Question

What is the molarity of the acetic acid solution and what is the percentage by mass, of acetic acid in the solution? Assume the density of the solution is 1 g/mL.

Answer 1

`0.12M`

Explanation:

We're asked to find the molar concentration of the `"CH"_3"COOH"` solution, with some known titration measurements.

Let's first write the chemical equation for this neutralization reaction:

`"NaOH"(aq) + "CH"_3"COOH" (aq) rarr "CH"_3"COONa" (aq) + "H"_2"O" (l)`

Let's find the number of moles of `"NaOH"` using the given molarity and volume (which we convert to liters):

`"mol NaOH" = (0.1047"mol"/(cancel("L")))(0.02865cancel("L")) = color(red)(0.003000` `color(red)("mol NaOH"`

Using the coefficients of the equation, let's now find the relative number of moles of `"CH"_3"COOH"` used up:

`color(red)(0.003000)cancel(color(red)("mol NaOH"))((1color(white)(l)"mol CH"_3"COOH")/(1cancel("mol NaOH")))`

`= 0.003000` `"mol CH"_3"COOH"`

Lastly, using the given volume of acetic acid solution (agin converting to liters), let's find the molarity of the acetic acid solution:

`"molarity" = "mol solute"/"L soln"`

`= (0.003000color(white)(l)"mol CH"_3"COOH")/(0.025color(white)(l)"L soln") = color(blue)(0.12M`

rounded to `2` significant figures.