1) How do you calculate pH when a weak acid is added to a weak base?

Tris (or Trizma Base) is a weak base.

H2O + (HOCH2)3CNH2 < = > (HOCH2)3CNH3+ + OH-, (pKa = 8.1), pKb = 5.9,
Kb = 1.3 x10-6

Boric acid is a weak acid.

B(OH)3 + H2O < = > H+ + B(OH)4-, pKa = 9.2, Ka = 6.3 x10-10

A common electrolyte used for electrophoresis is [1x] TBE which is 89
mM Tris, 89 mM Boric acid, and 2.5 mM EDTA
(ethylenediaminetetraacetic) which can be ignored.

Taken individualy,

[(HOCH2)3CNH2] = (89 x10-3 - X), [(HOCH2)3CNH3+] = X, [OH-] = X

Kb = 1.3 x10-6 = ([(HOCH2)3CNH3+] x [OH-]) / [(HOCH2)3CNH2] = X2 / (89
x10-3 - X), X = 3.4 x10-4

pOH = 3.5, pH = 10.5

and,

[B(OH)3] = (89 x10-3 - Y), [H+] = Y, [B(OH)4-] = Y

Ka = 6.3 x10-10 = ([H+] x [B(OH)4-]) / [B(OH)3] = Y2 / (89 x10-3 - Y),
Y = 7.5 x10-6

pH = 5.1

but when you add them together,

B(OH)3 + (HOCH2)3CNH2 + 2H2O < = > (HOCH2)3CNH3+ + B(OH)4-+  H+ + OH-

How do you calculate the pH (references say that the pH is 8.3)?

2) TBE is commonly called a buffer because I presume it can maintain
it?s pH as [H+] is increased at the anode and [OH-] is increased at
the cathode during electrophoresis due to the electrolysis of water.

4e- + 4H2O => 2H2 + 4OH-, cathode

2H2O => O2 + 4H+ + 4e-, anode

If [H+] and [OH-] are increased by 1mM how does the pH change?

For simplicity, I'll write "tris0" and "tris+" to denote the neutral
and protonated versions of tris.  Quantities in square brackets
indicate concentrations, as usual.


This problems has 6 unknowns:  [tris0], [tris+], [B(OH)3], [B(OH)4-],
[OH-] and [H+]

You have, however, 6 equations relating these 6 unknowns:

[tris+]*[OH-]/[tris0] = 10^-5.9

[B(OH)4-]*[H+]/[B(OH)3] = 10^-9.2

[B(OH)] + [B(OH)4-] = 89mM

[tris0] + [tris+] = 89mM

[H+]*[OH-] = 10^-14

[B(OH)4-] + [OH-] = [H+] + [tris+]

The only one of these equations that might not be obvious to you is
the last.  This equation is a statement of electrical neutrality in
the solution -- the sum of the concentrations of the negatively
charged species must equal the sum of the concentrations of the
positively charged species.

Given these six equations, you can solve for each of the unknowns,
including the concentration of H+.  The pH, of course, is the negative
logarithm of [H+].

It gets somewhat complicated (for me, at least) to solve those 6
simultaneous equations. (They are not all of the simple form, aX + bY
+ .... = c.)

Since you have a weak base and a weak acid, you can initially remove
the H+ and OH- concentrations from consideration. (The H+ and OH-
concentrations are much smaller than the concentrations of the weak
acid and weak base and their conjugates.)

You can initially work without the H+ and OH- by eliminating them from
your "added" equation:

B(OH)3 + (HOCH2)3CNH2 + 2H2O < = > (HOCH2)3CNH3+ + B(OH)4-+  H+ + OH-

whose equilibrium constant is (1.3 x 10-6) * (6.3 x 10-10).

Since [H+] [OH-] = 1 x 10-14, you can use this to convert your "added" equation to

B(OH)3 + (HOCH2)3CNH2 + H2O < = > (HOCH2)3CNH3+ + B(OH)4-

whose equilibrium constant is (1.3 x 10-6) * (6.3 x 10-10) / (1 x 10-14).

You can then let x concentration/amount of boric acid and of tris that
react and get:

x * x / {(.089 - x) *(.089 - x)} = (1.3 x 10-6) * (6.3 x 10-10) / (1 x 10-14)

Solving this quadratic equation will give you the values for boric
acid and borate; you can then use the equation for boric acid
dissociation to get the H+ concentration and the pH.

I didn't get a pH of 8.3, but the values I get satisfy the 6 equations
shown, except that the two halves of equation 6 differ by 4.55 x 10-6
(which is off by only 2 parts in 10,000 and within the two significant
figure error for the equilibrium constants).

If you want to post your answer, I'll let  you know if we have the same answer.
----
re: part 2:

I presume the solutions at the two electrodes are not allowed to mix
with each other. For each electrode, then, let the H+ or OH- react
with the tris or the boric acid and use the result as your starting
point. Your starting equation will look something like this:

(x + .001) * x /(.088-x) * (.089 -x) = K(eq)

Thanks for the comment hfshaw.  It hadn?t occurred to me to examine
this question as just a math problem, and the last equation that
balances the charge is clever.  Using substitution, I was left with a
quartic equation, but after a good struggle I gave up trying to solve
for the roots.  Instead, I asked a friend to solve it numerically and
these are the results.

[tris+] = 0.019570697 = 20mM
[OH-] = 4.46618 x10^-6
[tris0] = 0.069429303 = 69mM
[B(OH)4-] = 0.019566233 = 20mM
[H+] = 2.23905 10^-9
[B(OH)3] = 0.069433767 = 69mM

Therefore pH = 8.7.

Thanks for you comment brix24.  After giving up trying to solve the 6
equations by substitution I started on your suggestion.

Starting with the added equation

B(OH)3 + (HOCH2)3CNH2 + 2H2O < = > (HOCH2)3CNH3+ + B(OH)4-+  H+ + OH-

Keq = Kb x Ka

At first I was uncertain of your claim that the H+ and OH- could be
dropped because I thought that [(HOCH2)3CNH3+] = [OH-] and [B(OH)4-] =
[H+].  But then I realized the role of Kw which would have H+ and OH-
react to form H20 until [H+][OH-] = 10^-14 in which case they would be
present in such comparatively low concentrations that they could be
ignored.  Is this right?

Then, as you showed, the equation can be simplified as

B(OH)3 + (HOCH2)3CNH2 + 2H2O < = > (HOCH2)3CNH3+ + B(OH)4-

Keq = Kb x Ka / Kw

Where

[B(OH)3] = (.089-X), [(HOCH2)3CNH2] = (.089-X), [(HOCH2)3CNH3+] = X, [B(OH)4-] = X

X^2 / (.089-X)^2 = (1.3 x 10^-6) x (6.3 x 10^-10) / 10^-14, X /
(.089-X) = SQRT(.0819), X = .020

Therefore

[B(OH)3] = (.089-.020) = .069, [B(OH)4-] = .020

I wasn?t obvious to me what to do next, so your guidance was very
useful.  Plugging this back into the equation for the dissociation of
boric acid gives.

Ka = [B(OH)4-] x [H+] / [B(OH)3], 6.3 x10^-10 = .020 x [H+] / .069,
[H+] = 2.2 x10^-9, pH = 8.7

So, how about that, the same answer as the first method.  What do
references (Sigma catalog and others) know about pH (unless I?ve got
Ka and Kb wrong)!

I?m still working on the the part of my question that asked about the
buffering ability of TBE.  When I?m done I?ll post my answer.  I?d be
very thankful if you could comment on it, thanks for your
consideration.

Yes, I also got pH 8.7 (or 8.66) and 69 mM and 20 mM.

Your comment that the H+ and OH- are in such low concentration that
they can be ignored in determining the pH is correct.

Perhaps it would have been better had I said that the H+ and OH-
concentrations are determined by the weak acids and bases left after
the Tris and the boric acid react. The goal was first to get an
equation without H+ and OH- and use that to determine the equilibrium
contrations of Tris, Tris+, boric acid, and borate -- then to get the
H+ or OH- concentraion from either the Tris/Tris+ or boric acid/borate
equilibrium (which you did).

For the second part, the initial H+ or OH- concentraion is no longer
so negligible, so there is one preliminary step: reduce the H+ or OH-
concentration to near zero by assuming they react either with the Tris
or boric acid. Then the problem can be solved by the same method as in
part one, except that instead of starting with 89 mM Tris and 89 mM
boric acid, you are starting with 88 mM Tris, 1 mM Tris+, and 89 mM
boric acid (for 1 mM H+ present initially). This works because
starting either with (1mM H+, 89 mM Tris, and 89 mM boric acid) or (88
mM Tris, 1 mM Tris+, and 89 mM boric acid) will result in reaching the
same final equilibria - it's just that it's easier to solve starting
with the second combination.

Re the 8.3 -- 8.7 pH difference, I know that pH can depend on
temperature and ionic strength, but I don't know if that is enough
here to explain this particular difference.

Thank you once again brix24 for your comments.  Your suggestion that
the H+ and OH- concentrations are determined by the weak acids and
bases left after the Tris and the boric acid react really helped me
understand what I think is happening.  This is how I reasoned things
through.


1)  The strongest acid present, B(OH)3, reacts with the strongest base
present, tris

B(OH)3 + tris + H2O < = > tris+ + B(OH)4-

Use the initial concentrations and let

[B(OH)3] = (.089-X), [tris] = (.089-X), [tris+] = X, [B(OH)4-] = X

then substitute into

Keq = [tris+][B(OH)4-]/[B(OH)3][tris] = KbKa/Kw = 8.19x10^-2

to find that

[B(OH)3] = .069, [tris] = .069, [tris+] = .020, [B(OH)4-] = .020


2)  Then, the strongest acid that remains, B(OH)3, reacts with the
next strongest base present, H2O.

B(OH)3 + H2O < = > H+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = (.069-X), [H+] = X, [B(OH)4-] = (.020+X)

then substitute into

Ka = [H+][B(OH)4-]/[B(OH)3] = 6.3x10^-10

to find that

[B(OH)3] ~ .069, [B(OH)4-] ~ .020, [H+] = 2.17x10^-9, pH = 8.66

which is same result I got when I let

[B(OH)3] = .069, [H+] = X, [B(OH)4-] = .020

and ?plugged into? Ka without thinking about what was happening,
although it serves as a valid approximation since you would assume X
would be small enough that it could be ignored in the .069-X and
0.20+X terms.


I liked this reasoning because I can apply it when determining what
the pH at the anode (where H+ is generated) is when [H+] is increased
by 1 mM.


1)  First, the strongest acid present, H+, reacts with the strongest
base present, tris

H+ + tris = > tris+

Use the initial concentrations and let

[H+] = (.001-X), [tris] = (.089-X), [tris+] = X

then knowing that H+ is a strong acid and Keq is very large

[H+] ~ 0, [tris] = (.088), [tris+] = .001


2)  Then, the strongest acid that remains, B(OH)3, (since H+ is
depleted), reacts with the strongest base present, tris.

B(OH)3 + tris + H2O < = > tris+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = (.089-X), [tris] = (.088-X), [tris+] = (.001+X), [B(OH)4-] = X

then substitute into

Keq = [tris+][B(OH)4-]/[B(OH)3][tris] = KbKa/Kw = 8.19x10^-2

to find that

[B(OH)3] = .070, [tris] = .069, [tris+] = .020, [B(OH)4-] = .019


3)  Then, the strongest acid that remains, B(OH)3, reacts with the
next strongest base present, H2O.

B(OH)3 + H2O < = > H+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = (.070-X), [H+] = X, [B(OH)4-] = (.019+X)

then substitute into

Ka = [H+][B(OH)4-]/[B(OH)3] = 6.3x10^-10

to find that

[B(OH)3] ~ .070, [B(OH)4-] ~ .019, [H+] = 2.32x10^-9, pH = 8.63


As you can see, I?ve followed just what you suggested doing.  If you?d
like to comment on my reasoning behind this approach I?d be grateful. 
I couldn?t have thought this problem through without your guidance,
thanks.

Not to get carried away, but if you wanted to know how much H+ could
be generated at the anode to bring about a 50% increase in [H+], you
reverse the above steps.


Initially, [H+] = 2.17x10^-9, pH = 8.66.  A 50% increase is, [H+] =
3.26x10^-9, pH = 8.49.


From step (3)

B(OH)3 + H2O < = > H+ + B(OH)4-

Use the new concentration and let

[B(OH)3] = (.089-Y-3.26x10^-9), [H+] = 3.26x10^-9, [B(OH)4-] = (Y+3.26x10^-9)

then substitute into

Ka = [H+][B(OH)4-]/[B(OH)3] = 6.3x10^-10

to find that

[B(OH)3] ~ .075, [B(OH)4-] ~ .014


From step (2)

B(OH)3 + tris + H2O < = > tris+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = .075, [tris] = (.075-Z), [tris+] = (Z+.014), [B(OH)4-] = .014

then substitute into

Keq = [tris+][B(OH)4-]/[B(OH)3][tris] = KbKa/Kw = 8.19x10^-2

to find that

[tris] = .062, [tris+] = .027


From step (1)

H+ + tris = > tris+

Use the new concentrations and let

[H+] = X, [tris] = .062, [tris+] = .027

then knowing that H+ is a strong acid and Keq is very large

[H+] = .027


So, it would take the generation of [H+] = 27mM to lower the pH to 8.49.

I have been impressed by your obvious desire to understand, and now I
am impressed by your creativity in posing your additional, not
run-of-the-mill problem and solving it.

Re answer agreement: For part two of your original problem, I also get
pH 8.63, and, starting with the 75 mM boric acid, 14 mM borate
numbers, I also get 27 mM for the additional [H+] needed. However, my
method and reasoning differs somewhat from yours, so I'll have to
explain the difference.

This comment will be somewhat lengthy; it discusses some issues left
mostly implicit before and explains how my concept differs from what I
think yours is.

Let me begin with a previous statement of mine:

"Your comment that the H+ and OH- are in such low concentration that
they can be ignored in determining the pH is correct."

Taken out of context, this is subject to an interpretation I did not
intend, so I will bring in some more information here.

* All the acid-base reactions reach equilibrium together, that is, you
can?t have two reactions with a common reactant with one being in
equilibrium and the other not. If that were the case, the reaction not
at equilibrium would continue towards equilibrium and change the
concentration of the common reactant, changing the reaction at
equilibrium to one not at equilibrium. (This will be become more
relevant when getting to how I solved the most recent problems.)

* Water is such a weak acid and base that the hydrolysis reaction
functions primarily to keep the [H+] and [OH-] concentrations adjusted
so that [H+][OH-] = 10^-14. When a weak acid or base is added, they
are responsible for the production of [H+] or [OH-, with the
hydrolysis of water essentially serving to adjust either
(complementary) [OH-] or [H+] so that [H+][OH-] is 10-14.

* In the simple case of addition of a weak acid to water, both the
acid dissociation and the hydrolysis reaction reach equilibrium
together. For boric acid,

boric acid < = > H+ + borate

and 

H2O < = > H+ + OH-

both reactions reach equilibrium quickly. To solve for the H+
concentration, we determine the dissociation of the boric acid, with x
being the concentration that dissociates to reach equilibrium. Since
the hydrolysis of water produces so little [H+], we can ignore its
contribution, but it serves to lower the [OH-] to keep the 10^-14
value.

*If we now consider the reactions occurring in the Tris-boric acid
buffer, the following reactions all reach equilibrium together:

Tris + H2O < = > Tris+ + OH-
boric acid < = > H+ + borate
H2O < = > H+ + OH-

as well as the equations derived from these equations, that is,

Tris + boric acid < = > Tris+ + borate

Which is the sum of the first two equations minus the third equation.

All four of these reactions must all reach equilibrium together, so
that the one equilibrium concentration of Tris satisfies all four
reactions simultaneously - since all the reactions take place in the
same solution and face the same concentraion of Tris. The same holds
true for the equilibrium concentration of any other reactant. Since
these reactions must all reach equilibrium together, if we determine
the concentration of one reactant, say, Tris, for one reaction (in the
face of any other reactions using Tris), we have the equilibrium Tris
concentration for all the reactions. This concept will the basis for
the difference in our approaches for solving the most recent problems.

* In my previous comments about how to get an equilibrium
concentration of Tris, I focused on the following:

We have both a weak acid (boric acid) and a weak base (Tris) and
possibly their conjugates; the weak acid and the weak base will react
with one another and are the primary determinants of the H+
concentraion. The approach was to find a reaction (the combined one,
as it turned out) that reflected the reaction between these Tris and
boric acid and that didn't have other terms, such as H+ and OH-, that
would complicate solving for equilibrium concentraions. (Implicit was
the concept that the final H+ concentration was determined by the
concentrations of the conjugate acid / conjuate base couples left
after the weak acid and the weak base react.)

The combined equation was ideal for our purposes since it contained
Tris, boric acid, and their conjugates, but not any H+ or OH-. With
the combined equation, then, we could use x to represent the
concentration of Tris and boric acid that react with one another.
Solving for the Tris and boric acid concentrations then gives the
equilibrium concentrations for them in all the reactions.

Once we have the concentrations of Tris, boric acid, and their
conjugates from the combined equation, we can solve for the H+
concentration. These concentrations are equilibrium concentrations and
can simply be substituted in any of the appropriate remaining
equilibrium reactions to calculate the equilibrium concentrations of
H+ or OH-.

There is no need for a "minus x" in this calculation for H+ since
equilibrium has been achieved for all the equations and there is no
further change in concentrations. (One can use either the boric acid
equation or, successively, the Tris equation and the water hydrolysis
equation to get the H+ concentration.)

* Going back to your most recent comment, then, I would modify your
method, which you list as:

"3)  Then, the strongest acid that remains, B(OH)3, reacts with the
next strongest base present, H2O.

B(OH)3 + H2O < = > H+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = (.070-X), [H+] = X, [B(OH)4-] = (.019+X)?


Instead of using 

[B(OH)3] = (.070-X), [H+] = X, [B(OH)4-] = (.019+X)

I use

[B(OH)3] = .070, [H+] = X, [B(OH)4-] = .019

Mathematically, either method gives the same result since the x here
is so small. However, the combined equation used to determine the .070
M boric acid is really just a combination of the boric acid equation
and the other two equations. All the reactions are at equilibrium, and
the boric acid concentration will be the same for all of them, that
is, it is not necessary to use "minus x"

(There is an alternative, more cumbersome way to get the H+ without
using the boric acid equation. You can also the Tris reaction to
determine the OH- concentration and then the hydrolysis constant for
water to get the H+ concentration from the OH- concentration - again,
without having to use a "minus x.")

* Switching, now, to your very creative new problem, I would not use
the "minus 3.26x10^-9" or "minus z (even though the value of z is not
negligible here).

In solving for the H+ concentration, you wrote

"B(OH)3 + H2O < = > H+ + B(OH)4-

Use the new concentration and let

[B(OH)3] = (.089-Y-3.26x10^-9), [H+] = 3.26x10^-9, [B(OH)4-] = (Y+3.26x10^-9)

then substitute into

Ka = [H+][B(OH)4-]/[B(OH)3] = 6.3x10^-10"


I would use

[B(OH)3] = .089-Y, [H+] = 3.26x10^-9, [B(OH)4-] = Y

instead of 

[B(OH)3] = (.089-Y-3.26x10^-9), [H+] = 3.26x10^-9, [B(OH)4-] = (Y+3.26x10^-9)

at this point since the final ( aka equilibrium) concentration of H+
is given. A certain amount of H+ was added - some reacting with the
bases present and some staying as H+ to reach double the final H+
concentration. The equilibrium equation is

[borate] * 3.26x10^-9 = Ka * [boric acid]

The other constraint on the borate and boric acid concentrations is
that their sum must be .089, so one concentration must be .089 - y and
the other y.

Solving this gives you the equilibrium boric acid and borate
concentrations (.075 and .014); you can then substitute these into the
combined equation. though I used the ?z? unknown somewhat differently
than you di.

You have

"B(OH)3 + tris + H2O < = > tris+ + B(OH)4-

Use the new concentrations and let

[B(OH)3] = .075, [tris] = (.075-Z), [tris+] = (Z+.014), [B(OH)4-] = .014

then substitute into

Keq = [tris+][B(OH)4-]/[B(OH)3][tris] = KbKa/Kw = 8.19x10^-2

to find that

[tris] = .062, [tris+] = .027"


All the reactions should already be at equilibrium since you added H+
until the final H+ concentration doubles - with some of the added H+
reacting with borate, some with Tris, and the rest staying to raise
the H+ concentration, If any reaction were not at equilibrium, then it
would either generate or use up H+, but this is contrary to saying we
added H+ until the final H+ doubled.

Instead of 

[B(OH)3] = .075, [tris] = (.075-Z), [tris+] = (Z+.014), [B(OH)4-] = .014

I used

[B(OH)3] = .075, [tris] = .089 - z, [tris+] = z, [B(OH)4-] = .014

reasoning that the boric acid and borate concentrations (as you did)
are at equilibrium and that the sum of the Tris and Tris+
concentrations must be .089. There is really no difference in whether
this last equation is solved by your method or by the method that I
used. The methods are equivalent, but with different starting points.

My equation is clearly different in appearance than yours; why, then,
do we get the same answer? There are two ways to look at this:

Chemically: you are looking for equilibrium starting from one point of
view (.075 M Tris and .014 M Tris+) and I am looking for equilibrium
starting from .089 M Tris. In either case, their is only one set of
equilibrium concentrations for  Tris and Tris+ that will satisfy this
equation.

Mathematically: the (.075 - z), (0.14 + z) and the (0.089 - z), (0 +
z) are transforms (?) of one another:  You can change one equation
into the other by substituting (Q + .014) for z in your equation and
get the form of my equation.

If you?ve read this far, congratulations again on your enthusiasm for
understanding, on your obvious hard work, and on your creativity.