Question

Question #ef831

Answer 1

`"p"K_a - 0.30`

Explanation:

You're dealing with a strong acid-conjugate base buffer here, so right from the start, you know that you can use the Henderson - Hasselbalch equation to find its `"pH"`.

`"pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))`

Here

`"p"K_a = - log(K_a)`

with `K_a` being the acid dissociation constant of the weak acid.

In your case, acetic acid, `"CH"_3"COOH"`, is the weak acid and the acetate anion, `"CH"_3"COO"^(-)`, is its conjugate base. The acetate anions are delivered to the solution by the soluble sodium acetate in a `1:1` mole ratio, so you know that you have

`["CH"_3"COO"^(-)] = "0.125 M"`

Notice that your solution contains more weak acid than conjugate base. This tells you that the `"pH"` of the solution will be lower than the `"p"K_a` of the weak acid.

This is the case becase at equal concentrations of weak acid and conjugate base, the `"pH"` of the solution is actually equal to the `"p"K_a` of the weak acid. So if you have more acid than conjugate, the `"pH"` of the solution will fall below* the `"p"K_a` of the weak acid.

Plug in your values to find

`"pH"= "p"K_a + log (( 0.125 color(red)(cancel(color(black)("M"))))/(0.25color(red)(cancel(color(black)("M")))))`

`"pH" = "p"K_a + log(1/2)`

This is equivalent to

`"pH" = "p"K_a + overbrace(log(1))^(color(blue)(=0)) - log(2)`

`"pH" = "p"K_a - log(2)`

`"pH" = "p"K_a - 0.30`

Now all you have to do is to use the `"p"K_a` of acetic acid, which you can find listed here, to find the `"pH"` of the solution. You should round the answer to two decimal places, the number of sig figs you have for the concentration of acetic acid.

Notice that you have

`"pH" = "p"K_a - 0.30 " " < " " "p"K_a`

which is consistent with the fact that the buffer contains more acetic acid than acetate anions.