Question

Question #20bfa

Answer 1

`"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, `"NaOH"`, your `"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 `"20.0 g"` of sodium hydroxide for every `"100.0 g"` of solution. In your case, the `"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

`color(red)(2)"OH"_text((aq])^(-) + "H"_2"C"_2"O"_text(4(aq]) -> "C"_2"O"_text(4(aq])^(2-) + 2"H"_2"O"_text((l])`

Notice that you need `color(red)(2)` moles of hydroxide anions to neutralize `1` mole of oxalic acid. Since your `"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")`