I have had some trouble verifying the systematic treatment in a weak acid base weak balance. The great equation I must arrive at is this:

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and I just can't see clearly how to get that equation from these:

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  • $\begingroup$ doublecheck your expression for [A-] at the bottom. I think you have a Ka where you should have an [H+]. . .Also don't forget to change [OH-] to Kw/[H+] in your charge balance eq. $\endgroup$
    – Andrew
    Jan 4, 2021 at 20:11
  • $\begingroup$ have a look at chemistry.stackexchange.com/questions/60068/… for an alternative and simpler derivation you might find interesting. $\endgroup$
    – porphyrin
    Jan 5, 2021 at 9:39
  • 3
    $\begingroup$ It is a set of 4 nonlinear equations with 4 variables. Use a substitution method to eliminate 3 equations and 3 variables but [H+]. Use such a substitution where a variable is eliminated by an expression depending on still remaining variables. But as @porphyrin suggests, seriously consider simplified approach assuming some strong inequalities. $\endgroup$
    – Poutnik
    Jan 5, 2021 at 10:09

1 Answer 1


This sort of derivation is a just painful, but straightforward. As Poutnik mentioned it is just a series of equations where you eliminate variables by substitution.

So start with the basics.

$\ce{HA <=> H+ + A-}\tag{A}$

$\ce{BOH <=> B+ + OH-}\tag{B}$

$\ce{H2O <=> H+ + OH-}\tag{C}$


  • $\mathrm{C_A}$ is the concentration of the HA reagent
  • $\mathrm{V_A}$ is the volume of the HA reagent added to the solution
  • $\mathrm{C_B}$ is the concentration of the BOH reagent
  • $\mathrm{V_A}$ is the volume of the BOH reagent added to the solution

So there are six unknowns: $\ce{[HA], [A-], [BOH], [B+], [H+]}$ and $\ce{[OH-]}$

Thus we need 6 independent equations to solve the problem.

Step 1

$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-1}$

$\mathrm{K_b} = \dfrac{\ce{[B+][OH-]}}{\ce{[BOH]}}\tag{2-1}$

$\mathrm{K_w} = \ce{[H+][OH-]}\tag{3-1}$

$\ce{[H+] + [B+] = [OH-] + [A-]}\tag{4-1}$

$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-1}$

$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-1}$

So now for the 6 equations just eliminate equations one at a time.

Step 2

So for example we want to know the final equation to only have the unknown $\ce{[H+]}$. Rearrange equation (3-1) to:

$\ce{[OH-]} = \dfrac{\mathrm{K_w}}{\ce{[H+]}}\tag{3-2}$

Now in equations (1-1), (2-1), (4-1), (5-1), and (6-1) substitute for $\ce{[OH-]}$ and equation (3-1) is eliminated. So now you're left with 5 equations and 5 unknowns.

$\mathrm{K_a} = \dfrac{\ce{[H+][A-]}}{\ce{[HA]}}\tag{1-2}$

$\mathrm{K_b} = \dfrac{\ce{[B+]\cdot \mathrm{K_w}}}{\ce{[BOH][H+]}}\tag{2-2}$

$\ce{[H+] + [B+] = \dfrac{\mathrm{K_w}}{\ce{[H+]}} + [A-]}\tag{4-2}$

$\ce{([HA] + [A-])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_A\cdot C_A}}\tag{5-2}$

$\ce{([BOH] + [B+])\cdot \mathrm{(V_A + V_B)} = \mathrm{V_B\cdot C_B}}\tag{6-2}$

Obviously as you continue it gets more and more tedious, but the process is straightforward.

The point is that you have 6 unknowns and 7 equations below. So all of the equations can't be independent. With a problem that is going to be this tedious I like to write out everything explicitly so that:

(1) I don't get confused

(2) So that I can check my work step by step.

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