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.