Question #b2a5b
1 Answer
None of the above?
Explanation:
For starters, you should know that acetic acid has
#"p"K_a = 4.75 -># source
Now, the
#"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))#
Right from the start, the fact that you have
#"pH " > " p"K_a#
tells you that the buffer has a higher concentration of acetate anions, the conjugate base of acetic acid, than of acetic acid.
This implies that
#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) > 1#
This means that options (1) and (4) are eliminated.
To find the actual ratio that exists between the conjugate base and the weak acid, rearrange the Henderson - Hasselbalch equation
#6 = 4.75 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#
as
#log ((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 6 - 4.75#
This is equivalent to
#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^1.26#
which gives you
#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 17.8#
This means that in order to have an acetic acid-acetate anions buffer of
If you want, you can take options (2) and (3) and use the given ratios to find the
You will have
#(2): " "(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 10#
#"pH" = 4.75 + log(10) = 5.75#
#(3): " "(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 100#
#"pH" = 4.75 + log(100) = 6.75#
As you can see, a