It is widely known that we can add multiple equilibrium reactions to obtain a net reaction and a corresponding equilibrium constant for the net reaction. However, I am not sure of exactly when this method can be applied. For instance consider a diiprotic acid $\ce{H_2A}$

$$\ce{H_2A <=> HA^{-} + H^{+}} $$ $$\ce{HA^{-} <=> A^{2-} + H^{+}} $$

with hypothetical first and second ionization constants $10^{-5}$ and $10^{-10}$ respectively. Note that $K_{a_1} \gg K_{a_2}$ as is the case for real acids. If we start with $0.1~\rm M $ solution of the acid then the pH is straightforward to calculate since $\ce{H^{+}}$ from the second ionization is negligible we can write

$$\frac{[\ce{H^{+}}][\ce{HA^{-}}]}{[\ce{H_2A}]} = 10^{-5}$$

Since the concentration of the terms in the numerator are equal and the concentration of the un-dissociated acid remains approx constant we get

$$[\ce{H^{+}}] = 10^{-3} ~\mathrm M$$

If instead we take the net reaction with equilibrium constant $K_{a_1}\cdot K_{a_2} = 10^{-15}$

$$\ce{H_2A <=> A^{2-} + 2H^{+}}$$

and write the equation ($x$ is amount of acid dissociated):

$$\frac{(x)(2x)^2}{0.1-x} = 10^{-15}$$

This gives

$$2x = [\ce{H^{+}}] \approx 5.8 \times 10^{-6}$$

which is quite different from the result obtained using the first method.

Note that this is just one of the many cases I have encountered in which adding equations does not work.

Why does adding the equations not work in this case? Also, in general, when is adding equations to figure out equilibrium concentrations justified?


1 Answer 1


When you combine the reactions and the equilibrium constants like you do, you're calculating the amount of reaction which goes all the way to completion (the fraction which reacts in both steps).

The complication that you're seeing is that the protons are not just coming from the $\ce{H_2A}$ which completely dissociates, but also a (rather substantial) portion of the molecule which undergoes only partial dissociation.

To see this, think about what happens when the hydrogens in the molecule are distinguishable: for example, the dissociation of $\ce{HDA}$ (hemi-deuterated compound), where the reactions are

$$\ce{HDA <=> DA^{-} + H^{+}} $$ $$\ce{DA^{-} <=> A^{2-} + D^{+}} $$

(in reality this wouldn't be the obligate order, but just assume it is for a thought experiment).

The $10^{-15}$ equilibrium constant is only applicable for the formation of $\ce{D^{+}}$ - you still have the rest of the $\ce{H^{+}}$ from the first step, which will also contribute to the pH.

So your assumption in your last step that $\ce{[H^{+}] = 2 [A^{2-}]}$ is incorrect, thus resulting in an incorrect result.

Further explanation

Let's actually work through the $\ce{HDA}$ case, assuming (incorrectly) that the dissociation has an obligate order. You start off with a certain amount of $\ce{HDA}$ ($T = 0.1M$). Now, imagine that at equilibrium we get $x$ total turnovers for the first reaction. A certain amount of $\ce{DA^-}$ (call it $y$) is then going to go and dissociate further. So in the end at equilibrium, we have the following concentrations:

$$\ce{[HDA]} = T-x$$ $$\ce{[H^+]} = x$$ $$\ce{[DA^-]} = x-y$$ $$\ce{[D^+]} = y$$ $$\ce{[A^{2-}]} = y$$

Now we can set up our equilibrium equations:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} = \frac{(x)(x-y)}{T-x} = 10^{-5} M$$ $$\frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HD^{-}}]} = \frac{(y)(y)}{x-y}= 10^{-10} M$$

We have a system of two equations with two unknowns. We can thus solve for $x$ and $y$ and get $x=0.995\cdot10^{-3} M$ and $y = 3.15\cdot10^{-7} M$.

Now, let's do the combined equation. First off, we can see where it comes from:

$$\frac{[\ce{H^{+}}][\ce{DA^{-}}]}{[\ce{HDA}]} \cdot \frac{[\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{DA^{-}}]} = \frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = 10^{-5} M \cdot 10^{-10} M = 10^{-15} M^2$$

So we set up the equation:

$$\frac{[\ce{H^{+}}][\ce{D^{+}}][\ce{A^{2-}}]}{[\ce{HDA}]} = \frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

Um ... we have two unknowns, but only a single equation. We can't solve this without additional information. We can't assume that $[\ce{H^{+}}] = [\ce{D^{+}}]$, or that all the $\ce{HDA}$ which disappeared went completely to $\ce{A^{2-}}$ - some of it can (has to) form $\ce{DA^-}$ which doesn't react further. The equation is valid (if you substitute $x=0.995\cdot10^{-3} M$ into the equation, as determined above, you'll get the correct answer for $y$), but only if we have some way of knowing how much of the $\ce{HDA}$ reacted and didn't go completely to completion.

The picture is further complicated because we're not dealing with distinguishable $\ce{H^+}$ and $\ce{D^+}$ which dissociate in an obligate order, but a single species which is involved in both reactions. That is, instead of

$$\frac{(x)(y)(y)}{(T-x)} = 10^{-15} M^2$$

we would have

$$\frac{(x+y)(x+y)(y)}{(T-x)} = 10^{-15} M^2$$

Again, the combined equilibrium constant is perfectly correct, and accurately expresses the relationship between the concentrations at equilibrium. It's just that the additional assumptions about the relationships between the concentrations that you're making are not. Specifically, $[\ce{H^{+}}] \neq 2 \cdot [\ce{A^{2-}}]$. Note this isn't particularly specific to this sort of combined equilibrium. Any additional reaction you're not considering can change the concentrations and make your assumptions less than correct - not just formation of intermediates which don't go all the way to completion to side reactions that draw off substrates, reactants or products into other forms.

The equilibrium constant tells you the relationship between the species concentrations, but the concentrations as they actually are in solution, not what the reaction generates. The latter is only true when the reaction in question is the only one under consideration.

  • $\begingroup$ But the net reaction accounts for $H^{+}$ from both ionizations. Indeed, the net reaction in case of your hemi-deuterated compound will be $\ce{HDA <=> A^{2-} + D^{+} + H^{+}}$ ($K_a = 10^{-15}$) which includes both the proton and the deuteron. $\endgroup$
    – Gerard
    Apr 8, 2016 at 3:14
  • $\begingroup$ @Gerard - Right, there's a contribution from both reactions, but only a small fraction of the starting material will go through both reactions. There's a much larger fraction which will only go through the first reaction and stop. That's not accounted for in the combined reaction rate constant, but it still contributes to $\ce{[H^{+}]}$ $\endgroup$
    – R.M.
    Apr 8, 2016 at 15:21
  • $\begingroup$ I can understand what you mean intuitively. However, I don't see where the mathematics goes wrong. In adding the equations we have nowehrer $\endgroup$
    – Gerard
    Apr 9, 2016 at 17:41
  • $\begingroup$ @Gerard I expanded things greatly, I hope that addresses your concerns. $\endgroup$
    – R.M.
    Apr 9, 2016 at 20:47

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