Question #7f273
1 Answer
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,
#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
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 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
Now, let's say that
If we take
#["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
#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 = 28.52 - 52.48 * n#
#n * (52.48 + 1) = 28.52 implies n= 28.52/53.48 = 0.5333#
Therefore, you must add
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
#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!