# The algebraic problem in the equilibrium of weak acids and bases

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

• 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. Jan 4 at 20:11
• have a look at chemistry.stackexchange.com/questions/60068/… for an alternative and simpler derivation you might find interesting. Jan 5 at 9:39
• 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. Jan 5 at 10:09

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.

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

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

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

Given:

• $$\mathrm{C_A}$$ 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.