Question #b2a5b

1 Answer
Oct 13, 2017

Answer:

None of the above?

Explanation:

For starters, you should know that acetic acid has

#"p"K_a = 4.75 -># source

Now, the #"pH"# of a weak acid-conjugate base buffer can be calculated using the Henderson - Hasselbalch equation, which looks like this

#"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 #"pH" = 6#, the solution must contain #17.8# times more conjugate base than weak acid.

If you want, you can take options (2) and (3) and use the given ratios to find the #"ph"# of the buffer,

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 #"pH"# of #6# is much closer to a #10:1# conjugate base/weak acid ratio.