# Question e9b66

Mar 5, 2017

Here's what I got.

#### Explanation:

A generic weak acid $\text{HA}$ will dissociate in aqueous solution according to the following equilibrium

${\text{HA"_ ((aq)) rightleftharpoons "H"_ ((aq))^(+) + "A}}_{\left(a q\right)}^{-}$

Now, the acid dissociation constant, ${K}_{a}$, for this reaction can be written as

${K}_{a} = \left(\left[\text{H"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

Now, the $\text{p} {K}_{a}$ is calculated by taking the negative common log of the value of ${K}_{a}$

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{p} {k}_{a} = - \log \left({K}_{a}\right)}}}$

Plug the expression you have for the acid dissociation constant into the expression you have for the $\text{p} {K}_{a}$ 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{pH" = - log(["H}}^{+}}}}$

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

$\textcolor{w h i t e}{a}$

• $\underline{\left[\text{A"^(-)] = ["HA}\right]}$

In this case, you have

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

This will get you

$\text{pH" = "p} {K}_{a}$

The $\text{pH}$ of the solution is equal to the $\text{p} {K}_{a}$ of the acid because the buffer contains equal amounts of weak acid and conjugate base.

$\textcolor{w h i t e}{a}$

• $\underline{\left[\text{A"^(-)] > ["HA}\right]}$

In this case, you have

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

This will get you

$\text{pH" = "p"K_a + "something positive}$

which implies that

$\text{pH" > "p} {K}_{a}$

In this case, the $\text{pH}$ of the buffer is higher than the $\text{p} {K}_{a}$ of the acid because the buffer contains more conjugate base than weak acid.

$\textcolor{w h i t e}{a}$

• $\underline{\left[\text{A"^(-)] < ["HA}\right]}$

In this case, you have

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

This will get you

$\text{pH" = "p"K_a + "something negative}$

which implies that

$\text{pH" < "p} {K}_{a}$

In this case, the $\text{pH}$ of the buffer is lower than the $\text{p} {K}_{a}$ of the weak acid because the buffer contains more weak acid than conjugate base.