Consider soil water with a pH of 7 containing a weak organic acid, which we will designate HA, at a concentration of 1 × 10−4 M. If the apparent dissociation constant of the acid is 10−4.5, what fraction of the acid is dissociated?
Answer: We have two unknowns: the concentration of the dissociated and undissociated acid, and we have two equations: the equilibrium constant expression for dissociation and the mass balance equation. We will have to solve the two simultaneously to obtain the answer. Our two equations are:
Solving the dissociation constant expression for [A–], we have:
Then solving the conservation equation for [HA] and substituting, we have
Setting H+ to 10−7 and ΣHA to 10−4, we calculate [A–] as 3.16 × 10−5 M, so 31.6% of the acid is dissociated.
3.10.3 Electrical neutrality
There is an additional condition that electrolyte solutions must meet: electrical neutrality. Thus, the sum of the positive charges in solutions must equal the sum of the negative ones, or:
(3.99)
While this presents some experimental obstacles, for example, we cannot add only Na+ ion to an aqueous solution while holding other compositional parameters constant; it also allows placement of an additional mathematical constraint on the solution. It
is often convenient to rearrange eqn. 3.99 so as to place anions and cations on different sides of the equation:
(3.100)
As an example, consider water in equilibrium with atmospheric CO2 and containing no other species. The charge balance equation in this case is:
As Example 3.10 illustrates, the electrical neutrality constraint can prove extremely useful.
Example 3.10 Determining the pH of rainwater from its composition
Determine the pH of the two samples of rain in the adjacent table. Assume that sulfuric and nitric acid are fully dissociated and that the ions in the table, along with H+ and OH–, are the only ones present.
| Analysis of rainwater | ||
| Rain 1 (μM) | Rain 2 (μM) | |
| Na | 9 | 89 |
| Mg | 4 | 16 |
| K | 5 | 9 |
| Ca | 8 | 37 |
| Cl | 17 | 101 |
|
|
10 | 500 |
|
|
18 | 228 |
Answer: This problem is simpler than it might first appear. Given the stated conditions, there are no reactions between these species that we need to concern ourselves with. To solve the problem, we observe that this solution must be electrically neutral: any difference in the sum of cations and anions must be due to one or both of the two species not listed: OH– and H+.
We start by making an initial guess that the rain is acidic and that the concentration of H+ will be much higher than that of OH–, and that we can therefore neglect the latter (we will want to verify this assumption when we have obtained a solution). The rest is straightforward. We sum the product of charge times concentration (eqn. 3.99) for both cations and anions and find that anions exceed cations in both cases: the difference is equal to the concentration of H+. Taking the log of the concentration (having first converted concentrations to M from μM by multiplying by 10−6), we obtain a pH of 4.6 for the first sample and 3.14 for the second.
Now we need to check our simplifying assumption that we could neglect OH–. The equilibrium between OH– and H+ is given by:
From this we compute [OH–] as 10−10 in the first case and 10−11 in the second. Including these would not change the anion sum significantly, so our assumption was justified.
| Charge balance for rainwater | ||
| Rain 1 | Rain 2 | |
| Σ cations | 38 | 204 |
| Σ anions | 63 | 1057 |
| Δ | 25 | 853 |
| pH | 4.60 | 3.07 |
3.10.4 Equilibrium constant expressions
For each chemical reaction in our system, we can write one version of eqn. 3.85. This allows us to relate the equilibrium activities of the species undergoing reaction in our system to one another.
Solution of aqueous equilibria problems often hinge on the degree to which we can simplify the problem by minimizing the number of equilibrium constant expressions we must solve. For example, H2SO4 will be completely dissociated
