Question

Explain why the value of the heat of neutralization of strong acids by the strong base is constant?

Answer 1

Because it's really the same reaction. It's just clevetly disguised. See below for how to see through the disguise.

Explanation:

This is an example of net ionic equations. When there are ions in solution, it is often individual ions that react rather than the substances you originally dissolved.

Let us compare two neutralization reactions with strong acids and strong bases. Remember that strong acids and bases are completely dissociated into ions, and so are the salts they form.

Reaction 1:

`color(red)("HCl"+"NaOH"\rightarrow "NaCl"+"H"_2"O")`

Reaction 2:

`color(blue)("HNO"_3+"KOH"\rightarrow "KNO"_3+"H"_2"O")`

The strong acids and bases are dissociated into ions and so are the salts, so we should really render the equations in terms of ions:

`color(red)("H"^+(aq)+"Cl"^{-}(aq)+"Na"^+(aq)+"OH"^{-}(aq)\rightarrow "Na"^+(aq)+"Cl"^{-}(aq)+"H"_2"O")`

`color(blue)("H"^+(aq)+"NO"_3^{-}(aq)+"K"^+(aq)+"OH"^{-}(aq)\rightarrow "K"^+(aq)+"NO"_3^{-}(aq)+"H"_2"O")`

In the first reaction rendered in red, note that the sodium and chloride ions are the same on both sides; they are "spectators" that do not really react. Only the aqueous hydrogen and hydroxide ions, formed when the strong acid and base dissociated, really react. So we cancel out the spectator ions and are left with the real neutralization reaction:

`color(red)("H"^+(aq)+"OH"^{-}(aq)\rightarrow "H"_2"O")`

The same thing happens with the second reaction. The potassium and nitrate ions are spectators and canceling them out gives:

`color(blue)("H"^+(aq)+"OH"^{-}(aq)\rightarrow "H"_2"O")`

Both reactions are now the same in terms of the ions that really react; they both have the same net ionic equation. So they have the same heat of reaction.