# Question a2062

Sep 22, 2017

4%

#### Explanation:

The dissociation equilibrium for your weak acid $\text{HA}$ can be written as follows

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

By definition, the acid dissociation constant, ${K}_{a}$, is equal to--keep in mind that the concentrations of pure liquids are not added to the expression of the acid dissociation constant!

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

Now, you know that for every $1$ mole of $\text{HA}$ that dissociates, you get $1$ mole of hydronium cations and $1$ mole of ${\text{A}}^{-}$, the conjugate base of the acid.

This means that, at equilibrium, you have

$\left[\text{H"_3"O"^(+)] = ["HA}\right] = x$

The equilibrium concentration of the weak acid will be equal to

${\left[\text{HA"] = ["HA}\right]}_{0} - x$

This basically tells you that in order to have $x$ $\text{M}$ of hydronium cations and of ${\text{A}}^{-}$ in the solution, you need to consume $x$ $\text{M}$ of the weak acid.

Plug this into the expression of the acid dissociation constant to get--keep in mind that you have ["HA"]_0 = "0.50 M"

${K}_{a} = \frac{x \cdot x}{0.50 - x}$

${K}_{a} = \frac{{x}^{2}}{0.50 - x}$

Now, because the value of the acid dissociation constant is very small compared to the initial concentration of the acid, you can use the following approximation

$0.50 - x \approx 0.50$

This means that you have

$8 \cdot {10}^{- 4} = {x}^{2} / 0.50$

Rearrange to solve for $x$

$x = \sqrt{0.50 \cdot 8 \cdot {10}^{- 4}} = 0.02$

This means that, at equilibrium, this solution contains

["H"_3"O"^(+)] = "0.02 M"

At this point, you need to check the approximation by using

(["H"_3"O"^(+)])/(["HA"]_0) * 100% < color(red)(5%)

You will have

(0.02 color(red)(cancel(color(black)("M"))))/(0.50color(red)(cancel(color(black)("M")))) * 100% = 4% < color(red)(5%)

This means that the approximation holds.

Moreover, this represents the percent dissociation of the acid, so you can say that

color(darkgreen)(ul(color(black)("% dissociation" = 4%)))#

This means that for every $100$ molecules of $\text{HA}$ present in the solution only $4$ will dissociate to produce hydronium cations and ${\text{A}}^{-}$.