Question

Question #36d09

Answer 1

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

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

the acid dissociation constant, `K_a`, is defined as

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

As you know, the `"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

`["H"_3"O"^(+)] = 10^(-"pH")`

Plug this into the expression for `K_a` to get

`K_a = (10^(-"pH") * ["A"^(-)])/(["HA"])`

Now, notice that for every mole of `"HA"` that ionizes in aqueous solution, you get `1` mole of hydronium cations and `1` mole of `"A"^(-)`.

This means that if you take `["HA"]_0` to be the initial concentration of the weak acid, you can say that the equilibrium concentration of the weak acid, `["HA"]`, will be equal to

`["HA"] = ["HA"]_0 - ["A"^(-)]`

This means that in order for the ionization to produce `["A"^(-)]`, the initial concentration of the weak acid must decrease by `["A"^(-)]`.

Plug this into the expression for `K_a` to get

`K_a = (10^(-"pH") * ["A"^(-)])/(["HA"]_0 - ["A"^(-)])`

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 `"pH"` of the solution and the equilibrium concentration of `"A"^(-)`.

For example, if you start with `"0.1 M"` of `"HA"`, you will have

`K_a = (10^(-"pH") * ["A"^(-)])/(0.1 - ["A"^(-)])`