# Question 20bfa

Nov 19, 2015

$\text{1.75 L}$

#### Explanation:

Before focusing on the neutralization reaction that takes place here, your task is to determine exactly how many moles of sodium hydroxide, $\text{NaOH}$, your $\text{1.00-L}$ sample contains.

To do that, use the solution's density to determine its mass, then its percent concentration by mass to determine exactly how much sodium hydroxide it contains.

1.00color(red)(cancel(color(black)("L"))) * (1000color(red)(cancel(color(black)("mL"))))/(1color(red)(cancel(color(black)("L")))) * "1.40 g"/(1color(red)(cancel(color(black)("mL")))) = "1400 g"

Now, a 20%"w/w" solution means that you have $\text{20.0 g}$ of sodium hydroxide for every $\text{100.0 g}$ of solution. In your case, the $\text{1400-g}$ sample will contain

1400color(red)(cancel(color(black)("g solution"))) * "20.0 g NaOH"/(100color(red)(cancel(color(black)("g solution")))) = "280 g NaOH"

Finally, use sodium hydroxide's molar mass to determine how many moles you have in the sample

280color(red)(cancel(color(black)("g NaOH"))) * "1 mole NaOH"/(40.0color(red)(cancel(color(black)("g")))) = "7.00 moles NaOH"

Now you can focus on the neutralization reaction. The balanced chemical equation will looks like this - I'll use the net ionic equation

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

Notice that you need $\textcolor{red}{2}$ moles of hydroxide anions to neutralize $1$ mole of oxalic acid. Since your $\text{1.00-L}$ sample contains $7$ moles of sodium hydroxide, which as you know dissociates to produce sodium cations and hydroxide anions in a $1 : 1 : 1$ mole ratio, you will need

7color(red)(cancel(color(black)("moles OH"^(-)))) * ("1 mole H"_2"C"_2"O"_4)/(color(red)(2)color(red)(cancel(color(black)("moles OH"^(-))))) = "3.5 moles H"_2"C"_2"O"_4

All you have to do now is use the molarity of the oxalic acid solution to figure out what volume would contain this many moles

color(blue)(c = n/V implies V = n/c

V_"oxalic" = (3.5color(red)(cancel(color(black)("moles"))))/(2.0color(red)(cancel(color(black)("moles")))/"L") = color(green)("1.75 L")#