# Question #a2062

##### 1 Answer

#### Answer:

#### Explanation:

The *dissociation equilibrium* for your weak acid

#"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A"_ ((aq))^(-)#

By definition, the **acid dissociation constant**, *pure liquids* are **not** added to the expression of the acid dissociation constant!

#K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])#

Now, you know that **for every** **mole** of **mole** of hydronium cations and **mole** of

This means that, at equilibrium, you have

#["H"_3"O"^(+)] = ["HA"] = x#

The equilibrium concentration of the weak acid will be equal to

#["HA"] = ["HA"]_ 0 - x#

This basically tells you that in order to have *consume*

Plug this into the expression of the acid dissociation constant to get--keep in mind that you have

#K_a = (x * x)/(0.50 -x)#

#K_a = (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 ~~ 0.50#

This means that you have

#8 * 10^(-4) = x^2/0.50#

Rearrange to solve for

#x = sqrt(0.50 * 8 * 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