# Question a46f8

Jan 7, 2016

For part (a) $\text{50.0 mL}$

#### Explanation:

I'll show you how to solve part (a), and leave part (b) to you as practice.

You're dealing with two neutralization reactions, that take place between a strong acid and a strong base.

In both cases, a complete neutralization requires the consumption of both reactants. However, the products of the reaction will differ.

When a strong base and a strong acid undergo complete neutralization, the resulting solution will contain water, $\text{H"_2"O}$, and an aqueous salt.

As you know, strong acids and bases dissociate completely in aqueous solution to form ions

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

${\text{NaOH"_text((aq]) -> "Na"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

When these two solutions are mixed, the reaction vessel will contain - hydronium cations, ${\text{H"_3"O}}^{+}$, are represented here as hydrogen cations, ${\text{H}}^{+}$

At this point, the hydronium cations and the hydroxide anions will neutralize each other and form water

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

After the reaction is complete, your reaction vessel will contain

Notice that the two reactants react in a $1 : 1$ mole ratio, which mean that in order for the neutralization to be complete, you need equal numbers of moles of hydrochloric acid and sodium hydroxide.

Use the molarity and volume of the sodium hydroxide solution to determine how many moles it contains

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{N a O H} = \text{0.250 M" * 20.0 * 10^(-3)"L" = "0.00500 moles NaOH}$

Since both the strong acid and the strong base dissociate in a $1 : 1$ mole ratio with the hydronium cations and the hydroxide anions, you can say that since the solution contains $0.00500$ moles of ${\text{OH}}^{-}$, it must also contain $0.00500$ moles of ${\text{H"_3"O}}^{+}$.

This means that you will need a volume of

$\textcolor{b l u e}{c = \frac{n}{V} \implies V = \frac{n}{c}}$

V_(HCl) = (0.00500 color(red)(cancel(color(black)("moles"))))/(0.100color(red)(cancel(color(black)("moles")))/"L") = "0.0500 L"#

Expressed in milliliters, the answer will be

${V}_{H C l} = \textcolor{g r e e n}{\text{50.0 mL}}$

You can take the exact same approach for part (b). Start with the balanced chemical equation

${\text{HClO"_text(4(aq]) + "KOH"_text((aq]) -> "H"_2"O"_text((l]) + "KClO}}_{\textrm{4 \left(a q\right]}}$

Once again, use the $1 : 1$ mole ratio and the molarity and volume of the perchloric acid, ${\text{HClO}}_{4}$, solution to find the volume of potassium hydroxide needed here.