)(aq) -> "CaS"(s) + color(red)(2)"LiOH"(aq)" ")|)`

As usual, coefficients apply to the entire compound, and subscripts apply to the element directly preceding them.

Calcium sulfide is generally regarded as "slightly soluble" in water, but for the sake of a reaction occurring, we will regard it as "insoluble", i.e. as a solid, `(s)`.

The complete ionic equation separates the ions of aqueous species:

`barul(|stackrel(" ")(" "overbrace(color(red)(2)"Li"^(+)(aq) + "S"^(2-)(aq))^("Li"_color(red)(2)"S"(aq)) + overbrace("Ca"^(2+)(aq) + color(red)(2)"OH"^(-)(aq))^("Ca"("OH")_color(red)(2)(aq)) -> "CaS"(s) + overbrace(color(red)(2)"Li"^(+)(aq) + color(red)(2)"OH"^(-)(aq))^(color(red)(2)"LiOH"(aq))" ")|)`

(Remember that subscripts in compounds had indicated how many of that element there were in that compound. Thus, separating it out, you must rewrite that subscript as a coefficient.)

And the net ionic equation/reaction ignores species that did nothing. In other words, it focuses on the main action; we disregard ions that form themselves again, and call them spectator ions.

`overbrace(cancel(color(red)(2)"Li"^(+)(aq)) + "S"^(2-)(aq))^("Li"_color(red)(2)"S"(aq)) + overbrace("Ca"^(2+)(aq) + cancel(color(red)(2)"OH"^(-)(aq)))^("Ca"("OH")_color(red)(2)(aq)) -> "CaS"(s) + overbrace(cancel(color(red)(2)"Li"^(+)(aq)) + cancel(color(red)(2)"OH"^(-)(aq)))^(color(red)(2)"LiOH"(aq))`

As a result, the actual net ionic equation becomes:

`barul(|stackrel(" ")(" ""Ca"^(2+)(aq) + "S"^(2-)(aq) -> "CaS"(s)" ")|)`