# 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.

Jul 3, 2017

$0.12 M$

#### Explanation:

We're asked to find the molar concentration of the $\text{CH"_3"COOH}$ solution, with some known titration measurements.

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

$\text{NaOH"(aq) + "CH"_3"COOH" (aq) rarr "CH"_3"COONa" (aq) + "H"_2"O} \left(l\right)$

Let's find the number of moles of $\text{NaOH}$ 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 $\text{CH"_3"COOH}$ used up:

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

$= 0.003000$ $\text{mol CH"_3"COOH}$

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

$\text{molarity" = "mol solute"/"L soln}$

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

rounded to $2$ significant figures.