Here's what I got.
Well, you can't use an acid to neutralize another acid!
Let's say that you're neutralizing a
The balanced chemical equation that describes the reaction looks like this
#"HCl"_ ((aq)) + "NaOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "NaCl"_ ((aq))#
The net ionic equation looks like this
#"H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-) -> 2"H"_ 2 "O"_ ((l))#
The two reactants react in a
Now, a solution's molarity tells you the number of moles of solute present in
On the other hand, the sodium hydroxide solution has a molarity of
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 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
SIDE NOTE A mixture of hydrochloric acid and nitric acid in a