# Question 36d09

Oct 1, 2017

Here's what I got.

#### Explanation:

I think that you also need to include the initial concentration of the acid in the expression of the acid dissociation constant.

For your generic weak acid ionization equilibrium

${\text{HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A}}_{\left(a q\right)}^{-}$

the acid dissociation constant, ${K}_{a}$, is defined as

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["A"^(-)])/(["HA}\right]\right)$

As you know, the $\text{pH}$ of a solution is given by

"pH" = - log(["H"_3"O"^(+)])#

This implies that the equilibrium concentration of hydronium cations can be written as

$\left[\text{H"_3"O"^(+)] = 10^(-"pH}\right)$

Plug this into the expression for ${K}_{a}$ to get

${K}_{a} = \left({10}^{- \text{pH") * ["A"^(-)])/(["HA}}\right)$

Now, notice that for every mole of $\text{HA}$ that ionizes in aqueous solution, you get $1$ mole of hydronium cations and $1$ mole of ${\text{A}}^{-}$.

This means that if you take ${\left[\text{HA}\right]}_{0}$ to be the initial concentration of the weak acid, you can say that the equilibrium concentration of the weak acid, $\left[\text{HA}\right]$, will be equal to

$\left[{\text{HA"] = ["HA"]_0 - ["A}}^{-}\right]$

This means that in order for the ionization to produce $\left[{\text{A}}^{-}\right]$, the initial concentration of the weak acid must decrease by $\left[{\text{A}}^{-}\right]$.

Plug this into the expression for ${K}_{a}$ to get

${K}_{a} = \left({10}^{- {\text{pH") * ["A"^(-)])/(["HA"]_0 - ["A}}^{-}}\right)$

At this point, if you know the initial concentration of the acid, you can plug that in and get an expression for ${K}_{a}$ in terms of the $\text{pH}$ of the solution and the equilibrium concentration of ${\text{A}}^{-}$.

For example, if you start with $\text{0.1 M}$ of $\text{HA}$, you will have

${K}_{a} = \left({10}^{- {\text{pH") * ["A"^(-)])/(0.1 - ["A}}^{-}}\right)$