Addition of Potassium Hydroxide (KOH)
Given is a water sample with pH 7.9.1
- What is the pH after addition of 0.3 mmol/L KOH?
- What is the impact of calcite precipitation?
Click on Open and select the example water
calcite-3.sol. Then, click on Reac and enter for “KOH” the value 0.3 mmol/L.
Run the calculation with Start. The result will be displayed as shown in the right screenshot:
|pH = 9.03||(without calcite precipitation)|
|pH = 7.83||(with calcite precipitation)|
The two aqueous solutions result from two different calculations. The first ignores mineral phases at all (Output 1); the second considers phase equilibria and allows precipitation if, and only if, one or more minerals are supersaturated (Output 2). In the given example the final solution is supersaturated with calcite (CaCO3), thus calcite precipitates (and alters the pH value).
Amount of Precipitated Calcite
Close inspection of the output table (next window that opens after the above screenshot) reveals that the aqueous solutions differ in four parameters (concentration units in mM):
|parameter||input water||output 1||output 2|
|(no precipitation)||(with CaCO3 precipitation)|
In fact, the difference of the Ca values in the two last columns yields the molar amount of precipitated calcite:
|(1)||CaCO3 = 1.22 mM – 0.91 mM = 0.31 mM|
Exactly the same result is obtained from the DIC values:
|(2)||CaCO3 = 2.46 mM – 2.15 mM = 0.31 mM|
Calcium Carbonate Saturation State
Each water is characterized by a set of parameters controlling the CaCO3 saturation state. In the present case we would find the following information:
|SI_Calcite = 1.29 (saturation index)|
|Calcium Carbonate Precipitation Potential (CCPP): 0.30 mmol/L|
|29.7 mg/L calcite precipitates.|
The value of CCPP tells us how much calcite precipitates. But this value (0.30 mM) does not exactly coincide with the amount predicted in Eq.(1) or Eq.(2). What’s going on?
The answer is simple. The parameters of the CaCO3 saturation state are determined for the evaluation temperature Teval = 5.5°C which differs from the temperature T = 20°C of the water sample (both temperatures can be set in the input window separately). What we have is:
|(3a)||amount of precipitated calcite||at T = 20 °C:||0.31 mM|
|(3b)||amount of precipitated calcite||at T = 5.5 °C:||0.30 mM|
It’s easy to check this statement. Set Teval = 20°C in the input panel and rerun the calculation. You will get CCPP = 0.31 mM.
This water is an example shipped with the program as