Question #b2a5b

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.