# Algebraic treatment of equilibrium

This is from Laitinen and Harris, "Chemical Analysis, An Advanced Text & Reference" which describes a simple acid-base equilibrium. They derive a relation in the end directly. I haven't seen this expression in many books which teach solution equilibria. The authors cite an article from the Journal of Chemical Education for the expressions (4 & 5). I didn't find article very useful. How should we derive the equations 4 and 5 in a better way than the author's explanation in italics?

The authors start from very simple equations:

$$\ce{HA + H2O <=> H3O^+ + A-}\tag{1}$$

Conventionally,

$$\ce{HA <=> H+ + A^-}\tag{2}$$

and the ionization of water, $$\ce{H2O <=> H+ + OH-}\tag{3}$$

If $$C_{HA}$$ and $$C_{A^-}$$ are the analytical concentrations of $$\ce{HA}$$ and $$\ce{A^-}$$, and if $$\ce{[HA]}$$ and $$\ce{[A^-]}$$ are the equilibrium concentrations, the relations...

$$\ce{[HA]} = C_{HA} - \ce{([H^+] - [OH^-])}\tag{4}$$

$$\ce{[A^-]} = C_{A^-} + \ce{([H^+] - [OH^-])}\tag{5}$$

The authors explain ...

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

• ..in reaction (3) is wrong imho, it should be 1 or 2. Apr 12 '20 at 6:25

There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid $$\ce{HA}$$ results in equal amounts of hydronium and of conjugate base $$\ce{A-}$$. Unless a salt of the conjugate base is explicitly added then the acid is the only source of $$\ce{A-}$$. In absence of salt the total hydronium concentration in solution is therefore

$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$

Note that this equation reflects the charge balance condition.

Solving for $$\ce{[A-]}$$

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

We also know that the original amount of acid equals the sum of undissociated acid and conjugate base:

$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$

Note that this equation reflects the mass balance condition.

Solving for $$\ce{[HA]}$$

$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$

Substituting Eq. (1) into Eq. (2) we obtain

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$

The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution ($$C_\ce{A-}=0$$). If a completely dissociating salt $$\ce{MA}$$ is added at concentration $$C_\ce{A-}$$ then it is necessary to modify the mass balance expression:

$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$

and the charge balance expression:

$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$

or since $$\ce{[M+]}=C_\ce{A-}$$

$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$

Solving for $$\ce{[A-]}$$

$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$

Combining the two balance expressions leads once again to Eq. (3):

$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$

By the way, the quote you cite should read:

"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

It might be that "typo" ("3" written instead of "2") that led to the confusion.

• I'm confused. Shouldn't the second charge expression be $$\ce{[H+] + [M+]} = \ce{[OH-]} + C_\ce{A-}$$
– MaxW
Apr 12 '20 at 18:04
• @MaxW Maybe I should edit the equation as written but it is correct. The key point is that $[M^+] = C_{A-}$ Apr 12 '20 at 18:23
• I can't figure out $\ce{[H+] +[M+] =[H+]}+C_\ce{A-}$ since $\ce{[H+]}$ is on both sides.
– MaxW
Apr 12 '20 at 18:28
• @MaxW I edited in an attempt to clarify Apr 12 '20 at 18:30
• @AdnanAL-Amleh Yes, this is correct. May 4 '20 at 6:26