# Question 1a169

Feb 1, 2016

$\text{0.4 g}$

#### Explanation:

The idea here is that the neutralization reaction that takes place between sodium hydroxide, a strong base, and hydrochloric acid, a strong acid, can be represented as the reaction between hydronium and hydroxide ions

overbrace("OH"_text((aq])^(-))^(color(blue)("coming from NaOH")) + overbrace("H"_3"O"_text((aq])^(+))^(color(purple)("coming from HCl")) -> 2"H"_2"O"_text((l])

This means that you can ignore the presence of sodium sulfate, ${\text{Na"_2"SO}}_{4}$, in solution.

You could argue that the sulfate anion, ${\text{SO}}_{4}^{2 -}$, acts as a weak base in aqueous solution, but the concentration of hydroxide anions it produces is insignificant compared to what the strong base delivers.

Since both sodium hydroxide and hydrochloric acid dissociate in a $1 : 1$ mole ratio with the hydroxide ions and the hydronium ions, respectively, you will have

["H"_3"O"^(+)] = ["HCl"]" " and " "["OH"^(-)] = ["NaOH"]

Use the molarity and volume of the hydrochloric acid solution to find how many moles of hydronium ions were needed for the neutralization of the initial solution

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

${n}_{{H}_{3} {O}^{+}} = {\text{0.5 M" * 20 * 10^(-3)"L" = "0.01 moles H"_3"O}}^{+}$

According to the balanced chemical equation for the neutralization reaction, you have equal numbers of moles of hydronium and hydroxide ions reacting.

This means that your initial solution contained

${n}_{O {H}^{-}} = {\text{0.01 moles OH}}^{-}$

Use sodium hydroxide's molar mass to find the mass of sodium hydroxide used to make this solution

0.01 color(red)(cancel(color(black)("moles NaOH"))) * "40.0 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(green)("0.4 g")#

The answer will thus be 2. $\text{0.4 g}$