Читать книгу Geochemistry - William M. White - Страница 141

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