What is the net ionic equation for the reaction between HF (a weak acid) and NaOH? | Socratic

Or you want a quick look:

Hydrofluoric acid, #"HF"#, a weak acid, will react with sodium hydroxide, #"NaOH"#, a strong base, to produce aqueous sodium fluoride, #"NaF"#, and water.

#"HF"_ ((aq)) + "NaOH"_ ((aq)) -> "NaF"_ ((aq)) + "H"_ 2"O"_ ((l))#


Now, the important thing to remember here is that hydrofluoric acid is a weak acid, which implies that it does not dissociate completely in aqueous solution to form hydrogen cations, #"H"^(+)#, usually referred to as hydronium cations, #"H"_3"O"^(+)#, and fluoride anions, #"F"^(-)#.

Instead, an equilibrium will be established between the undissociated acid and the dissociated ions

#"HF"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "F"_ ((aq))^(-)#

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Sodium hydroxide, on the other hand, will dissociate completely in aqueous solution to produce sodium cations, #"Na"^(+)#, and hydroxide anions, #"OH"^(-)#

#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#

Here's what happens at this point. The hydroxide anions and the hydrogen cations will neutralize each other to produce water.

#"H"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "H"_ 2"O"_ ((l))#

This will consume hydrogen cations and cause the dissociation equilibrium of hydrofluoric acid to shift to the right #-># more of the molecules of acid will dissociate.

This will go on until all the molecules of acid have dissociated and all the hydrogen cations have reacted with the hydroxide anions #-># a complete neutralization took place.

Now, the complete ionic equation looks like this -- keep in mind that sodium fluoride is soluble in aqueous solution

#"HF"_ ((aq)) + "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> "Na"_ ((aq))^(+) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

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Notice that because hydrofluoric acid does not dissociate completely, you must represent it in molecular form.

Eliminate the spectator ions, which in this case are the sodium cations

#"HF"_ ((aq)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) -> color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

to get the net ionic equation that describes this neutralization reaction

#color(green)(|bar(ul(color(white)(a/a)color(black)("HF"_ ((aq)) + "OH"_ ((aq))^(-) -> "F"_ ((aq))^(-) + "H"_ 2"O"_ ((l)))color(white)(a/a)|)))#

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