Question #36d09

1 Answer
Oct 1, 2017

Answer:

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"^(-)])#