Question #7f273

1 Answer
Apr 7, 2017

Answer:

#"249 mL KOH"#

Explanation:

The idea here is that you can use the Henderson - Hasselbalch equation to figure out the ratio that must exist between the concentration of conjugate base and the concentration of weak acid in this buffer.

In your case, acetic acid, #"CH"_3"COOH"#, is the weak acid and the acetate anion, #"CH"_3"COO"^(-)#, is its conjugate base.

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

Now, acetic acid has

#"p"K_a = 4.76#

The buffer must have a pH of #6.48#, significantly higher than the #"p"K_a# of the acid. This tells you that the buffer will contain a lot more conjugate base than weak acid.

So right from the start, you should expect to see

#["CH"_3"COO"^(-)] > ["CH"_3"COOH"]#

which implies

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) > 1#

You start with #"875 mL"# of a #"0.621 M"# acetic acid solution. The number of moles of acetic acid present in the initial solution will be equal to

#875 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.621 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L")))) #

# = "0.5434 moles CH"_3"COOH"#

When you add potassium hydroxide, a strong base, the hydroxide anions delivered to the solution by the salt will react with the acid to form acetate anions.

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

Notice that it takes #1# mole of hydroxide anions, i.e. #1# mole of potassium hydroxide, to react with #1# mole of acetic acid in order to produce #1# mole of acetate anions. Keep this in mind.

Now, let's say that #x# represents the number of moles of acetic acid and #y# represents the number of moles of acetate anions present in the target solution.

If we take #V# to be the volume of potassium hydroxide solution, expressed in milliliters, added to the initial acetic acid solution, we can say that, at equilibrium, the buffer will contain

#["CH"_3"COOH"] = "x moles"/( (875 + V) * 10^(-3)"L") = ((10 ^ 3 * x)/(875 + V))color(white)(.)"M"#

#["CH"_3"COO"^(-)] = "y moles"/((875 + V) * 10^(-3)"L") = ( (10^3 * y)/(875 + V))color(white)(.)"M"#

This means that the ratio that must exist between the conjugate base and the weak acid in the buffer will be equal to

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = ( ( (color(red)(cancel(color(black)(10^3))) * y)/color(red)(cancel(color(black)(875 + V))))color(red)(cancel(color(black)("M"))))/( ((color(red)(cancel(color(black)(10^3))) * x)/color(red)(cancel(color(black)(875 + V))))color(red)(cancel(color(black)("M")))) = y/x#

According to the Henderson - Hasselbalch equation, you will have

#6.48 = 4.76 + log(y/x)#

This is equivalent to

#log(y/x) = 1.72#

#10^log(y/x) = 10^1.72#

which gets you

#y/x = 52.48#

Let's get back to the balanced chemical equation that describes the neutralization reaction. If we use #n# as the number of moles of hydroxide anions added to the initial solution, you can say that the buffer will contain

#x = 0.5434 - n -># the number of moles of acetic acid

#y = 0 + n -># the number of moles of acetate anions

You can thus say that

#y/x = n/(0.5434 - n) = 52.48#

Solve this for #n# to get

#n = 28.52 - 52.48 * n#

#n * (52.48 + 1) = 28.52 implies n= 28.52/53.48 = 0.5333#

Therefore, you must add #0.5333# moles of hydroxide anions to the initial acetic acid solution.

Use its molarity to figure out the volume that would contain this number of moles of potassium hydroxide

#0.5333 color(red)(cancel(color(black)("moles KOH"))) * "1 L solution"/(2.14 color(red)(cancel(color(black)("moles KOH")))) = "0.2492 L solution"#

Expressed in milliliters and rounded to three sig figs, the answer will be

#color(darkgreen)(ul(color(black)("volume KOH = 249 mL")))#

You can double-check the result by calculating the concentrations of the two species in the buffer

#["CH"_3"COOH"] = ((0.5434 - 0.5333)"moles")/((875 + 249) * 10^(-3)"L") = "0.008986 M"#

#["CH"_3"COO"^(-)] = "0.5333 moles"/((875 + 249) * 10^(-3)"L") = "0.4745 M"#

As you cans ee, the buffer does contain significantly more conjugate base than weak acid, just as we predicted by looking at the #"pH"# of the buffer and at the #"p"K_a# of the acid.

#6.48 = 4.76 + log ((0.4745 color(red)(cancel(color(black)("M"))))/(0.008986color(red)(cancel(color(black)("M")))))#

If we round the calculation, we will get

#6.48 = 4.76 + 1.72 " "color(darkgreen)(sqrt())#

or

#6.48 ~~ 4.76 + 1.7227 " " -> # close enough!