Weak Acid-Weak Base Equilibria

  • Thread starter ekhahniii
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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!
 

Borek

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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.
 

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