Question #3292c

Oct 10, 2015

Because you're going from an acidic solution and a basic solution, to a neutral solution.

Explanation:

A neutralization reaction takes place when an acid reacts with a base to produce a salt, which will exist in solution as cations and anions, and water.

Pure water, $\text{H"_2"O}$, undergoes self-ionization to produce hydronium ions, ${\text{H"_3"O}}^{+}$, and hydroxide ions, ${\text{OH}}^{-}$. At room temperature, the equilibrium reaction that takes place in water produces ${10}^{- 7} \text{M}$ of hydronium ions and ${10}^{- 7} \text{M}$ of hydroxide ions

$2 {\text{H"_2"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

So, an acidic solution is characterized by a concentration of hydronium ions that is greater than ${10}^{- 7} \text{M}$. This is equivalent to saying that an acidic solution has a pH that is smaller than $7$.

A basic solution is characterized by a concentration of hydroxide ions that is greater than ${10}^{- 7}$, which is equivalent to a concentration of hydronium ions that is smaller than ${10}^{- 7}$ and a pH that is bigger than $7$.

When you react sulfuric acid, ${\text{H"_2"SO}}_{4}$, which makes for a very acidic solution, with sodium hydroxide, $\text{NaOH}$, which makes for a very basic solution, in the right proportions, you will get a reaction between the hydronium and hydroxide ions

${\text{H"_2"SO"_text(4(aq]) + 2"NaOH"_text((aq]) -> "Na"_2"SO"_text(4(aq]) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

At an ionic level, the reaction looks like this

$2 {\text{H"_text((aq])^(+) + "SO"_text(4(aq])^(2-) + 2"Na"_text((aq])^(+) + 2"OH"_text((aq])^(-) -> 2"Na"_text((aq]) + "SO"_text(4(aq])^(2-) + 2"H"_2"O}}_{\textrm{\left(l\right]}}$

The ions that are present on both sides of the equation do not take part in the reaction, so you can say that

${\text{H"_text((aq])^(+) + "OH"_text((aq])^(-) -> "H"_2"O}}_{\textrm{\left(l\right]}}$

This shows you that the excess hydronium ions that are coming from the sulfuric acid will react with the excess hydroxide ions that are coming from the base to prodcue water.

In essence, the hydronium ions and the hydroxide ions are neutralizing each other, hence the name neutralization reaction.

The pH of the solution will once again be equivalent to the pH of water, characterized by a concentration of hydronium ions equal to ${10}^{- 7} \text{M}$.

This reaction is very exothermic, which is another way of saying that heat will be released.

Oct 10, 2015

Keep in mind the resulting cations and anions as well.

Explanation:

The cations of strong bases and the anions of strong acids are neutral in aqueous solution,

This means that if you take sulfuric acid's ionization to be complete, i.e. it donates both protons, then the sulfate anion, ${\text{SO}}_{4}^{2 -}$, will be neutral.

Likewise, sodium hydroxide, being a strong base, will produce sodium cations, ${\text{Na}}^{+}$, which are neutral.

It is worth noting that sulfuric acid ionizes in two steps, and that only only its first ionization is considered to be characteristic of a strong acid.

However, its conjugate base, the bisulfate anion, ${\text{HSO}}_{4}^{-}$, is a relatively strong weak acid in its own right, which is why, in practice, both ionizations are considered to be strong.

If you were to consider the second ionization as being that of a weak acid, then the resulting anion, ${\text{SO}}_{4}^{2 -}$, would be basic, but not by any significant amount.

That is why, for all intended purposes, the sulfate anion is considered neutral, which in turn adds to the fact that sodium sulfate, ${\text{Na"_2"SO}}_{4}$, is a neutral salt.