You have a #"250.0-mL"# sample of #"1.00-M"# acetic acid. Assuming no volume change, how much #"NaOH"# must be added to make the best buffer? Calculate the #"pH"# of the best buffer.
#K_a=1.8xx10^(-5) #
Thank you!
Thank you!
1 Answer
Explanation:
In this context, the "best buffer" is a buffer that contains equal concentrations of acetic acid, the weak acid, and of acetate anions, the conjugate base.
As you know, the
#"pH" = "p"K_a + log ( (["conjugate base"])/(["weak acid"]))#
Here
#"p"K_a = - log(K_a)#
Notice that when the buffer contains equal concentrations of the weak acid and of its conjugate base, the
#log ( (["conjugate base"])/(["weak acid"])) = log(1) = 0#
Your goal here will be to figure out how much sodium hydroxide must be added to your initial solution in order to get
#["CH"_3"COOH"] = ["CH"_3"COO"^(-)]#
Acetic acid and sodium hydroxide, which I'll represent as hydroxide anions,
#color(white)(overbrace(color(black)("CH"_ 3"COOH"_ ((aq))))^(color(blue)("1 mole consumed"))) + color(white)(overbrace(color(black)("OH"_ ((aq))^(-)))^(color(blue)("1 mole consumed"))) -> color(white)(underbrace(color(black)("CH"_ 3"COO"_ ((aq))^(-)))_ (color(blue)("1 mole produced"))) + "H"_ 2"O"_ ((l))#
So in order for the reaction to produce
Your initial solution contains
#250.0 color(red)(cancel(color(black)("mL solution"))) * ("1.00 moles CH"_ 3"COOH")/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.250 moles CH"_ 3"COOH"#
Now, after the reaction is complete, you want the resulting solution to contain equal numbers of moles of acetic acid and of acetate anions.
So if you start with
In order to consume
To convert this to grams, use the molar mass of sodium hydroxide.
#0.125 color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("5.00 g")))#
The answer is rounded to three sig figs, the number of sig figs you have for the concentration of the initial solution.
After the reaction is complete, the buffer will contain--keep in mind that we assume that the volume of the buffer will not change upon the addition of the sodium hydroxide
#["CH"_3"COOH"] = "0.125 moles"/(250.0 * 10^(-3) quad "L") = "0.500 mol L"^(-1)#
and
#["CH"_3"COO"^(-)] = "0.125 moles"/(250.0 * 10^(-3) quad "L") = "0.500 mol L"^(-1)#
The
#"pH" = "p"K_a#
which is equal to
#"pH" = - log(1.8 * 10^(-5)) = color(darkgreen)(ul(color(black)(4.74)))#
The answer is rounded to two decimal places because you have two sig figs for the acid dissociation constant.