Question #e9b66

1 Answer
Mar 5, 2017

Answer:

Here's what I got.

Explanation:

A generic weak acid #"HA"# will dissociate in aqueous solution according to the following equilibrium

#"HA"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "A"_ ((aq))^(-)#

Now, the acid dissociation constant, #K_a#, for this reaction can be written as

#K_a = (["H"^(+)] * ["A"^(-)])/(["HA"])#

Now, the #"p"K_a# is calculated by taking the negative common log of the value of #K_a#

#color(blue)(ul(color(black)("p"k_a = - log(K_a))))#

Plug the expression you have for the acid dissociation constant into the expression you have for the #"p"K_a# to get

#"p"K_a = - log ((["H"^(+)] * ["A"^(-)])/(["HA"]))#

This can be written as

#"p"K_a = - log ( ["H"^(+)] * (["A"^(-)])/(["HA"]))#

which, in turn, is equivalent to

#"p"K_a = -[ log(["H"^(+)]) + log((["A"^(-)])/(["HA"]))]#

#"p"K_a = - log(["H"^(+)]) - log((["A"^(-)])/(["HA"]))#

At this point, you should recognize that

#color(blue)(ul(color(black)("pH" = - log(["H"^(+)]))#

This means that you have

#"p"K_a = "pH" - log((["A"^(-)])/(["HA"]))#

which is equivalent to

#color(red)(ul(color(black)("pH" = "pK"_a + log((["A"^(-)])/(["HA"]))))) -># the Henderson - Hasselbalch equation

This equation can be used to calculate the pH of a buffer solution, which, as you know, contains a weak acid and its conjugate base or a weak base and its conjugate acid in comparable amounts.

Now, notice what happens when

#color(white)(a)#

  • #ul(["A"^(-)] = ["HA"])#

In this case, you have

#(["A"^(-)])/(["HA"]) = 1" "# and #" " log((["A"^(-)])/(["HA"])) = log(1) = 0#

This will get you

#"pH" = "p"K_a#

The #"pH"# of the solution is equal to the #"p"K_a# of the acid because the buffer contains equal amounts of weak acid and conjugate base.

#color(white)(a)#

  • #ul(["A"^(-)] > ["HA"])#

In this case, you have

#(["A"^(-)])/(["HA"]) > 1" "# and #" "log((["A"^(-)])/(["HA"])) > 0#

This will get you

#"pH" = "p"K_a + "something positive"#

which implies that

#"pH" > "p"K_a#

In this case, the #"pH"# of the buffer is higher than the #"p"K_a# of the acid because the buffer contains more conjugate base than weak acid.

#color(white)(a)#

  • #ul(["A"^(-)] < ["HA"])#

In this case, you have

#(["A"^(-)])/(["HA"]) < 1" "# and #" "log((["A"^(-)])/(["HA"])) < 0#

This will get you

#"pH" = "p"K_a + "something negative"#

which implies that

#"pH" < "p"K_a#

In this case, the #"pH"# of the buffer is lower than the #"p"K_a# of the weak acid because the buffer contains more weak acid than conjugate base.