Weak Acid-Weak Base Equilibria

In summary, the conversation discusses a problem involving 3 equilibria - a weak acid, a weak base, and water autoionization. The speaker asks for a reference for a rigorous analysis of this problem without approximations. Another person provides a link to a website that explains the problem in detail and provides equations for a full version of the problem. They also mention that there are better methods for solving the problem.
  • #1
ekhahniii
1
1
Such an equilibrium problem involves 3 equilibria: a weak acid, a weak base, and water autoionization.

I find in various texts that several assumptions are made in dealing with this class of equilibria problems (ignore water's autoionization, conversion of the weak acid is on the order of the conversion of the weak base, etc), and a more rigorous analysis is dismissed as unnecessary (4th order polynomials scare the authors of my chemistry texts :redface:). Perhaps the dismissal is warranted. But that just sparks my curiosity.

Would someone please be kind enough to provide a reference in which this problem is dealt with rigorously, without approximations?

Thanks!
 
  • Like
Likes SilverSoldier
Chemistry news on Phys.org
  • #2
Take a look here:

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

In most cases explanations start with a full version of the problem, approximations and simplifications are shown later. Take a look here:

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

equation 11.16 - that's kind of thing you will need to work with if you want a full version. It is not in a polynomial form, that would require some additional rearranging, boring and easy to make a mistake, but nothing extraordinary in terms of techniques needed.

If you are still curious - here are coefficients of 10th degree polynomial describing mixture of acid (concentration Ca) and base (Cb), both with 4 dissociation steps:

Code:
a10 = +Kb4;
a9  = +(4*Cb*Kb4+Ka1*Kb4+Kb3*KW); 
a8  = -(Ca*Ka1*Kb4-Cb*(4*Ka1*Kb4+3*Kb3*KW)-Ka1*Kb3*KW-Ka2*Kb4-KW*(Kb2*KW-Kb4)); 
a7  = -(Ca*(Ka1*Kb3*KW+2*Ka2*Kb4)-Cb*(3*Ka1*Kb3*KW+4*Ka2*Kb4+2*Kb2*KW*KW)+Ka1*KW*(Kb4-Kb2*KW)-Ka2*Kb3*KW-Ka3*Kb4-KW*KW*(Kb1*KW-Kb3)); 
a6  = -(Ca*(Ka1*Kb2*KW*KW+2*Ka2*Kb3*KW+3*Ka3*Kb4)-Cb*(2*Ka1*Kb2*KW*KW+3*Ka2*Kb3*KW+4*Ka3*Kb4+Kb1*KW*KW*KW)+Ka1*KW*KW*(Kb3-Kb1*KW)+Ka2*KW*(Kb4-Kb2*KW)-Ka3*Kb3*KW-Ka4*Kb4+KW*KW*KW*(Kb2-KW)); 
a5  = -(Ca*(Ka1*Kb1*KW*KW*KW+2*Ka2*Kb2*KW*KW+3*Ka3*Kb3*KW+4*Ka4*Kb4)-Cb*(Ka1*Kb1*KW*KW*KW+2*Ka2*Kb2*KW*KW+3*Ka3*Kb3*KW+4*Ka4*Kb4)+KW*(Ka1*KW*KW*(Kb2-KW)+Ka2*KW*(Kb3-Kb1*KW)+Ka3*(Kb4-Kb2*KW)-Ka4*Kb3+Kb1*KW*KW*KW)); 
a4  = -KW*(Ca*(Ka1*KW*KW*KW+2*Ka2*Kb1*KW*KW+3*Ka3*Kb2*KW+4*Ka4*Kb3)-Cb*(Ka2*Kb1*KW*KW+2*Ka3*Kb2*KW+3*Ka4*Kb3)+Ka1*Kb1*KW*KW*KW+Ka2*KW*KW*(Kb2-KW)+Ka3*KW*(Kb3-Kb1*KW)+Ka4*(Kb4-Kb2*KW)+KW*KW*KW*KW); 
a3  = -KW*KW*(Ca*(2*Ka2*KW*KW+3*Ka3*Kb1*KW+4*Ka4*Kb2)-Cb*(Ka3*Kb1*KW+2*Ka4*Kb2)+Ka1*KW*KW*KW+Ka2*Kb1*KW*KW+Ka3*KW*(Kb2-KW)+Ka4*(Kb3-Kb1*KW)); 
a2  = -KW*KW*KW*(Ca*(3*Ka3*KW+4*Ka4*Kb1)-Cb*Ka4*Kb1+Ka2*KW*KW+Ka3*Kb1*KW+Ka4*(Kb2-KW)); 
a1  = -KW*KW*KW*KW*(4*Ca*Ka4+Ka3*KW+Ka4*Kb1); 
a0  = -Ka4*KW*KW*KW*KW*KW;

If memory serves me well KW is water ionic product, Kxy is yth overall dissociation constant of x (acid or base). a10 is a coefficient of [H+]10 and so on.

There are better ways of dealing with the problem.
 
  • Informative
Likes SilverSoldier

1. What is a weak acid?

A weak acid is an acid that does not completely dissociate in aqueous solution. This means that only a small percentage of the acid molecules will release hydrogen ions (H+) into the solution. Examples of weak acids include acetic acid (found in vinegar) and citric acid (found in citrus fruits).

2. What is a weak base?

A weak base is a base that does not completely dissociate in aqueous solution. This means that only a small percentage of the base molecules will accept hydrogen ions (H+) from the solution. Examples of weak bases include ammonia and ammonium hydroxide.

3. How do weak acid-weak base equilibria occur?

Weak acid-weak base equilibria occur when a weak acid is mixed with a weak base in aqueous solution. The acid and base will react with each other to form their respective conjugate base and conjugate acid. This reaction will reach a point of equilibrium, where the concentrations of the acid, base, conjugate base, and conjugate acid remain constant.

4. What is the dissociation constant (Ka) of a weak acid?

The dissociation constant (Ka) of a weak acid is a measure of the acid's ability to dissociate in aqueous solution. It is equal to the concentration of the products (conjugate base and hydrogen ions) divided by the concentration of the reactants (weak acid). A higher Ka value indicates a stronger acid.

5. How do you calculate the pH of a weak acid-weak base equilibrium?

The pH of a weak acid-weak base equilibrium can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([conjugate base]/[weak acid]). The pKa value can be found on a table, or it can be calculated using the Ka value. The concentrations of the conjugate base and weak acid can be determined from the initial concentrations and the equilibrium constant (Ka).

Similar threads

Replies
4
Views
3K
Replies
4
Views
7K
  • Chemistry
Replies
4
Views
2K
Replies
4
Views
4K
Replies
2
Views
3K
  • Chemistry
Replies
1
Views
2K
Replies
3
Views
4K
Replies
1
Views
2K
  • Biology and Chemistry Homework Help
Replies
1
Views
4K
Back
Top