## Water Sample with Incomplete Data

Problem

What is the equilibrium composition and alkalinity of a water sample with incomplete analytical data? The only information we have is that it is a potable water with:

 pH 7.5 T 5 °C DIC 2.6 mM

Irrespective of the incomplete analytical data we are able to deduce some valuable hydrochemical information.

First of all, the above dataset does not contain any cations. Therefore, we assume the existence of Ca as a major ion (which is surely present in any potable water). We set, for example, 2 mM Ca as an initial value,1 and let the program determine the exact value.

The procedure: Start with pure water (button H2O) and activate the checkbox Mol. Then, enter the following parameters:

 pH 7.5 T 5 °C DIC 2.6 mmol/L Ca 2 mmol/L

After clicking on Start, the program immediately displays a Charge Balance Error of 25.66. Select the parameter Ca to establish charge balance, and click on next which yields:

 Ca: 2.0   ⇒   1.188 mM

Thus, our first-choice value of 2 mM (= 80.16 mg/L) was decreased by an amount of 0.812 mM down to 1.188 mM. The obtained solution is now perfectly charge-balanced.

Go on with next to the main output table, then click on the upper button Ions, which offers the complete carbonate speciation of the input solution (raw data) and of two equilibrium solutions (one with and one without mineral precipitation). In this example, the last two columns are equal since there is no precipitation of minerals. The results are:

 CO2 = 0.229 mM HCO3- = 2.349 mM CO3-2 = 0.002 mM

This information is presented in a subsequent window (button next) together with alkalinities and buffer capacities:

 ANC to pH 4.3 = 2.41 mM ANC to pH 8.2 < 0 BNC to pH 4.3 < 0 BNC to pH 8.2 = 0.20 mM M alkalinity = 2.38 meq/L

In fact, the values of HCO3-, of alkalinity, and of ANC are nearly the same.

Remarks

1. The result is independent of the initial choice for Ca; you can take Ca = 0.001 mM or 1000 mM or any other value.