Question #12b86

1 Answer
Apr 16, 2017

Answer:

Here's what I got.

Explanation:

The Henderson - Hasselbalch equation will be your tool of choice here

#"pH" = "p"K_a + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

Your buffer contains acetic acid, #"CH"_3"COOH"#, a weak acid, and the acetate anion, #"CH"_3"COO"^(-)#, its conjugate base, delivered to the solution by the sodium acetate.

Look up the #"p"K_a# of acetic acid

#"p"K_a = 4.75#

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Now, notice that your initial solution contains more weak acid than conjugate base. This should tell you that the #"pH"# of the buffer will be lower than the #"p"K_a# of the acid.

Looking at the initial concentrations of weak acid and conjugate base, you can say that the initial #"pH"# will be slightly lower than the #"p"K_a# of the acid.

You will have

#"pH" = 4.75 + log( (0.115 color(red)(cancel(color(black)("M"))))/(0.125color(red)(cancel(color(black)("M"))))) #

#color(darkgreen)(ul(color(black)("pH" = 4.71)))#

Now, you are adding #0.020# moles of hydrochloric acid, a strong acid, to the buffer. Right from the start, you should expect the #"pH"# of the resulting solution to be lower than the #"pH"# of the initial solution.

This is the case because the buffer will convert the strong acid to a weak acid by consuming acetate anions and producing acetic acid. So the #"pH"# will drop, but not by the same factor as you would see if you added hydrochloric acid to pure water.

So, you have

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

The hydronium cations react with the acetate anions in a #1:1# mole ratio and produce acetic acid in a #1:1# mole ratio.

Calculate the number of moles of acetic acid and of acetate anions present in the initial solution

#450.0 color(red)(cancel(color(black)("mL"))) * ("0.125 moles CH"_3"COOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.05625 moles"# #"CH"_3"COOH"#

#450.0 color(red)(cancel(color(black)("mL"))) * ("0.115 moles CH"-3"COO"^(-))/(10^3color(red)(cancel(color(black)("mL")))) = "0.05175 moles"# #"CH"_3"COO"^(-)#

Now, notice that you have fewer moles of hydronium cations than moles of acetate anions. This tells you that the former will act as a limiting reagent. The hydronium cations will be completely consumed before all the moles of acetate anions will get the chance to react.

The resulting solution will contain

#n_ ("H"_ 3"O"^(+)) = "0 moles" -># completely consumed by the reaction

#n_ ("CH"_ 3"COO"^(-)) = "0.05175 moles" - "0.020 moles" = "0.03175 moles"# #"CH"_3"COO"^(-)#

#n_ ("CH"_3"COOH") = "0.05625 moles" + "0.020 moles" = "0.07625 moles"# #"CH"_3"COOH"#

Assuming that the volume of the buffer does not change upon the addition of the strong acid, you can say that the new concentrations of acetic acid and of acetate anions will be

#["CH"_3"COOH"] = "0.07625 moles"/(450.0 * 10^(-3)"L") = "0.1694 M"#

#["CH"_3"COO"^(-)] = "0.03175 moles"/(450.0 * 10^(-3)"L") = "0.07056 M"#

As you can see, the strong acid decreased the concentration of the conjugate base and increased the concentration of the weak acid #-># the #"pH"# of the resulting solution will be #< 4.71#.

More specifically, you will have

#"pH" = 4.75 + log((0.07056 color(red)(cancel(color(black)("M"))))/(0.1694color(red)(cancel(color(black)("M")))))#

#color(darkgreen)(ul(color(black)("pH" = 4.37)))#

The answers are rounded to two decimal places, the number of sig figs you have for the number of moles of hydrochloric acid added to the initial buffer.