# Does the acidity constant depend on the solvent?

Today I came across $$pK_a$$ values that depend on the solvent. For the hypothetical reaction $$HA \rightarrow H^+ + A^-$$, I had in mind that the acidity constant is defined simply as $$K_a = \frac{\{H^+\}\{A^-\}}{\{HA\}}$$. Therefore, no reference to the solvent seems to appear.

Of course, you wouldn't find protons by themselves, but would it still be a "pure" acidity constant if you used $$H_3O^+$$ and $$H_2O$$? In this case you would then have $$HA + H_2O\rightarrow H_3O^+ + A^-$$. As water is a liquid $$\{H_2O\} = 1$$. Thus, for the constants to be equal $$\{H^+\} = \{H_3O^+\}$$. But that should not hold for all solvents since different values come up, right?

• The solvent acts here as an implicit Broensted-Lawry base. A weaker solvent base means lower acid dissociation. Acid dissociation is neutralization reaction HA + B <=> A- + BH+, where BH+/B is conjugate acid/base pair of the solvent. If as a solvent is used acetic acid, bases get much stronger and acids much weaker. And vice versa, if some suitable organic base is used as a solvent. This can be used for titration of very weak acids or bases. If the solvent is a mixture of more solvents (e.g. 95/5 vol% of ethanol/water), parallel equilibrii are established with all solvent components. Sep 14, 2022 at 8:45
• Right! The acidity is an equilibrium with the solvent so each solvent will have its set of equilibrium conditions and the appropriate equilibrium constant. Oct 5, 2022 at 20:41

The hypothetical reaction you're referencing is an over-simplified version of the actual reaction that is taking place. In strict terms, the hydronium ion does no exist as "$$[H^+]$$", but rather as "$$[H_3O^+]$$". The complete reaction (if the solvent is water) would be:

$$\ce{HA(aq) + H_2O(l) <=> H_3O^+(aq) + A^-(aq)}$$

And the equilibrium constant for this reaction would be:

$$K_{eq}=\frac{[H_3O^+][A^-]}{[HA][H_2O]}$$

However, because the solvent (in this case, water) is in excess, its concentration is considered a constant value. This means $$[H_2O]$$ can be grouped up with $$K_{eq}$$ (another constant value):

$$[H_2O]*K_{eq}=\frac{[H_3O^+][A^-]}{[HA]}$$

This product is defined as:

$$K_a=[H_2O]*K_{eq}$$

So we have:

$$K_a=\frac{[H_3O^+][A^-]}{[HA]}$$

Even though using $$[H_2O]=1$$ leads to $$K_a = K_{eq}$$, that's not the case because $$K_a ≠ K_{eq}$$

The actual value of $$[H_2O]$$ at 25°C is:

$$[H_2O]=\frac{\rho_{H_2O}}{M_{H2O}}=\frac{1000 g/L}{18 g/mol}=55.56 mol/L$$

So, if we repeat the reaction using a different solvent "$$S$$":

$$\ce{HA(aq) + S(l) <=> H_3O^+(aq) + A^-(aq)}$$

$$K_{eq}=\frac{[H_3O^+][A^-]}{[HA][S]}$$

$$[S]*K_{eq}=\frac{[H_3O^+][A^-]}{[HA]}$$

$$K_a=[S]*K_{eq}$$

$$[S]=\frac{\rho_{S}}{M_{S}}$$

In other words, $$K_a$$ is a function of $$[S]$$

Since "S" in this case isn't water, it follows that:

$$[S]≠[H_2O]$$

In conclusion:

(1) For the same reaction at the same temperature, $$K_a$$ calculated with water as a solvent is different than $$K_a$$ calculated with a different solvent.

(2) $$K_a$$ depends on the solvent used.