Step 1. Write the balanced equation
"CH"_3"COOH" + "NaOH" → "CH"_3"COONa" + "H"_2"O"CH3COOH+NaOH→CH3COONa+H2O
Step 2. Calculate the moles of NaOH
"Moles of NaOH" = "0.030 84" color(red)(cancel(color(black)("dm"^3color(white)(l) "NaOH"))) × "0.128 mol NaOH"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"NaOH")))) = "0.003 948 mol NaOH"
Step 3. Calculate the moles of "CH"_3"COOH"
"Moles of CH"_3"COOH" = "0.003 948" color(red)(cancel(color(black)("mol NaOH"))) × ("1 mol CH"_3"COOH")/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.003 948 mol CH"_3"COOH"
Step 4. Calculate the mass of the "CH"_3"COOH"
"Mass of CH"_3"COOH" = "0.003 948" color(red)(cancel(color(black)("mol CH"_3"COOH"))) × ("60.05 g CH"_3"COOH")/(1 color(red)(cancel(color(black)("mol CH"_3"COOH")))) = "0.2370 g CH"_3"COOH"
Step 5. Calculate the mass percent of acetic acid.
"Mass percent" = "mass of acetic acid"/"mass of vinegar" × 100 % = (0.2370 color(red)(cancel(color(black)("g"))))/(5.441 color(red)(cancel(color(black)("g")))) × 100 %
= 4.36 %
