Question

Question #97ac9

Answer 1

Here's what I got.

Explanation:

!! VERY LONG ANSWER !!

The first thing you need to do here is to look up the `"p"K_a` of acetic acid, which you'll find listed as

`"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, when you mix acetic acid, a weak acid, and sodium acetate, the salt of the acetate anion, which is the conjugate base of acetic acid, you will get a buffer solution.

Right from the start, the fact that you have

`"pH" > "p"K_a`

should tell you that the buffer contains more conjugate base than weak acid.

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

`color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"])))))`

In your case, you will have

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

Use the pH of the solution to calculate the ratio between the concentration of the conjugate base and the concentration of the weak acid in the buffer

`4.85 = 4.75 + log ((["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))`

`log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 4.85 - 4.75`

This will be equivalent to

`10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.10`

which will get you

`(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.26`

Now, let's say that `V_1` represents the volume of the acetic acid solution and `V_2` represents the volume of the sodium acetate.

The total volume of the buffer will be equal to `V_1 + V_2`, which means that you have

`" "["CH"_ 3"COOH"] = n_"acetic acid"/(V_1 + V_2)" "` and `" "["CH"_ 3"COO"^(-)] = n_"acetate"/(V_1 + V_2)`

Consequently, you can say that

`(["CH"_ 3"COO"^(-)])/(["CH"_ 3"COOH"]) = (n_"acetate"/color(red)(cancel(color(black)(V_1 + V_2))))/(n_"acetic acid"/color(red)(cancel(color(black)((V_1 + V_2))))) = 1.26`

This is equivalent to

`n_"acetate" = 1.26 * n_"acetic acid"" " " "color(darkorange)("(*)")`

Now, use the concentrations of the two solutions to write

`n_"acetate" = "0.15 mol" color(red)(cancel(color(black)("L"^(-1)))) * V_2 color(red)(cancel(color(black)("L")))`

`n_"acetate" = (0.15 * V_2)color(white)(.)"moles CH"_3"COO"^(-)`

Do the same for the acetic acid

`n_"acetic acid" = "0.10 mol" color(red)(cancel(color(black)("L"^(-1)))) * V_2 color(red)(cancel(color(black)("L")))`

`n_"acetic acid" = (0.10 * V_2)color(white)(.)"moles CH"_3"COOH"`

You can now rewrite equation `color(darkorange)("(*)")` as

`(0.15 * V_2) color(red)(cancel(color(black)("moles"))) = 1.26 * (0.10 * V_1)color(red)(cancel(color(black)("moles")))`

which will get you

`0.15 * V_2 = 0.126 * V_1`

You know that

`V_1 + V_2 = "0.020 L"`

which means that you now have a system of two equations with two unknowns.

`V_1 = "0.020 L" - V_2`

Plug this into the first equation to get

`0.15 * ("0.020 L" - V_2) = 0.126 * V_2`

`"0.003 L" - 0.15 * V_2 = 0.126 * V_2`

You will end up with

`V_2 = "0.003 L"/(0.126 + 0.15) = "0.0109 L"`

This means that

`V_1 = "0.020 L" - "0.0109 L" = "0.0091 L"`

Therefore, you can say that in order to make `"20 mL"` of an acetic acid/sodium acetate buffer that has a pH equal to `4.85`, you must mix

`"9.0 mL " -> " 0.10 M"` acetic acid solution

`"11 mL " -> " 0.15 M"` sodium acetate solution

I'll leave the answers rounded to two sig figs, but keep in mind that you only have one significant figure for the volume of the buffer.