Question #becd4

1 Answer
Feb 8, 2018

Answer:

#"pH" = 4.1#

Explanation:

The first thing that you need to do here is to calculate the number of moles of acetic acid present in that sample.

You know that this solution is #5%# by volume--acetic acid solutions are usually described by their volume by volume percent concentration--, which means that every #"100 mL"# of this solution contains #"5 mL"# of acetic acid.

In your case, the sample contains

#30 color(red)(cancel(color(black)("mL solution"))) * ("5 mL CH"_3"COOH")/(100color(red)(cancel(color(black)("mL solution")))) = "1.5 mL CH"_3"COOH"#

Next, use the density of acetic acid to convert the volume to grams #-># see ((here** for the density of acetic acid.

#1.5 color(red)(cancel(color(black)("mL CH"_3"COOH"))) * "1.049 g"/(1color(red)(cancel(color(black)("mL CH"_3"COOH")))) = "1.5735 g"#

To find the number of moles of acetic acid present in the sample, use the compound's molar mass.

#1.5735 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "0.0262 moles CH"_3"COOH"#

To find the number of moles of sodium hydroxide, use the volume of the solution and its molarity.

#10 color(red)(cancel(color(black)("mL solution"))) * "0.5 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0050 moles NaOH"#

Now, acetic acid and sodium hydroxide react in a #1:1# mole ratio to produce acetate anions and water.

The balanced chemical equation that describes this neutralization reaction looks like this--I won't add the sodium cations because they are not relevant here.

#"CH"_ 3"COOH"_ ((aq)) + "OH"_ ((aq))^(-) -> "CH"_ 3"COOH"_ ((aq))^(-) + "H"_ 2"O"_ ((l))#

Since you're mixing #0.0050# moles of sodium hydroxide and #0.0262# moles of acetic acid, you can say that acetic acid wil be in excess here, i.e. sodium hydroxide is the limiting reagent.

This means that all the moles of hydroxide anions delivered to the reaction by the sodium hydroxide will be consumed. Since each mole of sodium hydroxide consumes #1# mole of acetic acid and produces #1# mole of acetate anions, you can say that after the reaction is complete, the solution will contain

#"0.0050 moles " - " 0.0050 moles" = "0 moles OH"^(-)#

All the moles of hydroxide anions are consumed by the reaction.

#"0.0262 moles " - " 0.0050 moles" = "0.0212 moles CH"_3"COOH"#

The reaction consumes the same number of moles of acetic acid and of hydroxide anions, so the number of moles of acetic acid will decrease.

#"0 moles " + " 0.0050 moles" = "0.0050 moles CH"_3"COO"^(-)#

The reaction produces the same number of moles of acetate anions as the number of moles of the two reactants it consumes.

The total volume of the resulting solution will be

#"30 mL + 10 mL = 40 mL"#

The concentrations of acetic acid and of acetate anions in the resulting solution will be--remebemr to use the volume of the solution in liters!

#["CH"_3"COOH"] = "0.0212 moles"/(40 * 10^(-3) quad "L") = "0.53 M"#

#["CH"_3"COO"^(-)] = "0.0050 moles"/(40 * 10^(-3) quad "L") = "0.125 M"#

Now, the resulting solution contains acetic acid, a weak acid, and acetate anions, its conjugate base, which means that you're dealing with a buffer solution.

The #"pH"# of a weak acid-conjugate base buffer can be calculated using the Henderson - Hasselbalch equation

#"pH" = "p"K_a = log( (["conjugate gbase"])/(["weak acid"]))#

Here

#"p"K_a = - log(K_a)#

with #K_a# being the acid dissociation constant of the weak acid.

Plug in your values to find

#"pH" = - log(1.8 * 10^(-5)) + log ((0.125 color(red)(cancel(color(black)("M"))))/(0.53color(red)(cancel(color(black)("M")))))#

#color(darkgreen)(ul(color(black)("pH" = 4.1)))#

The answer is rounded to one decimal place, the number of sig figs you have for your values.

Notice that you have

#"p"K_a = - log(1.8 * 10^(-5)) = 4.74#

The fact that the solution contains more weak acid than conjugate base implies that its #"pH"# will be lower than the #"p"K_a# of the weak acid.