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.

1 Answer
Jul 3, 2017

Answer:

#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.