# Question #e9b66

##### 1 Answer

#### Answer:

Here's what I got.

#### Explanation:

A generic *weak acid*

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

Now, the *acid dissociation constant*,

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

Now, the **negative common log** of the value of

#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 = - 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"]))))) -># theHenderson - 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

#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

#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 **higher** than the

#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 **lower** than the