Question

Question #a2062

Answer 1

`4%`

Explanation:

The dissociation equilibrium for your weak acid `"HA"` can be written as follows

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

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

Now, you know that for every `1` mole of `"HA"` that dissociates, you get `1` mole of hydronium cations and `1` mole of `"A"^(-)`, the conjugate base of the acid.

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 `x` `"M"` of hydronium cations and of `"A"^(-)` in the solution, you need to consume `x` `"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 = (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`

`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 `100` molecules of `"HA"` present in the solution only `4` will dissociate to produce hydronium cations and `"A"^(-)`.