Question cf0ae

Jul 6, 2016

Here's what I got.

Explanation:

Well, you can't use an acid to neutralize another acid!

Hydrochloric acid, $\text{HCl}$, and nitric acid, ${\text{HNO}}_{3}$, are both strong acids, so you need a base, either weak or strong, to neutralize them.

Let's say that you're neutralizing a $\text{6 M}$ hydrochloric acid solution with a $\text{3 M}$ sodium hydroxide solution, $\text{NaOH}$.

The balanced chemical equation that describes the reaction looks like this

${\text{HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaCl}}_{\left(a q\right)}$

The net ionic equation looks like this

${\text{H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2 "O}}_{\left(l\right)}$

The two reactants react in a $1 : 1$ mole ratio, which means that a complete neutralization requires equal numbers of moles of hydronium cations, ${\text{H"_3"O}}^{+}$, delivered to the solution by the hydrochloric acid, and of hydroxide anions, ${\text{OH}}^{-}$, delivered to the solution by the sodium hydroxide.

Now, a solution's molarity tells you the number of moles of solute present in $\text{1 L}$ of a given solution. In your case, the hydrochloric acid solution has a molarity of $\text{6 M}$, which means that $\text{1 L}$ of this solution contains $6$ moles of hydronium cations.

On the other hand, the sodium hydroxide solution has a molarity of $\text{6 M}$, which means that $\text{1 L}$ of this solution contains $6$ moles of hydroxide anions.

In other words, the hydrochloric acid solution contains twice as many moles of hydronium cations than the sodium hydroxide solution contains moles of hydroxide anions.

In order to deliver equal numbers of moles of hydronium cations and hydroxide anions to the reaction, you must use a volume of sodium hydroxide solution that is twice that of the hydrochloric acid solution.

Let's say that you have $V$ liters of hydrochloric acid solution. This sample would contain

V color(red)(cancel(color(black)("L solution"))) * ("6 moles H"_ 3"O"^(+))/(1color(red)(cancel(color(black)("L solution")))) = (6 * V)color(white)(a)"moles H"_ 3"O"^(+)

This is exactly how many moles of hydroxide anions you need for a complete neutralization. This means that the volume of the sodium hydroxide solution must be

(6 * V) color(red)(cancel(color(black)("moles OH"^(-)))) * "1 L solution"/(3 color(red)(cancel(color(black)("moles OH"^(-))))) = (2 * V)color(white)(a)"L"#

Therefore, for a volume $V$ of $\text{6 M}$ hydrochloric acid solution you need a volume $2 V$ of $\text{3 M}$ sodium hydroxide solution.

SIDE NOTE A mixture of hydrochloric acid and nitric acid in a $3 : 1$ ratio is called aqua regia because it and can dissolve gold and platinum.