Question

Question #e9b66

Answer 1

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.