Question

Question #6cce5

Answer 1

Here's what I got.

Explanation:

Start by writing down the balanced chemical equation that describes this neutralization reaction--keep in mind that sodium hydroxide is a strong base, so you can represent it as hydroxide anions in aqueous solution

`"CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COO"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

The acid and the base react in a `1:1` mole ratio, which means that if you know the number of moles of hydroxide anions needed for a complete neutralization, you also know the number of moles of acetic acid present in the `"20-mL"` sample.

Use the molarity and the volume of the sodium hydroxide solution to determine the number of moles of hydroxide anions it contained

`15.6 color(red)(cancel(color(black)("mL"))) * "0.100 moles OH"^(-)/(10^3color(red)(cancel(color(black)("mL")))) = "0.00156 moles OH"^(-)`

SIDE NOTE The problem provides you with the formality of the sodium hydroxide solution, but that is equivalent to its molarity in this context.

So, you know that the `"20-mL"` sample contained `0.00156` moles of acetic acid. To convert this to grams, use the compound's molar mass

`0.00156 color(red)(cancel(color(black)("moles CH"_ 3"COOH"))) * "60.05 g"/(1color(red)(cancel(color(black)("mole CH"_ 3"COOH")))) = "0.0937 g"`

Now, you know that this `"20-mL"` sample came from the `"250-mL"` sample, so use its known composition to determine the mass of acetic acid present in the `"250-mL"` sample

`250 color(red)(cancel(color(black)("mL solution"))) * ("0.0937 g CH"_3"COOH")/(20 color(red)(cancel(color(black)("mL solution")))) = "1.17 g CH"_3"COOH"`

The `"250-mL"` sample was made by diluting `"25 mL"` of the initial acetic acid solution by adding `"225 mL"` of water, which implies that the initial acetic acid solution contained `"1.17 g"` of solute.

Use the density of the initial solution to find its mass

`25 color(red)(cancel(color(black)("mL solution"))) * "1.02 g"/(1color(red)(cancel(color(black)("mL solution")))) = "25.5 g"`

Since this solution contains `"1.17 g"` of acetic acid, you can say that its percent concentration by mass will be equal to

`"% m/m" = (1.17 color(red)(cancel(color(black)("g"))))/(25.5 color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)("4.6% CH"_3"COOH")))`

I'll leave the answer rounded to two sig figs, but keep in mind that only have one significant figure for the volume of the `"20-mL"` sample used in the titration.