Question #e749d

Your starting point here will be the balanced chemical equation for this neutralization reaction

#"NaOH"_text((aq]) + "CH"_3"COOH"_text((aq]) -> "CH"_3"COONa"_text((aq]) + "H"_2"O"_text((l])#

Notice that you have a #1:1# mole ratio between sodium hydroxide, #"NaOH"#, and acetic acid, #"CH"_3"COOH"#.

This tells you that the reaction will consume equal numbers of moles of strong base and weak acid.

Now, you know the molarity and volume of the sodium hydroxide solution, which means that you can determine how many moles of sodium hydroxide were needed to neutralize the acid.

#color(blue)(c = n/V implies n = c * V)#

#n_(NaOH) = "0.1025 M" * 33.13 * 10^(-3)"L" = "0.0033958 moles NaOH"#

Since we've already established that a complete neutralization would require equal numbers of moles of both reactants, you can say that

#n_(CH_3COOH) = n_(NaOH) = "0.0033958 moles"#

You know that the mass of vinegar used in this reaction is equal to #"5.111 g"#, and that its density is equal to #"1.018 g/mL"#.

As you know, a substance's density tells you the mass of that substance per unit of volume. In this case, every #"1 mL"# of vinegar will have a mass of #"1.018 g"#. This means that your sample will have a volume of

#5.111 color(red)(cancel(color(black)("g"))) * "1 mL"/(1.018 color(red)(cancel(color(black)("g")))) = "5.0206 mL"#

So, the concentration of acetic acid in moles per liter will be equal to

#["CH"_3"COOH"] = "0.0033958 moles"/(5.0206 * 10^(-3)"L") = color(green)("0.6764 M")#

To get the concentration in grams per liter, use acetic acid's molar mass

#0.0033958 color(red)(cancel(color(black)("moles"))) * "60.052 g"/(1color(red)(cancel(color(black)("mole")))) = "0.20392 g"#

Therefore, you have

#["CH"_3"COOH"] = "0.20392 g"/(5.0206 * 10^(-3)"L") = color(green)("40.62 g/L")#