# Question #97ac9

##### 1 Answer

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 = 4.75#

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

The **total volume** of the buffer will be equal to

#" "["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

#(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

#"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.