[Cheese_Burgers]

I need to figure out the theoretical pH of a solution where the concentration of diethylamine is 0.3M (pkb 10.73) and the concentration of a carboxylic acid of pKa 4.85 is 0.2M.

This is the first step in interpreting some stability data of a formulation I have. Since it is impractical to titrate weak acid with weak base, I'm finding it hard to figure this out via the internet or any textbooks I have. I'm only finding stuff relating to conjugate acid/base buffers and strong base weak acid titrations.

My end goal is to 1) determine the initial molar ratio of DEA/COOH  that gives a solution with pH of 8.2 and 2) determine how a side reaction of the DEA with formaldehyde (which introduces a basic species with a lower conjugate acid pKa than DEA's conjugate acid) may lower the pH of the solution by shifting the equilibrium...

The only thing I found online was this person's yahoo answer (below). **It gives me a final pH of 2.263 but, when I solve for higher initial concentrations of DEA my final pH gets lower... which doesn't make sense.

Any help or references are much appreciated.

"
B + HA <---> BH+ + A-

K = [BH+][A-] / [ B][HA]

We have 2 unknowns, K and x, but we can find K if we know that:

Kb = [OH-][BH+] / [ B]
Ka = [H+][A-] / [HA]

If we do a little rearranging, then we can have:

[BH+]/[ B] = Kb/[OH-]
[A-]/[HA] = Ka/[H+]

The top formula can be written as:

K = [BH+][A-] / [ B][HA] ---> K = ([BH+]/ [ B])([A-]/[HA])

So K = KbKa / [OH-][H+]

Ka = 10^-4.57
Kb = 10^-6.77
[OH-][H+] = kw; kw = 1E-14

K = 10^-11.34 / 10^-14 = 457.1

We should expect a somewhat large K constant for this reaction because acids are proton donators and bases are proton acceptors.

So going back to the reaction at the top, both HA and B will form their conjugate pairs, so:

... ..B. . . +. . HA <---> BH+ + A-
I..0.0200. .0.0200. . .. .0. . . . .0
E.0.0200-x.0.0200-x. . .x . . . .x

K = [BH+][A-] / [ B][HA]

457.1 = x^2 / (0.02-x)^2

21.38 = x / 0.02-x

x = [BH+] = [A-] = 0.0191 M

So you have:

[BH+] = 0.0191 M
[A-] = 0.0191 M
[ B] = 0.0200 - 0.0191 = 0.0009 M
[HA] = 0.0009 M

The pH can be found by adding up the [H+] generated by HA and B, with the formula:

pH = -log ( [H+]HA + [H+]B )

However, knowing that bases aren't really a great source of [H+], the pH is just the [H+] from the acid.

The acid dissociates by reacting with water to produce its conjugate base form. Using the values that we currently have of [HA] and [A-], we can find [H+]:

....HA . . . .<-----> H+ + A-
I. . 0.0009. . . . . .0. . . .0.0191
E. 0.0009-x. . . . .x. . . .0.0191+x

ka = [H+][A-] / [HA]

10^-4.57 = x(0.0191+x) / (0.0009-x)

x = [H+] = 1.27E-6 M

pH = 5.90


I don't know if I'm right, but then again I rarely ever encounter weak base-weak acid titrations; it's just impractical."

Sincerly,

Cheese_Burgers

[AWK]

As the first, and quite good approximation, treat you mixture as a buffer.
Write down a balanced neutralization reaction and find concentration of salt and base in excess.

Note you wrote pK_a of DEA, not pK_b.

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-solution

Much easier approach is to use a dedicated calculator:

http://www.chembuddy.com/?left=Buffer-Maker&right=pH-calculator

or a simpler one:

http://www.chembuddy.com/?left=BATE&right=pH-calculator

[Cheese_Burgers]

As the first, and quite good approximation, treat you mixture as a buffer.
Write down a balanced neutralization reaction and find concentration of salt and base in excess.

Note you wrote pK_a of DEA, not pK_b.

Ahh I see my mistake writing pKa of DEA instead of Pkb! Thank you!