Question #ef831

1 Answer
Dec 26, 2017

Answer:

#"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.